polynomial is: - Algebra Class E-course 8 Polynomials.pdf · A polynomial is: ... 1 Side 1 =x 2 – 3x -4 Side 2: x 2 + 2x -3 2 3 Side 3: 5x -2 Side 4: 5x -3 4 Step 1: Rewrite vertically

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Unit Unit Unit Unit 8: Polynomials8: Polynomials8: Polynomials8: Polynomials

Lesson 1: Adding Polynomials

Adding Monomials

The MOST Important Rule for Adding Monomials is:

When adding monomials you may only add terms that are ____________________________

A polynomial is:______________________________________________________________________________



Example 1

Example 2

Example 3

(3y2 + 5y – 6) + (7y

2 – 9)

(3 + 2a + a2) + ( a

2 – 8a + 5)

Find the perimeter of the triangle, where the measurements of the sides are as follows:

1 2 Side 1: 4s -1

Side 2: 3s2 + 2s -1

3 Side 3: 8s2 -8s + 5



Lesson 1: Adding Polynomials Practice

Part 1: For each problem below, add the polynomials.

1. (6x2 + 2x – 7) + (4x2 – 4x -8) 6. (t2 – 7t – 9) + (6t – 2t2 + 2)

2. (4y3 – 8y2 + 9y -1) + (2y3 – 9y +2) 7. (b3 – 3b + 2) + (3b3 – 2b2 – 3b -10)

3. (9a3 –a2 +6) + (2a3 – 3a2 – a -1) 8. (3x – 8x2 -5) + (4 – 2x – 8x2)

4. (x – 4x2 + 5) + (3x2 -6x + 3) 9. (2x3 – 6x + 3x2 -9) + (6x -2x3 +4x2 +9)

5. (16y3 – 4y2 – 2y + 10) + (3y2 + 2y – 2) 10. (9d2 – 7) + (2d3 +4d)

Part 2: Find the missing polynomial.

1. (2x2 – 6x + 5) + (?) = 4x2 -7x + 10 4. (3y3 – 6y +2) + (?) = 2y3 + 4y2 -9y -6

2. (3x3 – 6x + 9) + (?)= 7x2 – 10x -1 5. (4a2 – 4) + (?) = -2a2 -8

3. (2x2 – 9) + (?) = 3x3 + x2 -2x -1 6. (3b2 – 4b3 – 2b + 1) +(?) = 5b3 +b2 -4b

Part 3: Find the perimeter of each shape.

1. 1 Side 1 =x2 – 3x -4 Side 2: x2 + 2x -3

2 3 Side 3: 5x -2 Side 4: 5x -3

4

2. This is an equilateral triangle with a base of:

3x2 - 4x + 7

1. Which is the sum of y4 + y2 + y? (1 point) 3. Find perimeter. (3 points)

a. y7 c. 3y7

b. y6 d. None of the above. 6x - 6

2. Simplify the expression. (2 points) 2x2 +2x -1

(3x4 + 2x2 + 4x – 9) + (2x3 + x2 – 3x +7)



Lesson 1: Adding Polynomials Practice – Answer Key

Part 1: For each problem below, add the polynomials.

1. (6x2 + 2x – 7) + (4x2 – 4x -8) 6. (t2 – 7t – 9) + ( 6t – 2t2 + 2)

2. (4y3 – 8y2 + 9y -1) + (2y3 – 9y +2) 7. (b3 – 3b + 2) + (3b3 – 2b2 – 3b -10)

3. (9a3 –a2 +6) + (2a3 – 3a2 – a -1) 8. (3x – 8x2 -5) + ( 4 – 2x – 8x2)

Step 1: Rewrite vertically with like terms together.

6x2 + 2x – 7

+ 4x2 – 4x -8

10x2 -2x – 15

Final Answer: 10x2 -2x – 15


Remember to take the sign in front with each term.

t2 – 7t – 9

+ -2t2 + 6t + 2

-t2 -1t - 7

Final Answer: -t2 – t - 7


Remember to use 0 as a place holder if you are missing a

term.

4y3 – 8y2 + 9y -1

+ 2y3 +0y2 – 9y +2

6y3 - 8y2 + 0y +1

Final Answer: 6y3 – 8y2 + 1



term.

b3 + 0b2 -3b + 2

+ 3b3 – 2b2 – 3b - 10

4b3 – 2b2 – 6b - 8

Final Answer: 4b3 – 2b2 – 6b - 8



term.

9a3 –a2 + 0a +6

+ 2a3 – 3a2 – a - 1

11a3 – 4a2 – a + 5

Final Answer: 11a3 – 4a2 – a + 5



-8x2 +3x – 5

+ -8x2 – 2x + 4

-16x2 +1x - 1

Final Answer: -16x2 +x - 1



4. (x – 4x2 + 5) + (3x2 -6x + 3) 9. (2x3 – 6x + 3x2 -9) + (6x -2x3 +4x2 +9)

5. (16y3 – 4y2 – 2y + 10) + (3y2 + 2y – 2) 10. (9d2 – 7) + (2d3 +4d)


1. (2x2 – 6x + 5) + (?) = 4x2 -7x + 10 4. (3y3 – 6y +2) + (?) = 2y3 + 4y2 -9y -6



-4x2 + x + 5

+ 3x2 – 6x + 3

-1x2 -5x + 8

Final Answer: -x2 – 5x + 8



2x3 + 3x2 – 6x – 9

+ -2x3 + 4x2 + 6x +9

0x3 + 7x2 + 0x+ 0

Final Answer: 7x2



16y3 – 4y2 – 2y + 10

+ 0y3 + 3y2 + 2y - 2

16y3 - 1y2 + 0y + 8

Final Answer: 16y3 – y2 + 8



0d3 + 9d2 + 0d– 7

+ 2d3 + 0d2 + 4d + 0

2d3 + 9d2 + 4d - 7

Final Answer: 2d3 + 9d2 + 4d - 7



2x2 – 6x + 5

+ 2x2 – 1x + 5

4x2 – 7x + 10

Final Answer: 2x2 – x + 5

What can we add in

order to arrive at the

correct answer?



3y3 + 0y2– 6y + 2

+ -1y3 + 4y2 – 3y - 8

2y3 + 4y2 – 9y - 6

Final Answer: -y3 + 4y2- 3y - 8

What can we add in

order to arrive at

the correct answer?



2. (3x3 – 6x + 9) +( ?)= 7x2 – 10x -1 5. (4a2 – 4) + (?) = -2a2 -8

3. (2x2 – 9) + (?) = 3x3 + x2 -2x -1 6. (3b2 – 4b3 – 2b + 1) +(?) = 5b3 +b2 -4b

Part 3: Find the perimeter of each shape.

1. 1 Side 1 =x2 – 3x -4 Side 2: x2 + 2x -3

2 3 Side 3: 5x -2 Side 4: 5x -3

4



3x3 + 0x2+ – 6x + 9

+ -3x3 + 7x2 - 4x - 10

0x3 + 7x2 – 10x - 1

Final Answer: -3x3 + 7x2+ - 4x - 10



4a2 + 0a – 4

+ -6a2 + 0a - 4

-2a2 + 0a - 8

Final Answer: -6a2 - 4

What can we add in

order to arrive at

the correct answer?

What can we add

in order to arrive

at the correct

answer?



0x3 + 2x2 + 0x – 9

+ 3x3 – 1x2 – 2x +8

3x3 + x2 – 2x - 1

Final Answer: 3x3 – x2 – 2x + 8



-4b3 + 3b2 – 2b + 1

+ 9b3 -2b2 -2b - 1

5b3 + b2 – 4b

Final Answer: 9b3 – 2b2 – 2b - 1

The perimeter is the measurement of the “outside” of the shape. We must add all sides of the shape

together in order to calculate the perimeter.

Side 1: x2 – 3x - 4

Side 2: x2 + 2x - 3

Side 3: 5x - 2

Side 4: + 5x – 3

2x2 + 9x -12

The perimeter of the shape is: 2x2 + 9x - 12



2. This is an equilateral triangle with a base of:

3x2 - 4x + 7

An equilateral triangle is a triangle that has 3 “equal” sides. Therefore, if the base is 3x2 – 4x + 7,

then all sides are 3x2 – 4x + 7.

There are two ways to solve this problem. The first way is to add all three sides.

Side1: 3x2 – 4x + 7

Side 2: 3x2 – 4x + 7

Side 3: 3x2 – 4x + 7

9x2 – 12x + 21

The second way is to multiply. Since all three sides are the same, we can multiply the base times 3.

We would need to use the distributive property to calculate this:

3(3x2 – 4x + 7)

9x2 – 12x + 21

The perimeter of this equilateral triangle is: 9x2 – 12x + 21

1. Which is the sum of y4 + y2 + y? (1 point) 3. Find perimeter. (3 points)

a. y7 c. 3y7

b. y6 d. None of the above. 6x - 6

2x2 +2x -1

2. Simplify the expression. (2 points)

(3x4 + 2x2 + 4x – 9) + (2x3 + x2 – 3x +7)

Since none of the y variables have the same

exponents, we do not have any like terms to

combine. Therefore, the sum is: y4 + y2 + y

D. None of the above is the correct

answer.

Write like terms together:

3x4 + 2x

3 + 2x

2 + x

2 +4x – 3x – 9 +7

Combine like terms:

3x4 +2x

3 +3x

2 +x – 2 Final Answer

Since this is a rectangle, we know that

the 2 widths are 6x – 6 and the 2

lengths are 2x2 +2x – 1.

Multiply each length and width by 2 and

add together.

2(6x-6) + 2(2x2 +2x – 1)

12x – 12 + 4x2 + 4x – 2 Distribute

4x2 +12x+4x -12-2 Like terms

together

4x2 +16x -14 Combine

is the perimeter.



Lesson 2: Subtracting Polynomials

Example 1

Example 2 Example 3

When Subtracting Polynomials, think of the phrase:

_______________________________________________________

Rewrite the subtraction problem as an _____________________________________________.

(3 + 2a + a2) – (a

2 + 4a – 8)

The triangle has a perimeter of: 4x2 – 6x + 2

Side 1 = x2 – 3x + 5 Side 2 = 2x

2 + 6

Find the measurement of side 3.

1 3

2



Lesson 2: Subtracting Polynomials Practice

Part 1: Simplify each expression by subtracting.

1. (4x2 + 2x – 2) - (4x2 – 3x -8) 5. (6t2 – 9t – 1) - ( 2t – 7t2 + 3)

2. (7y3 – 2y2 + 9y -9) - (y3 – 5y +1) 6. (b3 – 2b + 8) - (8b3 – 2b2 – 3b -17)

3. (a3 –2a2 +6) - (2a3 – 4a2 – a -1) 7. (3x + 3x2 -6) - ( 9 – 2x – 2x2)

4. (x – 4x2 + 9) - (x2 -2x + 5) 8. (2x3 – 3x + 3x2 -7) - (2x -x3 +6x2 +1)


1. (2x2 – 6x + 5) - (?) = x2 +3x + 1 4. (y3 – 7y +2) - (?) = 5y3 + 4y2 -8y -2

2. (3x3 – 6x + 2) -( ?)= 3x2 – 3x -1 5. (4a2 – 4) - (?) = -7a2 -2

3. (2x2 – 4) - (?) = 2x3 + 5x2 -2x -1 6. (b2 – 4b3 – 7b + 1) -(?) = 2b3 +3b2 -4

Part 3: Find the missing side of each shape.

1. 1 Side 1 =x2 – 3x -4 Side 2: x2 + 2x -3

2 3 Side 3: 5x -2 Side 4: ?

4 Perimeter = 3x2 +4x - 4

1 2 Side 1 = 5x-5 Side 2: ?

2. 3 4 5 Side 3 = 2x2 – 4x -2 Side 4 = 6x -7

Side 5 = 3x +2 Perimeter= 4x2-2x+8

1. (16y3 – 4y2 – 2y + 10) - (3y2 + 2y – 2) 2. (7d2 – 2) - (d3 +3d)

1

3. Find the missing side of the shape. 2 3

( 3 points) 4

Directions: Simplify each expression.

(2 points each)

Side 1: 2x2 + 3x - 1

Side 2: 5x-3

Side 3: x2 +5

Perimeter: 6x2+10x-4



Lesson 2: Subtracting Polynomials Practice – Answer Key

Part 1: Simplify each expression by subtracting.

1. (4x2 + 2x – 2) - (4x2 – 3x -8) 5. (6t2 – 9t – 1) - ( 2t – 7t2 + 3)

2. (7y3 – 2y2 + 9y -9) - (y3 – 5y +1) 6. (b3 – 2b + 8) - (8b3 – 2b2 – 3b -17)

3. (a3 –2a2 +6) - (2a3 – 4a2 – a -1) 7. (3x + 3x2 -6) - ( 9 – 2x – 2x2)

Step 1: Rewrite the problem as an addition

problem.

(4x2 + 2x – 2) + (-4x2 + 3x +8)

Step 2: Rewrite vertically and add.

4x2 + 2x – 2

+ -4x2 + 3x + 8

0x2 +5x +6

Final Answer: 5x +6


problem.

(6t2 – 9t – 1) + (- 2t + 7t2 - 3)


6t2 – 9t – 1

+ 7t2 -2t -3

13t2 – 11t - 4

Final Answer: 13t2 – 11t - 4


problem.

(7y3 – 2y2 + 9y -9) + (-y3 + 5y -1)


7y3 – 2y2 + 9y -9

+ -y3 + 0y2 + 5y – 1

6y3 – 2y2 + 14y - 10

Final Answer: 6y3 – 2y2 + 14y - 10


problem.

(b3 – 2b + 8) + (-8b3 + 2b2 + 3b +17)


b3 + 0b2 – 2b + 8

+ -8b3 + 2b2 + 3b +17

-7b3 + 2b2 + b + 25

Final Answer: -7b3 + 2b2 + b + 25


problem.

(a3 –2a2 +6) + (-2a3 + 4a2 + a +1)


1a3 – 2a2 + 0a + 6

+ -2a3 + 4a2 + a + 1

-1a3 +2a2 + a + 7

Final Answer: -a3 + 2a2 + a + 7


problem.

(3x + 3x2 -6) + (- 9 + 2x + 2x2)


3x2 + 3x – 6

+2x2 + 2x - 9

5x2 + 5x - 15

Final Answer: 5x2 + 5x - 15



4. (x – 4x2 + 9) - (x2 -2x + 5) 8. (2x3 – 3x + 3x2 -7) - (2x -x3 +6x2 +1)



problem.

(x – 4x2 + 9) + (-x2 +2x - 5)


-4x2 + x + 9

+ -x2 + 2x – 5

-5x2 + 3x + 4

Final Answer: -5x2 + 3x + 4


problem.

(2x3 – 3x + 3x2 -7) + (-2x +x3 -6x2 -1)


2x3 + 3x2 – 3x – 7

+ x3 – 6x2 – 2x – 1

3x3 – 3x2 – 5x - 8

Final Answer: 3x3 – 3x2 – 5x - 8

This set of problems is a little tricky. Think of it as solving for the “missing variable”. Let’s take a look at a problem

that we are familiar with. Let’s say we have 5 – x = 10. What would you do to solve for x?

You might: Subtract 5 from both sides and then multiply by -1 to make x positive.

5-5 –x = 10 – 5

-x = 5

(-1)(-x) = -1(5)

x= -5

Or you might: Add x to both sides and subtract 10 from both sides. This would make x positive too.

5-x+x = 10 +x

5 = 10 + x

5-10 = 10-10 + x -5 = x

This is exactly how the next set of problems is set up. So, we can

create a rule based on this simple equation: Subtract the right

side from the left side of the equation in order to find our

answer. (See the video for a better explanation)



1. (2x2 – 6x + 5) - (?) = x2 +3x + 1 4. (y3 – 7y +2) - (?) = 5y3 + 4y2 -8y -2

F

2. (3x3 – 6x + 2) -( ?)= 3x2 – 3x -1 5. (4a2 – 4) - (?) = -7a2 -2

Step 1: This of this problem as:

(2x2 – 6x + 5) – p = x2 + 3x + 1

(Assign a different variable for the missing

polynomial.) To solve for p, we can add p to

both sides and then subtract (x2 + 3x + 1) from

(2x2 – 6x + 5)

(2x2 – 6x + 5) – (x2 + 3x + 1)

Rewrite as an addition problem.

(2x2 – 6x + 5) + (-x2 -3x – 1)

Rewrite vertically and add:

2x2 – 6x + 5

+ -x2 – 3x – 1

x2 - 9x + 4 Final Answer


(y3 – 7y + 2) – p = 5y3 + 4y2 – 8y - 2



both sides and then subtract (5y3 + 4y2 – 8y - 2)

from (y3 – 7y + 2)

(y3 – 7y + 2) - (5y3 + 4y2 – 8y - 2)


(y3 – 7y + 2) + (-5y3 - 4y2 + 8y + 2)


y3 + 0y2 – 7y + 2

+-5y3 - 4y2 + 8y + 2

-4y3 – 4y2 + y + 4 Final Answer


(3x3 – 6x + 2) – p = 3x2 – 3x - 1



both sides and then subtract (3x2 – 3x - 1) from

(3x3 – 6x + 2)

(3x3 – 6x + 2) - (3x2 – 3x - 1)


(3x3 – 6x + 2) + (-3x2 + 3x + 1)


3x3 + 0x2 – 6x + 2

-3x2 + 3x + 1

3x3 – 3x2 – 3x + 3 Final Answer


(4a2 - 4) – p = -7a2 - 2



both sides and then subtract (-7a2 - 2) from

(4a2 - 4)

(4a2 - 4) - (-7a2 - 2)


(4a2 - 4) + (7a2 + 2)


4a2 – 4

+ 7a2 + 2

11a2 - 2 Final Answer



3. (2x2 – 4) - (?) = 2x3 + 5x2 -2x -1 6. (b2 – 4b3 – 7b + 1) -(?) = 2b3 +3b2 -4

Part 3: Find the missing side of each shape.

1. 1 Side 1 =x2 – 3x -4 Side 2: x2 + 2x -3

2 3 Side 3: 5x -2 Side 4: ?

4 Perimeter = 3x2 +4x – 4


(2x2 - 4) – p = 2x3 + 5x2 – 2x - 1



both sides and then subtract (2x3 + 5x2 – 2x - 1)

from (2x2 - 4)

(2x2 – 4) - (2x3 + 5x2 – 2x - 1)


(2x2 – 4) + (-2x3 - 5x2 + 2x + 1)


0x3 + 2x2 + 0x – 4

+-2x3 - 5x2 + 2x + 1

-2x3 – 3x2 + 2x - 3 Final Answer


(-4b3 + b2 – 7b + 1) – p = 2b3 + 3b2 - 4



both sides and then subtract (2b3 + 3b2 - 4) from

(-4b3 + b2 – 7b + 1)

(-4b3 + b2 – 7b + 1) - (2b3 + 3b2 - 4)


(-4b3 + b2 – 7b + 1) + (-2b3 - 3b2 + 4)


-4b3 + b2 – 7b + 1

+-2b3 - 3b2 + 0b + 4

-6b3 – 2b2 – 7b + 5 Final Answer

In order to solve for the missing side, we will need to add the 3 sides that we know (Side 1 + Side 2+

Side3). Then figure out what we need to add in order to equal the perimeter. This will tell us the missing

side.

Step 1: Add the 3 sides that you know. Step 2: Find the missing addend.

Side 1: x2 – 3x – 4 2x2 +4x – 9 (3 sides that we know)

Side 2: x2 + 2x – 3 + 1x2 +0x +5 (Missing Side)

Side 3: 5x – 2 3x2+ 4x – 4 (Equals the perimeter)

2x2 4x -9

Side 4 is equal to: x2 + 5



1 2 Side 1 = 5x-5 Side 2: ?

2. 3 4 5 Side 3 = 2x2 – 4x -2 Side 4 = 6x -7

Side 5 = 3x +2 Perimeter= 4x2-2x+8

1. (16y3 – 4y2 – 2y + 10) - (3y2 + 2y – 2) 2. (7d2 – 2) - (d3 +3d)

Step 1: Add the four sides that you know.

Side 1: 0x2 + 5x – 5

Side 3: 2x2+ - 4x – 2

Side 4: 0x2 + 6x – 7

Side 5: 0x2 + 3x + 2

2x2 + 10x -12 (The sum of the four sides that we know.)

Step 2: Find the missing addend.

2x2 + 10x – 12 (Sum of the four sides that we know)

+ 2x2 - 12x +20 (The missing side)

4x2 - 2x + 8 (The perimeter of the object)

The measurement of the missing side is: 2x2 - 12x +20

Directions: Simplify each expression.

(2 points each)


problem.

(16y3 – 4y2 – 2y + 10) + (-3y2 - 2y + 2)


16y3 – 4y2 – 2y + 10

+ 0y3 – 3y2 – 2y + 2

16y3 – 7y2 – 4y + 12

Final Answer: 16y3 – 7y2 – 4y + 12


problem.

(7d2 – 2) + (-d3 -3d)


0d3 + 7d2 + 0d – 2

+ -d3 + 0d2 – 3d + 0

-d3 + 7d2 – 3d - 2

Final Answer: -d3 + 7d2 – 3d - 2



1

3. Find the missing side of the shape. 2 3

( 3 points) 4

Side 1: 2x2 + 3x - 1

Side 2: 5x-3

Side 3: x2 +5

Perimeter: 6x2+10x-4

Step 1: Add the three sides that you know.

Side 1: 2x2 + 3x - 1

Side 2: 5x – 3

Side 3: x2 + 0x + 5

3x2 + 8x +1 (The sum of the three sides that we know.)

Step 2: Find the missing addend.

3x2 + 8x +1 (Sum of the three sides that we know)

+ 3x2 +2x -5 (The missing side)

6x2 + 10x - 4 (The perimeter of the object)

The measurement of the missing side is: 3x2 + 2x - 5

Note: You could also subtract the perimeter – the sum of the 3 sides:

(6x2 +10x -4) – (3x2 +8x +1)

6x2 +10x – 4 – 3x2 -8x -1

6x2 – 3x2 +10x -8x -4 -1 Write like terms together

3x2 +2x – 5 Combine like terms



Lesson 3: Multiplying Polynomials

Multiplying a Polynomial by a Monomial

Example 1 Example 2

Remember the Distributive Property?

Quick Review of Multiplying Monomials

6x (3x2 + 2x – 1) 3x

2 (6x

3 + 2x

2 – 1) - 2x(x

2-3x+1)



Lesson 3: Multiplying Polynomials Practice Problems

Part 1: Simplify each expression by multiplying.

1. 5(x2 + 2x – 3) 5. 9x2(x3 - 4x +3)

2. -3y(y2 – 7y +2) 6. -7p(p3 – 4p2 + 2p -1)

3. 9b(b3 – 2b +8) 7. 2x(y2 – 4y)

4. -2x2(x2 + 6x – 2) 8. 8x2(x2 -2xy + 3y2)

Part 2: Solve each word problem.

1. Find the area. 3. Suppose the width of a rectangle

is w inches long. Write an

3x expression for the area if the length is

2 less than 5 times its width.

• Suppose the width is 3 in. find

the area.

(2x2 + 5x)

2. Suppose the width of a floor 4.

tile is w inches long. Write an

expression for the area if the length

is 8 more than twice its width.

Part 3: Simplify. You may need to use more than one operation.

1. 3x(x2 – 3x -9) + 2x(x -3) 3. -x(x2 + 8x -2) – 2x(x2+4x – 1)

2. -6x2 (3x-5) + 2x(2x2 – 5x + 1) 4. 3y(y2 – 5y +1) – y(2y +5)

Part 4: Write and solve an equation for each problem.

1. A rectangle has a length of 2x-2 and a width of x+1. Write an expression for the perimeter of the

rectangle. Suppose x = 12 cm. What is the perimeter of the rectangle?

Write an expression for the volume of a box

if the width is w cm, the length is 5 more

than twice its width, and the height is 3

times its width.

• Suppose the width is 10cm, find the

volume of the box.



1. 3x (7x -9) + 4x(x2 – 2x +7) 2. 8s2(2s2 + 3s -1) – (s3 – 5s +9)

3. Suppose the width of a rectangle is w inches long. Write an

expression for the area if the length is 3 more than twice its width.

• Suppose the width is 3 in. find the area. (3 points)

Simplify each expression. (2 points each)



Lesson 3: Multiplying Polynomials – Answer Key

Part 1: Simplify each expression by multiplying.

1. 5(x2 + 2x – 3) 5. 9x2(x3 - 4x +3)

2. -3y(y2 – 7y +2) 6. -7p(p3 – 4p2 + 2p -1)

3. 9b(b3 – 2b +8) 7. 2x(y2 – 4y)

Step 1: Distribute the 5 to each term inside of

the parenthesis.

5(x2 + 2x – 3) = 5x2 + 10x – 15

5●x2 5 ● 2x 5●-3


Step 1: Distribute 9x2 to each term inside of the

parenthesis.

9x2(x3 - 4x +3) = 9x5 – 36x3 + 27x2

9x2 ●x3 9x2

● (-4x) 9x2●3

**Remember that when you multiply powers, you

add the exponents.

Final Answer: 9x5 – 36x3 + 27x2

Step 1: Distribute -3y to each term inside of the

parenthesis.

-3y(y2 – 7y +2) = -3y3 + 21y2 – 6y

-3y ●y2 -3y●-7y -3y ●2


add the exponents.

Final Answer: -3y3 + 21y2 – 6y

Step 1: Distribute -7p to each term inside of the

parenthesis.

-7p(p3 – 4p2 + 2p - 1) = -7p4 +28p3 -14p2 +7p

-7p ●p3 -7p●(-4p

2) -7p ●2p -7p●-1


add the exponents.

Final Answer: 7p4 +28p3 -14p2 +7p

Step 1: Distribute 9b to each term inside of the

parenthesis.

9b(b3 – 2b +8) = 9b4 -18b2 +72b

9b ●b3 9b●-2b 9b ●8


add the exponents.

Final Answer: 9b4 – 18b2 + 72b

Step 1: Distribute 2x to each term inside of the

parenthesis.

2x(y2 – 4y)= 2xy2 – 8xy

2x●y2 2x●-4y

Final Answer: 2xy2 – 8xy



4. -2x2(x2 + 6x – 2) 8. 8x2(x2 -2xy + 3y2)

Part 2: Solve each word problem.

1. Find the area. 3. Suppose the width of a rectangle

is w inches long. Write an

3x expression for the area if the length is

2 less than 5 times its width.

• Suppose the width is 3in. Find the

area of the rectangle.

(2x2 + 5x)

Step 1: Distribute -2x2 to each term inside of the

parenthesis.

-2x2(x2 + 6x – 2) = -2x4 – 12x3 +4x2

-2x2 ●x2 -2x2 ●6x -2x2● -2

Final Answer: -2x4 – 12x3 + 4x2

Step 1: Distribute 8x2 to each term inside of the

parenthesis.

8x2(x2 -2xy + 3y2) = 8x4 – 16x3y + 24x2y2

8x2●x2 8x2 ● -2xy 8x2 ●3y2

Final Answer: 8x4 – 16x3y + 24x2y2

The area formula for a rectangle is:

A = l x w. (Area = length times width)

We know: L = 2x2 + 5x W = 3x

Step 1: Substitute for the length and width.

A = L W

A = (2x2 + 5x)3x or 3x(2x2 + 5x)

Step 2: Distribute to simplify.

3x(2x2 + 5x) = 6x3 + 15x2

The area of this rectangle can be written as:

A = 6x3 + 15x2

Step 1: Identify your variables and write the

equation.

We know that the width is defined as w. This is

the variable that we will be working with.

We know the length is 2 less than 5 times its

width. This can be written as: L = 5w-2

A = LW L= 5w-2 W = w

A = (5w-2)w or w(5w -2)

A = w(5w-2)


A = 5w2 – 2w

If the width is 3in, then we can substitute 3 for w.

A = 5w2 – 2w

A = 5(3)2 – 2(3)

A = 39 in2 is the area of the rectangle.



2. Suppose the width of a floor 4.

tile is w inches long. Write an

expression for the area if the length

is 8 more than twice its width.

Write an expression for the volume of a box

if the width is w cm, the length is 5 more

than twice its width, and the height is 3

times its width.

• Suppose the width is 10 cm, find the

volume of the box.


equation.

We know that the width is defined as w. This is


We know the length is 8 more than twice its width.

This can be written as: L = 2w + 8

A = LW L= 2w + 8 W = w

A = (2w + 8)w or w(2w +8)

A = w(2w + 8)


A = 2w2 + 8w

The area for this floor can be written as:

A = 2w2 + 8w


equation.

We know the width is defined as w. This is


We also know the length is 5 more than twice

its width. This is written as: 2w + 5

The height is 3 times its width. This is written

as: 3w

The volume formula for a rectangular prism is:

V = lwh (Volume equals length times width

time height)

V = lwh L = 2w+5 W = w H= 3w

V = lwh

V = (2w+5)(w)(3w)

Let’s multiply (w)(3w) first.

V = (2w+5)(w)(3w)

V = (2w+5)(3w2) or 3w2 (2w+5)

V = 3w2 (2w+5) Distribute 3w2

V = 6w3 + 15w2

If the width is 10 cm, then we can substitute

10 for w.

V = 6w3 + 15w2

V = 6(10)3 + 15(10)2

V = 7500 cm3 is the volume of the box.



Part 3: Simplify. You may need to use more than one operation.

1. 3x(x2 – 3x -9) + 2x(x -3) 3. -x(x2 + 8x -2) – 2x(x2+4x – 1)

2. -6x2 (3x-5) + 2x(2x2 – 5x + 1) 4. 3y(y2 – 5y +1) – y(2y +5)

Step 1: Utilize the distributive property.

3x(x2 – 3x -9) + 2x(x -3)

(3x3 – 9x2 – 27x) +( 2x2 – 6x)

Step 2: Write the addition problem

vertically and add.

3x3 – 9x2 – 27x

+ 2x2 – 6x

3x3 -7x2 – 33x

Final Answer: 3x3 -7x2 – 33x


-x(x2 + 8x -2) – 2x(x2+4x – 1)

(-x3 – 8x2 + 2x) + (-2x3 – 8x2 + 2x)

Step 2: Write the addition problem vertically

and add.

-x3 – 8x2 + 2x

+ -2x3 – 8x2 + 2x

-3x3 – 16x2+ 4x

Final Answer: -3x3 – 16x2+4x

Distribute -2x


-6x2 (3x-5) + 2x(2x2 – 5x + 1)

(-18x3 + 30x2) + (4x3 – 10x2 +2x)


vertically and add.

-18x3 + 30x2

+ 4x3 – 10x2 + 2x

-14x3 + 20x2 + 2x

Final Answer: -14x3 + 20x2 + 2x


3y(y2 – 5y +1) – y(2y +5)

(3y3 – 15y2 + 3y)+ (-2y2 – 5y)


and add.

3y3 – 15y2 + 3y

+ -2y2 – 5y

3y3 – 17y2 – 2y

Final Answer: 3y3 – 17y2 – 2y

Distribute -1y



Part 4: Write and solve an equation for each problem.

1. A rectangle has a length of 2x-2 and a width of x+1. Write an expression for the perimeter of the

rectangle. Suppose x = 12 cm. What is the perimeter of the rectangle?

1. 3x (7x -9) + 4x(x2 – 2x +7) 2. 8s2(2s2 + 3s -1) – (s3 – 5s +9)

The formula for the perimeter of a rectangle is : 2L + 2W = P (2 times the length + 2 times the width)

Let’s substitute: 2L + 2W = P

2(2x-2) + 2(x+1) = P

4x – 4 + 2x + 2 = P

Combine like terms:

4x+2x – 4 + 2 = P

6x – 2 = P

Since x = 12 cm, we can substitute 12 for x.

6(12) – 2 = p

72 – 2 = P

70 = P

The perimeter of the rectangle is 70 cm.


3x (7x -9) + 4x(x2 – 2x +7)

(21x2 – 27x) + (4x3 – 8x2 + 28x)


vertically and add.

0x3 + 21x2 – 27x

+ 4x3 – 8x2 + 28x

4x3 + 13x2 + x

Final Answer: 4x3 + 13x2 + x


8s2(2s2 + 3s -1) – (s3 – 5s +9)

(16s4 + 24s3 – 8s2)+ (-s3 + 5s – 9)


and add.

16s4 + 24s3 – 8s2

+ s3 + 0s2 + 5s – 9

16s4 +23s3 – 8s2 + 5s – 9

Final Answer: 16s4 +23s3 – 8s2 + 5s – 9

Rewrite as addition



3. Suppose the width of a rectangle is w inches long. Write an

expression for the area if the length is 3 more than twice its width.

• Suppose the width is 3 in. find the area. (3 points)

Step 1: Identify your variables and write the equation.

We know that the width is defined as w. This is the variable that we will be working with.

We know the length is 3 more than twice its width. This can be written as: L = 2w + 3

A = LW L= 2w + 3 W = w

A = (2w + 3)w or w(2w +3)

A = w(2w + 3)


A = 2w2 + 3w

The area for this rectangle can be written as:

A = 2w2 + 3w

• If the width is 3in, we can substitute 3 for w in order to find the area.

A = 2w2 + 3w

A = 2(3)2 + 3(3)

A = 27 in2

The area of the rectangle is 27 in2.


Unit Unit Unit Unit 8: 8: 8: 8: PolynomialsPolynomialsPolynomialsPolynomials

Lesson 4: Multiplying Binomials (Using the FOIL Method)

The Extended Distributive Property

Example 1 FOIL Method

To multiply two sums, multiply each term in the __________________________ by_______________

_______________________________________.

Multiplication Counting Principle

If one term has a terms and another has b terms, there will be a● b terms in the product.

(However, always remember to simplify the product by combining like terms)

Exploring a New Method of Multiplying

Multiplying two digit numbers Extended Distributive Property

(x +4)(x+5) F

O

I

L



Example 2

Example 3

(3x + 4) (2x – 7)

John is building a sidewalk around his garden. The maximum length for the garden and sidewalk is 15 feet. The

maximum width for the garden and sidewalk is 13 feet. The width of the sidewalk is represented by x on all sides.

X

13 ft

15 ft

GARDEN

X

X

• Write an expression that represents the

length of the garden.


width of the garden.

• Write an expression that could be used to

find the area of the garden.

• If the width of the sidewalk is 5 feet, find

the area of the garden.



Lesson 4: Multiplying Binomials (Using the FOIL Method) Practice

Part 1: Multiply.

1. (x-5)(x-8) 5. (2w -8)2

2. (t +6) (t-4) 6. (x –y) (x-y)

3. (3y +8) (5y +2) 7. (2x +y) (8x –y)

4. (9x -9) (x +2) 8. (4x +2y)(6x -4y)

Part 2: Applying Your Knowledge

Find the expression that represents

the area.

1.

Let the length = x +6

Let the width = x-3

2. Find the area of a square

whose side is 8x-5

5. Emily is framing a picture. She would like to place a mat around the photo. The length of the

photo is 12 inches. The width of the photo is 8 inches. The width of the mat on each side is

represented by x.

3. What do the letters FOIL stand for?

4. Fill in the missing blanks.

(x -5) (x____) = x2 _____ - 35

12 in.

8 in

X

X


length of the mat in terms of x.


width of the mat in terms of x.


area of the mat. (This is tricky –

remember there is hole in the center

where the picture is placed.)

• Suppose the width of the mat on all

sides (x) is 3 inches, what is the area of

just the mat?

PICTURE



(4x+1)

3.

Directions: Simplify (2 points each)

1. (a +9) (a -9) 2. (9m -2n)(m-2n)

Photo Mat

3x-2

Photo

To the left is a picture frame that includes matting. This picture

frame is a perfect square and the photo inside is a perfect square.

(4 points)

• Find the area of the photo mat (green section). **There is a

square cut out for the photo insertion.***

• Explain step by step how you arrive at your final answer.



Lesson 4: Multiplying Binomials (Using the FOIL Method-Answer Key

Part 1: Multiply.

1. (x-5)(x-8) 5. (2w -8)2

2. (t +6) (t-4) 6. (x –y) (x-y)

F irst (x)(x) = x2

Outer (x)(-8) = -8x

Inner (-5)(x) = -5x

Last (-5)(-8) = 40

x2 – 8x – 5x + 40 Write all terms together

x2 – 13x + 40 Combine x terms.

Final Answer: x2 – 13x + 40

(2w-8)(2w-8)

F irst (2w)(2w) = 4w2

Outer (2w)(-8) = -16w

Inner (-8)(2w) = -16w

Last (-8)(-8) = 64

4w2 – 16w –16w + 64 Write all terms together

4w2 – 32w + 64 Combine w terms.

Final Answer: 4w2 – 32w + 64

F irst (t)(t) = t2

Outer (t)(-4) = -4t

Inner (6)(t) = 6t

Last (6)(-4) = -24

t2 – 4t + 6t -24 Write all terms together

t2 +2t - 24 Combine t terms.

Final Answer: t2 + 2t - 24

F irst (x)(x) = x2

Outer (x)(-y) = -xy

Inner (-y)(x) = --xy

Last (-y)(-y) = y2

x2 – xy – xy + y2 Write all terms together

x2 – 2xy+y2 Combine xy terms.

Final Answer: x2 – 2xy+y2



3. (3y +8) (5y +2) 7. (2x +y) (8x –y)

4. (9x -9) (x +2) 8. (4x +2y)(6x -4y)

F irst (3y)(5y) = 15y2

Outer (3y)(2) = 6y

Inner (8)(5y) = 40y

Last (8)(2) = 16

15y2 +6y +40y + 16 Write all terms together

15y2 +46y + 16 Combine y terms.

Final Answer: 15y2 +46y + 16

F irst (2x)(8x) =16x2

Outer (2x)(-y) = -2xy

Inner (y)(8x) = 8xy

Last (y)(-y) = -y2

16x2 – 2xy +8xy –y2 Write all terms together

16x2 + 6xy – y2 Combine xy terms.

Final Answer: 16x2 + 6xy – y2

F irst (9x)(x) = 9x2

Outer (9x)(2) = 18x

Inner (-9)(x) = -9x

Last (-9)(2) = -18

9x2 +18x – 9x - 18 Write all terms together

9x2 + 9x - 18 Combine x terms.


F irst (4x)(6x) = 24x2

Outer (4x)(-4y) = -16xy

Inner (2y)(6x) = 12xy

Last (2y)(-4y) = -8y2

24x2 – 16xy +12xy -8y2 Write all terms together

24x2 – 4xy – 8y2 Combine xy terms.

Final Answer: 24x2 – 4xy – 8y2



Part 2: Applying Your Knowledge

Find the expression that represents

the area.

1.

Let the length = x +6

Let the width = x-3

2. Find the area of a square

whose side is 8x-5

3. What do the letters FOIL stand for?

4. Fill in the missing blanks.

(x -5) (x____) = x2 _____ - 35

The formula for area of a rectangle is: A = lw

We know the length is; x+6 the width is: x-3

Substitute:

A = lw

A = (x+6)(x-3)

A = (x)(x) +(x)(-3)+ 6(x) +(6)(-3) Multiply

A = x2 -3x +6x - 18

A = x2 +3x – 18 Combine -3x +6x

The expression that represents the area is:

x2 +3x – 18

First

Outer

Inner

Last

A square has 4 equal sides, so the length and

width of this square is: 8x-5.

A = lw

A = (8x-5)(8x-5) Substitute for l and w.

A = (8x)(8x) + (8x)(-5) +(-5)(8x) +(-5)(-5)

A = 64x2 – 40x – 40x + 25

A = 64x2 -80x + 25 Combine -40x- 40x

The area of the square is: 64x2 – 80x + 25

(x-5) (x + 7 ) = x2 +2x - 35

You have to first think of what number and

sign we can fill into the parenthesis. Since

the constant is -35, we have to think about

our last terms. -5 times what will give you -

35. -5 times 7, so we must fill in the

parenthesis with + 7.

Now we have: (x-5)(x+7)

(x)(x) + (x)(7) + (-5)(x) + (-5)(7)

x2 + 7x – 5x - 35

x2 + 2x – 35 Combine 7x – 5x

+2x goes in the last blank



5. Emily is framing a picture. She would like to place a mat around the photo. The length of the

photo is 12 inches. The width of the photo is 8 inches. The width of the mat on each side is

represented by x.

Length of mat

12 in.

8 in

X

X


length of the mat in terms of x.


width of the mat in terms of x.


area of the mat. (This is tricky –

remember there is hole in the center

where the picture is placed.)

• Suppose the width of the mat on all

sides (x) is 3 inches, what is the area of

just the mat?

PICTURE

• The length of the mat is 12in (the length of the picture) plus a measurement of x on both sides.

Therefore, the length can be written as 2x + 12

• The width of the mat is 8 in (the width of the picture) plus a measurement of x on both sides.

Therefore, the width can be written as 2x + 8

• The area of the mat if the mat were solid, would be A = lw

A = (2x+12) (2x+8) but we also have to remember that the middle is cut out for the picture.

So, what’s the area of the picture? A = lw

A = 12(8)

A = 96 in2

So, we have to find the expression for the area using the formula above and then subtract the

area of the picture which is 96 in2

A = lw

A = (2x+12)(2x+8)

A = 2x(2x) + (2x)(8) + 12(2x) +12(8)

A = 4x2 + 16x + 24x + 96

A = 4x2 + 40x + 96 This would be the area of the mat if it did not contain the cut-out for the

picture. Now let’s subtract the area of the picture (which is the same size as the cut-out). The

area of the picture is 96in2.

A = 4x2 + 40x +96 – 96

A = 4x2 + 40x is the area of the mat.

• If x = 3 (the width of the mat on all sides), then we can substitute to find the area.

A = 4x2 + 40x

A = 4(3)2 + 40(3)

A = 156in2 The area of just the mat is: 156 in2

x 12 in x

x

x

8in



5. (a +9) (a -9) 10. (9m -2n)(m-2n)

Directions: Simplify (2 points each)

1. (a +9) (a -9) 2. (9m -2n)(m-2n)

F irst (a)(a) = a2

Outer (a)(-9) = -9a

Inner (9)(a) = 9a

Last (9)(-9) = -81

a2 -9a + 9a -81 Write all terms together

a2 - 81 Combine a terms.

Final Answer: a2 - 81

F irst (9m)(m) = 9m2

Outer (9m)(-2n) = -18mn

Inner (-2n)(m) = -2mn

Last (-2n)(-2n) = 4n2

9m2 -18mn – 2mn +4n2 Write all terms together

9m2 + -20mn +4n2 Combine mn terms.

Final Answer: 9m2 - 20mn +4n2

In order to find the area of the green mat, we will need to find the area of the large green square and subtract

the area of the small white square.

Step 1: Find the area of the large green square.

A = s2

A = (4x+1)2

A =(4x+1)(4x+1)

A = (4x)(4x) +4x(1)+1(4x)+1(1)

A = 16x2 +4x +4x +1

A = 16x2 + 8x +1 Area of large green square

Step 3: Subtract the area of the small white square from the area of the large green square to find the area

of just the mat.

(16x2 +8x +1) – (9x2 – 12x + 4) Subtract

16x2 + 8x + 1 – 9x2 + 12x – 4 Distribute the minus sign ( a negative) throughout parenthesis

16x2 – 9x2 + 8x + 12x + 1 – 4 Write like terms together

7x2 + 20x -3 is the area of the green photo mat. Combine like terms.

To the left is a picture frame that includes matting. This picture

frame is a perfect square and the photo inside is a perfect square.

(4 points)

• Find the area of the photo mat (green section). **There is a

square cut out for the photo insertion.***

• Explain step by step how you arrive at your final answer.

Step 2: Find the area of the small white

square.

A = s2

A = (3x-2)2

A = (3x-2)(3x-2)

A = (3x)(3x) +3x(-2)+-2(3x)+-2(-2)

A = 9x2 -6x – 6x +4

A = 9x2 – 12x + 4 Area of small white

square


Unit Unit Unit Unit 8888: Polynomials: Polynomials: Polynomials: Polynomials

Polynomials – Quiz #1

Part 1: Simplify each expression.

1. (3x2 + 2x -7) + (8x2 + 2x -5) 6. (10x2 -9x +2) + (3x +4)

2. (2x2 – 4x +1) – (x +3) 7. (9x -5) (2x -3) + (3x-2)(x+2)

3. 8x2 (3x2 -9x +1) 8. -9x2 (8x +2) – (x+2) (x-1)

4. (6x3 -5x +2) – (3x3 +7x2 -8x +1) 9. (10x3 -6x2 +5x -9) – (3x3 +2x2 -2x)

5. (x+4)(x-8) 10. (2x-y) (5x +4y) – (3x+y) (2x-y)

Part 2: Applying Your Knowledge.

1. A rectangular picture has the following dimensions (in inches):

Length = x +3 Width = 3x -5

• Write an expression that represents the perimeter of the picture.

• If the perimeter is 44 inches, find the width of the picture.

• Justify your answer.

2. Use the same picture and dimensions as problem #1.

• Write an expression that represents the area of the picture.

• If x = 6 inches, find the area of the picture.


3. How much greater is the area of a square with side (x+4) than the area of a square with side (x-

2)?

• Justify your answer using substitution. (Choose any number for x and show that your answer

is correct.



Polynomials – Quiz #1 - Answer Key

Part 1: Simplify each expression. (1 point each)

1. (3x2 + 2x -7) + (8x2 + 2x -5) 6. (10x2 -9x +2) + (3x +4)

2. (2x2 – 4x +1) – (x +3) 7. (9x -5) (2x -3) + (3x-2)(x+2)

3. 8x2 (3x2 -9x +1)

Step 1: Write vertically and add.

3x2 + 2x – 7

+ 8x2 + 2x - 5

11x2 + 4x – 12

Final Answer: 11x2 + 4x – 12


10x2 – 9x + 2

+ 3x + 4

10x2 -6x + 6

Final Answer: 10x2 -6x + 6

Step 1: Rewrite as an addition problem.

(2x2 – 4x + 1) + (-x – 3)


2x2 – 4x + 1

+ -x – 3

2x2 - 5x - 2

Final Answer: 2x2 - 5x - 2

Step 1: Multiply using FOIL

(9x – 5)(2x-3) + (3x – 2) (x+2)

9x(2x) +9x(-3) +(-5)(2x)+ (-5)(-3) + 3x(x)+3x(2)+-2(x)+-2(2)

18x2 – 27x – 10x +15 + 3x2 + 6x – 2x – 4

(18x2 – 37x + 15) + ( 3x2 + 4x – 4)


18x2 – 37x + 15

+ 3x2 + 4x – 4

21x2 – 33x + 11

Final Answer: 21x2 – 33x + 11

Step 1: Distribute 8x2

8x2 (3x2 -9x +1) = 24x4 – 72x3 + 8x2

24x4 – 72x3 + 8x2 There are no like

terms, therefore, this is the final

answer.




4. (6x3 -5x +2) – (3x3 +7x2 -8x +1) 8. -9x2 (8x +2) – (x+2) (x-1)

5. (x+4)(x-8) 9. (10x3 -6x2 +5x -9) – (3x3 +2x2 -2x)

Step 1: Distribute -9x2 throughout the first set of

parenthesis

. -9x2 (8x +2) – (x+2) (x-1)

-72x3 – 18x2 – (x+2)(x-1)

Step 2: Use foil for the second part of the problem.

-72x3 – 18x2 – [x(x) + x(-1) + 2(x) +2(-1)]

-72x3 – 18x2 – (x2 –x +2x – 2)

-72x3 – 18x2 – (x2 +x – 2)

Step 3: Distribute a -1 throughout the second

parenthesis.

-72x3 – 18x2 – x2 – x +2

Step 4: Organize and combine like terms.

Final Answer: -72x3 – 19x2 – x +2

(6x3 -5x +2) – (3x3 +7x2 -8x +1)

Step 1: Distribute a -1 throughout the

second parenthesis.

(6x3 -5x +2) + (-3x3 – 7x2 + 8x – 1)


6x3 + 0x2 -5x +2

+ -3x3 – 7x2 + 8x – 1

3x3 – 7x2 + 3x + 1

Final Answer: 3x3 – 7x2 + 3x + 1

Step 1: Use foil to multiply.

(x+4)(x-8)

(x)(x) + x(-8) + 4(x) + 4(-8)

x2 – 8x + 4x - 32

x2 – 4x – 32

Final Answer: x2 – 4x - 32

(10x3 -6x2 +5x -9) – (3x3 +2x2 -2x)

Step 1: Distribute a -1 throughout the second set of

parenthesis.

(10x3 -6x2 +5x -9) + (-3x3 – 2x2 + 2x)


10x3 – 6x2 + 5x – 9

+ -3x3 - 2x2 + 2x

7x3 - 8x2 + 7x - 9

Final Answer: 7x3 - 8x2 + 7x - 9



10. (2x-y) (5x +4y) – (3x+y) (2x-y)

Part 2: Applying Your Knowledge. (3 points each)

1. A rectangular picture has the following dimensions (in inches):

Length = x +3 Width = 3x -5

• Write an expression that represents the perimeter of the picture.

Step 1: Use foil to multiply the first set of parenthesis and the second set of parenthesis.

(2x-y) (5x +4y) – (3x+y) (2x-y)

2x(5x) +2x(4y) +-y(5x) +-y(4y) - 3x(2x) + 3x(-y) + y(2x) + y(-y)

10x2 + 8xy – 5xy – 4y2 - 6x2 – 3xy + 2xy – y2

10x2 + 3xy – 4y2 - 6x2 – xy – y2+

Now we have: (10x2 + 3xy – 4y2) – (6x2 – xy – y2)

Step 2: Distribute -1 throughout the second set of parenthesis.

(10x2 + 3xy – 4y2) + (-6x2 + xy + y2)

Step 3: Rewrite with like terms together.

10x2 – 6x2 + 3xy + xy – 4y2 + y2

Step 4: Combine like terms.

4x2 + 4xy – 3y2

Final Answer: 4x2 + 4xy – 3y2

Perimeter = 2(length) + 2(width) (2 times the length + 2 times the width)

Let’s substitute for length and width.

P = 2(x+3) + 2(3x-5)

Let’s distribute and simplify

P = 2x+6 + 6x – 10

P = 8x – 4

The perimeter is equal to: P = 8x - 4



• If the perimeter is 44 inches, find the width of the picture.


2. Use the same picture and dimensions as problem #1.

• Write an expression that represents the area of the picture.

Using the same formula: P = 8x -4, we will substitute 44 for P.

44 = 8x – 4

Solve for x.

44+4 = 8x – 4 + 4 (Add 4 to both sides)

48 = 8x

48/8 = 8x/8 (Divide by 8 on both sides)

6 = x

We know the width is equal to: 3x – 5 and x = 6, so we’ll substitute.

Width: 3(6) – 5

Width: 13 inches.

We can justify our answer by substituting into the original formula.

We were given: Length = x + 3 Width = 3x-5 Perimeter = 44

We found x = 6 Length: 6+ 3 = 9 Width: 3(6) – 5 = 13 Perimeter = 44

P = 2l + 2w now let’s substitute and see if the problem works out.

44 = 2(9) +2(13)

44 = 18 + 26

44 = 44

Since our substitution works, we know that our solutions are correct.

Area is equal to length times width. A = lw

We know: length = x+3 width = 3x-5

A = (x+3)(3x-5)

We will use FOIL to find the Area expression.

(x+3)(3x-5)

x(3x) +x(-5) +3(3x) + 3(-5)

3x2 – 5x + 9x – 15

A = 3x2 + 4x - 15



• If x = 6 inches, find the area of the picture.


If x = 6, we can substitute to find the area of the picture.

A = 3x2 + 4x -15

A = 3(6)2 + 4(6) – 15

A = 117 in2

The area of the picture is: 117 in2

We can justify our answer by substituting.

We know: length = x+3 Width = 3x-5 x = 6

Length: 6+3 = 9 Width: 3(6) -5 = 13

A = lw

A = 9(13)

A = 117 in2 which is the same answer that we found in the problem above. Since they are the same

answer, we know our solution was correct.



3. How much greater is the area of a square with side (x+4) than the area of a square with side (x-

2)?

• Justify your answer using substitution. (Choose any number for x and show that your answer

is correct.

Area = lw

The Area of a Square with side (x+4) is:

(x+4)(x+4)

Use foil to multiply:

x(x) + x(4) + 4(x) + 4(4)

x2 + 4x + 4x + 16

x2 + 8x + 16

.When you want to find how much greater something is than something else, you subtract.

Think of an example that you are familiar with: How much greater is 5 than 2? You would subtract to

find your answer of 3.

Therefore, we must subtract the two expressions above in order to find how much greater.

(x2 + 8x + 16) – (x2 – 4x + 4)

Step 1: Distribute -1 throughout the second set of parenthesis.

(x2 + 8x + 16) + (-x2 + 4x – 4)

Now we can write vertically and add (or combine like terms in your head ☺)

x2 + 8x + 16

+ -x2 + 4x - 4

0 + 12x +12

The square with side x+ 4 is 12x + 12 units greater than the square with side x - 2 .

Justify:

Let x = 3 in (You can pick any number)

Side of Sq 1 = x +4 (3+4= 7) Side of Sq 2 = x – 2 (3-2=1)

Area of Square 1: Area of Square 2:

A = 7(7) A = 1(1)

A = 49 in2 A = 1 in2

49 – 1 = 48 is the difference. So, square 1’s area is 48 in2 greater than square 2’s area.

If 12x + 12 = 48 in then our solution is correct.

12(3) + 12 = 48 Therefore our solution was correct!

The area of the square with side (x-2) is:

(x-2)(x-2)

Use FOIL to multiply:

x(x) + x(-2) + -2(x) + -2(-2)

x2 – 2x – 2x + 4

x2 – 4x + 4

This Quiz is worth 19 points.



Lesson 5: Special Binomials

The Square of a Sum

Example 1 Example 2

Example 3

(a+b)2 (x +2)

2

The Square of a Sum

(2x+6)2



The Square of a Difference

Example 1 Example 2

Example 3

(a-b)2 (x -2)

2


(2x-6)2



The Difference of Two Squares

Example 1 Example 2

Example 3

(a+b) (a-b) (x +2) (x-2)


(2x+6) (2x-6)



Lesson 5: Special Binomials Practice

Part 1: State the rule for each of the special binomials listed in the box.

Part 2: Multiply and simplify.

1. (x + 5)2 7. (8x+2)2

2. (x – 5)2 8. (4+3y)(4-3y)

3. (x+5)(x-5) 9. (9x-1)2

4. (2y -1)2 10. (8y-2)2

5. (7y +2)2 11. (10x+3)2

6. (2x -5)(2x+5) 12. (8x -2)(8x+2)

Part 3: Extending Your Knowledge

1. Find the area of a square with a side of length n+4.

• Justify your answer by letting n = 5

2. Find the volume of a rectangular box that has the following dimensions:

Length: 3x-2 Width: 3x-2 Height: 5x2 +1

• Justify your answer by letting x = 2.

The Square of a Sum

(a+b)2 =


(a-b)2 =


(a+b)(a-b) =



1. Simplify: 3(7x+3)2 + 2(3x-8)2 (2 points)

2. Justify that (a+b)2 ≠ a2 + b2 by substituting numbers for a and b. (2 points)

8x+4

3. Simplify: 5(2x-1)2

4. Simplify: (3x+1)2 + (x-2)2

5. Simplify: (a+b)2 + (a-b)2

6. Simplify: 6(2x-1)(2x+1)

7. You have a square box with a side of y-6. You also have a rectangular box with the following

dimensions: Length: y – 6 Width: y +6 Height: y -2

• Find the volume for each box.

• Suppose y = 12. Which box has the greater volume?


3. You would like to build a pool with a deck

around the perimeter. The dimensions are 5x + 6

given in the drawing to the right.

(4 points) 8x-4

• Find the area of the large

shaded rectangle (deck area)

(Area of the entire rectangle, including pool)

• Find the area of the white rectangle (represents the area of the pool)

• Find the area of the area of just the deck portion surrounding the pool. (Shaded area

that you see in the diagram.

• Suppose that x = 3 feet. What is the area of just the deck portion surrounding the

pool?

POOL

5x-6



Lesson 5: Special Binomials Practice

Part 1: State the rule for each of the special binomials listed in the box.


1. (x + 5)2 7. (8x+2)2

2. (x – 5)2 8. (4+3y)(4-3y)

The Square of a Sum

(a+b)2 = a2 + 2ab + b2


(a-b)2 = a2 – 2ab + b2


(a+b)(a-b) = a2 – b2

(a+b)2 = a2 + 2ab + b2

(x+5)2 = x2 +2(x)(5) + 52

(x+5)2 = x2 +10x + 25

(Substitute x for a and 5 for b. Then simplify)

Final Answer: x2 +10x + 25

(a+b)2 = a2 + 2ab + b2

(8x+2)2 = (8x)2 +2(8x)(2) + 22

(8x+2)2 = 64x2 +32x + 4

(Substitute 8x for a and 2 for b. Then simplify)

Final Answer: 64x2 +32x + 4

(a-b)2 = a2 - 2ab + b2

(x-5)2 = x2 -2(x)(5) + 52

(x-5)2 = x2 -10x + 25


Final Answer: x2 -10x + 25

(a+b)(a-b) = a2 – b2 a = 4, b = 3y

(4+3y)(4-3y) = 42 – (3y)2

(4+3y)(4-3y) = 16 – 9y2

(Substitute 4 for a and 3y for b. Then simplify)

Final Answer: 16 – 9y2



3. (x+5)(x-5) 9. (9x-1)2

4. (2y -1)2 10. (8y-2)2

5. (7y +2)2 11. (10x+3)2

6. (2x -5)(2x+5) 12. (8x -2)(8x+2)

(a+b)(a-b) = a2 – b2 a = x, b = 5

(x+5)(x-5) = x2 – 52

(x+5)(x-5) = x2 – 25


Final Answer: x2 – 25

(a-b)2 = a2 - 2ab + b2

(9x-1)2 = (9x)2 -2(9x)(1) + 12

(9x-1)2 = 81x2 -18x + 1


Final Answer: 81x2 -18x + 1

(a-b)2 = a2 - 2ab + b2

(2y-1)2 = (2y)2 -2(2y)(1) + 12

(2y-1)2 = 4y2 -4y + 1

(Substitute 2y for a and 1 for b. Then simplify)

Final Answer: 4y2 -4y + 1

(a-b)2 = a2 - 2ab + b2

(8y-2)2 = (8y)2 -2(8y)(2) + 22

(8y-2)2 = 64y2 -32y + 4

(Substitute 8y for a and 2 for b. Then simplify)

Final Answer: 64y2 -32y + 4

(a+b)2 = a2 + 2ab + b2

(7y+2)2 = (7y)2 +2(7y)(2) + 22

(7y+2)2 = 49y2 +28y + 4

(Substitute7y for a and 2 for b. Then simplify)

Final Answer: 49y2 +28y + 4

(a+b)2 = a2 + 2ab + b2

(10x+3)2 = (10x)2 +2(10x)(3) + 32

(10x+3)2 = 100x2 +60x + 9

(Substitute10x for a and 3 for b. Then simplify)


(a+b)(a-b) = a2 – b2 a = 2x, b = 5

(2x-5)(2x+5) = (2x)2 – 52

(2x-5)(2x+5) = 4x2 – 25


Final Answer: 4x2 – 25

(a+b)(a-b) = a2 – b2 a = 8x, b = 2

(8x-2)(8x+2) = (8x)2 – 22

(8x-2)(8x+2) = 64x2 – 4





Part 3: Extending Your Knowledge

1. Find the area of a square with a side of length n+4.

• Justify your answer by letting n = 5

Area of a square = side (squared) or A = s2 s = n+4

A = (n+4)2

Use the definition for a perfect square: (a+b)2 = a2 + 2ab + b2 where a = n b = 4

A = n2 + 2(n)(4) +42

A = n2 + 8n + 16

Justify by letting n = 5

We know that a square with side 9 ( 5+4 = 9) has an area of 81 units 2 (92 = 81)

Let’s justify our answer: n2 + 8n + 16 where n = 5

A = 52 + 8(5) + 16

A = 81 units2 We know our solution is correct!



2. Find the volume of a rectangular box that has the following dimensions:

Length: 3x-2 Width: 3x-2 Height: 5x2 +1

• Justify your answer by letting x = 2.

Volume of a rectangular box is: V = lwh (length times width times height)

Substitute:

V = lwh

V = (3x-2)(3x-2) (5x2 + 1)

Step 1: We have a perfect square. Let’s apply our rule.

V = (3x-2)2 (5x2 +1)

V = (9x2 -12x +4) (5x2 +1)

Now continue multiplying using the distributive property. ( I will rewrite it with the binomial first)

V = (5x2 +1) (9x2 – 12x + 4)

V = 5x2(9x2) + 5x2(-12x) + 5x2 (4) +(1)9x2 + 1(–12x) +1(4)

V = 45x4 – 60x3 + 20x2 +9x2 – 12x + 4

Combine like terms: 20x2 + 9x2

V = 45x4 – 60x3 + 29x2 – 12x + 4

The volume is: 45x4 – 60x3 + 29x2 – 12x + 4

Now let’s justify:

We know: Length: 3x-2 Width: 3x-2 Height: 5x2 +1 Use our formula:

If x = 2: L =3(2)-2 = 4 W=4 H=5(2)2 +1 = 21 V =45x4 – 60x3 + 29x2 – 12x + 4

V = 4(4)(21) V =45(2)4 -60(2)3 +29(2)2 -12(2) +4

V = 336 units3 V = 720 – 480 + 116 – 24 +4

V = 336 units3

The volumes are the same; therefore, our solution is correct.



3. Simplify: 5(2x-1)2

4. Simplify: (3x+1)2 + (x-2)2

5. Simplify: (a+b)2 + (a-b)2

Remember the order of operations? Parenthesis, Exponents, Multiply/Divide, Add/Subtract

Even with polynomials, we must follow the order of operations. In this problem, we have multiplication

and exponents. Therefore, we must complete (2x-1)2 first, then multiply that answer by 5.

Step 1: 5(2x-1)2 Expand the exponent using our special rules for binomials. (Perfect Square)

5(4x2 -4x +1)

Step 2: Distribute the 5.

5(4x2 -4x +1) = 20x2 – 20x + 5


Step 1: We must expand the exponents first. They are both perfect squares, so we can use our

special binomial rule for perfect squares.

(3x+1)2 + (x-2)2

(9x2 +6x +1) + (x2 – 4x +4)

Step 2: Add the two polynomials. (Write vertically or add like terms in your head!)

9x2 + 6x + 1

+ x2 – 4x + 4

10x2 +2x + 5


Step 1: We must expand the exponents first. They are both perfect squares.

(a+b)2 + (a-b)2

(a2 +2ab +b2) + (a2 – 2ab +b2)

Step 2: Add the two polynomials. (Write vertically or add like terms in your head!)

a2 +2ab +b2

+ a2 – 2ab + b2

2a2 +0ab + 2b2

Final Answer: 2a2 + 2b2



6. Simplify: 6(2x-1)(2x+1)

7. You have a square box with a side of y-6. You also have a rectangular box with the following

dimensions: Length: y – 6 Width: y +6 Height: y -2

• Find the volume for each box.

In this problem, we have all multiplication. Normally you would start at the left and multiply, but we have

a special binomial rule that we can apply that will make this problem easier. Notice that (2x-1)(2x+1) is

the difference of two squares. So we can apply that rule first and then distribute the 6.

Step 1: Use the difference of two squares rule to multiply (2x-1)(2x+1)

6(2x-1)(2x+1)

6 (4x2 -1)

Step 2: Distribute the 6 throughout the parenthesis.

6 (4x2 -1) = 24x2 – 6


Square Box

Each side = y-6. Since all sides are the same

measurement the formula for volume is: V = s3

V = s3

V = (y-6)3 or (y-6)(y-6)(y-6)

(y-6) (y-6)2

Let’s use the perfect square rule first.

(y-6) (y-6)2

(y-6)(y2 -12y+36)

Distribute each term in (y-6) throughout the

parenthesis.

(y-6)(y2 -12y +36)

y(y2) +y(-12y)+y(36) + -6(y2)+ -6(-12y) + -6(36)

y3 – 12y2 + 36y -6y2 + 72y - 216

The volume for the box is: (Combine like terms)

y3 – 18y2 +108y - 216

Rectangular Box

L = y – 6 W = y+6 H = y-2

V = LWH (Length ● Width ● Height)

V = (y-6)(y+6) (y-2)

We’ll use the difference of two squares rule first to

multiply.

V = (y-6)(y+6) (y-2)

V = (y2 – 36) (y-2) or (y-2)(y2 – 36)

Continue multiplying using FOIL:

(y-2)(y2 – 36)

y(y2) + y(-36) + -2(y2) + -2(-36)

y3 -36y -2y2 + 72

The volume for the rectangular box is:

y3 – 2y2 – 36y + 72



• Suppose y = 12. Which box has the greater volume?


1. Simplify: 3(7x+3)2 + 2(3x-8)2

If y = 12, then the following would be the volumes for each box:

Square Box Rectangular Box

V = y3 – 18y2 +108y - 216 V = y3 – 2y2 – 36y + 72

V = (12)3 – 18(12)2 +108(12) – 216 V = (12)3 – 2(12)2 – 36(12) + 72

V = 216 units3 V = 1080 units3

The rectangular box has the greater volume.

Square box with sides y – 6

Substitute 12 for y. 12-6 = 6 units

Volume = s3

V = 63

V = 216 units 3 This agrees with our

answer above, therefore our solution was

correct.

Rectangular box with Length: y – 6 Width: y + 6

Height: y - 2

Substitute 12 for y. L = 6 W = 18 H = 10

Volume = LWH

V = 6(18)(10)

V = 1080 units3 This agrees with our answer above,

therefore, our solution was correct.

Step 1: We must expand the exponents first. There are 2, and they are both perfect squares, so we can

utilize our special binomial rule.

3(7x+3)2 + 2(3x-8)2

3(49x2 +42x +9) + 2 (9x2 – 48x + 64)

Step 2: Distribute the 3 throughout the first parenthesis and distribute the 2 throughout the

second parenthesis.

3(49x2 +42x +9) + 2 (9x2 – 48x + 64)

(147x2 + 126x + 27) + (18x2 – 96x + 128)

Step 3: Add like terms. Set up vertically and add or add like terms in your head)

147x2 + 126x + 27

+ 18x2 – 96x + 128

165x2 +30x + 155 Final Answer: 165x2 +30x + 155



2. Justify that (a+b)2 ≠ a2 + b2 by substituting numbers for a and b. (2 points)

8x + 4

3. You would like to build a pool with a deck

around the perimeter. The dimensions are 5x + 6

given in the drawing to the right.

(4 points) 8x-4

• Find the area of the large

shaded rectangle (deck area)

(Area of the entire rectangle, including pool)

• Find the area of the white rectangle (represents the area of the pool)

Let’s let a = 1 and b = 2 (You can choose any numbers for a and b, but I would suggest using small

numbers to make the calculations easier.)

(a+b)2 a2+ b2

(1+2)2 12 + 22

(3)2 = 9 1 + 4 = 5

Since 9 ≠ 5 we can verify that (a+b)2 ≠ a2+b2

POOL

5x-6

The length of the shaded rectangle is: 8x+4 The width of the shaded rectangle is 8x-4.

A = L● W

A = (8x+4) (8x-4) Notice how this is the difference of two squares – so we can use our rules!

A = 64x2 -16

The length of the white rectangle is: 5x+6 The width is 5x-6

A = L● W

A = (5x+6)(5x-6) Notice how this is the difference of two squares - so we can use our special rules!

A = 25x2 - 36



• Find the area of the area of just the deck portion surrounding the pool. (Shaded area

that you see in the diagram.

• Suppose that x = 3 feet. What is the area of just the deck portion surrounding the

pool?

In order to find the area of just the deck portion, we must take the area of the shaded rectangle and

subtract the area of the white rectangle. This leaves just the shaded portion around the pool, which

would be considered the deck.

Area of the entire shaded portion – Area of the white rectangle

(64x2 -16) - (25x2 – 36)

Rewrite as an addition problem: (64x2 – 16) + (-25x2 + 36)

Rewrite with like terms together: 64x2 – 25x2 – 16 + 36

The area of deck portion: 39x2 +20

A = 39x2 + 20 Let x = 3

A = 39(3)2 + 20

A = 371 ft3

The area of the deck portion surrounding the pool is 371ft2.



Lesson 6: More on Multiplying Polynomials

Example 1

Example 2

(2x -1) (3x2 + x -1)

(2x+y)2 (2x +2y -1)

The Extended Distributive Property

Multiply _____________________________________ by ______________________

___________________________________________________________________



Lesson 6: More Multiplying Polynomials Practice Problems


1. 7x (4x3 – 2x2 + 5x -1) 5. (a2 – 2a+1) (2a2+3a -2)

2. (2x3 -2x2 + 6x +4) 9x2 6. (y2 –y +8) (y2+4y -6)

3. (y2 – 2y + 6) (y+3) 7. (a-b) (2a – 3b -1)

4. (a-2)2 (a3 – 2a2+7) 8. (2a -3b + 2) (a +3b +4)

Part 2: Thinking Problems

9. Multiply: (2x-4y)3

10. Simplify: ( 2r-2)2 – (r+1)2

11. How much greater is the volume of a cube with sides of length 2x+4 than a rectangular box with

a length of x+2, a width of 3x+2, and a height of x+9?

12. You a creating a circular flower bed with a radius of 3x-5. Find the area of the flower bed.

• Justify your answer by substituting a number for x.

#3 on the following page


1. (b +5)2 (b2 – 2b + 8) 2. (x+y)(x-y) (2x – 4y + 2)



3. A sketch of a festival is shown below. The festival includes a theatre, food court, crafts area,

and kids play area. (Dimensions are in blue and orange) (4 points)

• Write a polynomial that represents the area of the entire festival.

• Write a polynomial that represents the combined area of theatre and kids play area.

• Suppose x = 8 feet and y = 12feet. Find the area of the entire festival, the combined area of

the theatre and kids play area and the combined area of the food court and crafts area.

• Explain how you determined your answers.

10x

10x

Theatre

Food Court

Kids Play Crafts Area

Area

Y

y



Lesson 6: More Multiplying Polynomials Practice Problems - Answers


1. 7x (4x3 – 2x2 + 5x -1) 5. (a2 – 2a+1) (2a2+3a -2)

2. (2x3 -2x2 + 6x +4) 9x2 6. (y2 –y +8) (y2+4y -6)

3. (y2 – 2y + 6) (y+3)

7. (a-b) (2a – 3b -1)

Step 1: Distribute 7x throughout the

parenthesis.

7x (4x3 – 2x2 + 5x -1)

7x(4x3) + 7x(-2x2) + 7x(5x) + 7x(-1)

28x4 – 14x3 + 35x2 – 7x

Final Answer: 28x4 – 14x3 + 35x2 – 7x

Distribute each term in the 1st parenthesis, throughout the 2

nd

parenthesis.

a2(2a

2)+a

2(3a)+a

2(-2)+ -2a(2a

2)+ -2a(3a)+ -2a(-2)+1(2a

2)+1(3a)+1(-2)

2a4 +3a

3 - 2a

2 -4a

3 -6a

2 + 4a +2a

2 + 3a - 2

Rewrite with like terms together: (Take the sign in front of each term)

2a4 + 3a

3 – 4a

3 – 2a

2 – 6a

2 + 2a

2 +4a +3a – 2

Combine like terms: 2a4 –a

3 -6a

2 + 7a – 2

Final Answer: 2a4 –a

3 -6a

2 + 7a – 2

You can rewrite as: 9x2(2x3 -2x2 +6x +4)

Distribute 9x2 throughout the

parenthesis.

9x2(2x3 -2x2 +6x +4)

9x2(2x3) +9x2(-2x2)+ 9x2(6x)+ 9x2(4)

18x5 -18x4 +54x3 +36x2

Final Answer:18x5 – 18x4 + 54x3 + 36x2


nd

parenthesis.

y2(y

2)+y

2(4y)+y

2(-6)+ -y(y

2)+ -y(4y)+ -y(-6)+8(y

2)+8(4y)+8(-6)

y4 +4y

3 - 6y

2 -y

3 -4y

2 + 6y +8y

2 +32y -48

Rewrite with like terms together: (Take the sign in front of each term)

y4 + 4y

3 – y

3 – 6y

2 –4y

2 + 8y

2 +6y +32y – 48

Combine like terms: y4 +3y

3 -2y

2 + 38y – 48

Final Answer: y4 +3y

3 -2y

2 + 38y – 48

You can rewrite as (y+3)(y2 – 2y + 6)

Distribute each term in (y+3) throughout the

2nd

parenthesis.

y(y2)+y(-2y)+y(6)+3(y

2)+3(-2y)+3(6)

y3 - 2y

2 +6y +3y

2 - 6y + 18

Rewrite like terms together:

y3 – 2y

2 + 3y

2 + 6y -6y + 18

Final Answer: y3 +y

2 +18


nd

parenthesis.

a(2a)+a(-3b)+a(-1) + -b(2a)+ -b(-3b)+ -b(-1)

2a2 -3ab -a -2ab +3b

2 + b

Rewrite like terms together:

2a2 -3ab -2ab –a + b +3b

2

Final Answer: 2a2 -5ab –a +b +3b

2



4. (a-2)2 (a3 – 2a2+7)

8. (2a -3b + 2) (a +3b +4)

Multiply the perfect square first.

(a-2)2 (a

3 – 2a

2+7)

(a2 -4a+4) (a

3 -2a

2 +7)

Distribute each term in the 1st parenthesis.

a2(a

3)+a

2(-2a

2)+a

2(7)+ -4a(a

3)+ -4a(-2a

2) + -4a(7)+ 4(a

3) +4(-2a

2) +4(7)

a5 -2a

4 +7a

2 -4a

4 +8a

3 -28a +4a

3 - 8a

2 + 28

Rewrite with like terms together:

a5 -2a

4 – 4a

4 +8a

3 + 4a

3 + 7a

2 – 8a

2 -28a + 28

Combine like terms to get your final answer:

Final answer: a5 -6a4 +12a3 –a2 -28a + 28


2a(a) + 2a(3b) + 2a(4) + -3b(a) +-3b(3b) +-3b(4) + 2(a) + 2(3b) + 2(4)

2a2 +6ab +8a -3ab -9b

2 -12b +2a +6b +8


2a2 +8a +2a +6b -12b +6ab -3ab -9b

2 +8


Final Answer: 2a2 +10a -6b +3ab -9b

2 +8



Part 2: Thinking Problems

9. Multiply: (2x-4y)3

10. Simplify: ( 2r-2)2 – (r+1)2

Rewrite this problem as: (2x-4y)2(2x-4y)

Multiply the perfect square using special binomial rules:

(2x-4y)2 (2x-4y)

(4x2 -16xy + 16y2) (2x-4y) Rewrite with the binomial first: (2x-4y) (4x2 – 16xy +16y2)

Distribute the binomial throughout the second set of parenthesis.

2x(4x2) + 2x(-16xy) +2x(16y2) + -4y(4x2) + -4y(-16xy) + -4y(16y2)

8x3 - 32x2y + 32xy2 -16x2y + 64xy2 - 64y3


8x3 -32x2y – 16x2y +32xy2 + 64xy2 – 64y3

Combine like terms for your final answer:

8x3 – 48x2y + 96xy2 – 64y3

Multiply each perfect square first, using the special binomial rules.

( 2r-2)2 – (r+1)2

(4r2 – 8r + 4) - (r2+ 2r +1)

Change to an addition problem by distributing a -1 throughout the second parenthesis.

(4r2 – 8r + 4) - (r2+ 2r +1)

(4r2 – 8r + 4) + (-r2- 2r -1)

Add by combining like terms or rewriting vertically.

4r2 -8r + 4

+ -r2 - 2r -1

3r2 -10r +3

Final Answer: 3r2 -10r +3



11. How much greater is the volume of a cube with sides of length 2x+4 than a rectangular box with

a length of x+2, a width of 3x+2, and a height of x+9?

We must first find the volume of the cube and the volume of the rectangular box.

Volume of the cube: V = s3

V = (2x+4)3 (Substitute 2x+4 for s.)

V = (2x+4)2 (2x+4) Rewrite so that you have a perfect square.

V = (4x2 + 16x + 16) (2x+4) Use your special binomial rule to multiply (2x+4)

2

V = (2x+4)(4x2 + 16x + 16) Rewrite with the binomial first. (Makes distributing easier)

V = 2x(4x2) + 2x(16x) + 2x(16) + 4(4x

2) + 4(16x) + 4(16) Distribute the binomial.

V = 8x3 + 32x

2 + 32x + 16x

2 + 64x + 64

V = 8x3 + 32x

2 + 16x

2 +32x + 64x + 64 Rewrite with like terms together.

V = 8x3 + 48x

2 + 96x + 64 Combine like terms.

Volume of the cube: V = 8x3 + 48x

2 + 96x + 64

Now we must find the volume of the rectangular box so that we can compare them.

length = x+2, a width = 3x+2, height = x+9

V = lwh (length x width x height)

V = (x+2) (3x+2) (x+9) Substitute for l, w, and h.

V = x(3x) +x(2) +2(3x) + 2(2) (x+9) Multiply (x+2)(3x+2)

V = 3x2 +2x +6x +4 (x+9)

V = ( 3x2 + 8x + 4) (x+9) Multiply your answer by (x+9)

V = (x+9) (3x2+8x +4) Rewrite with the binomial in front.

V = x(3x2) +x(8x) +x(4) + 9(3x

2) + 9(8x) + 9(4) Distribute the binomial.

V = 3x3 + 8x

2 + 4x + 27x

2 + 72x + 36

V = 3x3 + 8x

2 + 27x

2 + 4x + 72x + 36 Rewrite with like terms together.

V = 3x3 + 35x

2 + 76x + 36 The volume of the rectangular box.

Since the problem says “How much greater is the cube than the rectangular box, we must subtract the volume of the

box from the volume of the cube.

(8x3 + 48x

2 + 96x + 64) – (3x

3 + 35x

2 + 76x + 36) 8x

3 + 48x

2 + 96x + 64 The difference in volumes

(8x3 + 48x

2 + 96x + 64) +(-3x

3 - 35x

2 - 76x - 36) + -3x

3 -35x

2 - 76x - 36

Change to addition 5x3 +13x

2 +20x + 28



12. You a creating a circular flower bed with a radius of 3x-5. Write an expression for the area of

the flower bed.

• Justify your answer by substituting a number of x.

Since the flower bed is in the shape of a circle, we must know the formula for area of a circle.

Area of a circle = π r2 (Pi x radius (squared)) For pi we’ll use 3.14

A = π r2

A = π (3x-5)2 Substitute 3x-5 for r since it’s the radius.

A = π (9x2 – 30x +25) Use your special binomial rules to multiply (3x-5)2

A = 28.26x2 -94.2x + 78.5 Distribute 3.14 throughout the parenthesis.

The expression for the area of the flower bed is: 28.26x2 -94.2x + 78.5

Justification:

In order to justify we will choose a number for x and prove that our expression is correct. Let x = 2 (We

can’t have a negative number for a radius and 3(2) -5 = 1.

By substituting 2 for x, we know that the radius is 1. Therefore, the area is: π(1)2 = 3.14

If we substitute 2 for x into our expression for area, then we should end up with an answer of 3.14.

A = 28.26x2 -94.2x + 78.5

A = 28.26(2)2 -94.2(2) + 78.5

A = 3.14 units2 I know that my expression is correct, because when x = 2, the area equals 3.14.



1. (b +5)2 (b2 – 2b + 8)

2. (x+y)(x-y) (2x – 4y + 2)

Multiply the perfect square first.

(b+5)2 (b

2 – 2b+8)

(b2 +10b + 25) (b

2 – 2b+8)


b2(b

2) + b

2(-2b) + b

2(8) + 10b(b

2) +10b(-2b) +10b(8) + 25(b

2) + 25(-2b) + 25(8)

b4 - 2b

3 + 8b

2 +10b

3 -20b

2 + 80b +25b

2 -50b +200


b4 – 2b

3 + 10b

3 + 8b

2 - 20b

2 + 25b

2 +80b – 50b + 200


Final Answer: b4 +8b

3 +13b

2 +30b +200


Multiply the difference of two squares first.

(x+y)(x-y) (2x – 4y + 2)

(x2 – y

2) (2x-4y +2)


x2(2x) + x

2(-4y) + x

2(2) + -y

2(2x) + -y

2 (-4y) + -y

2(2)

2x3 -4x

2y +2x

2 -2xy

2 +4y

3 -2y

2

There are no like terms in this expression:

2x3 +2x

2 – 4x

2y – 2xy

2 -2y

2 + 4y

3



3. A sketch of a festival is shown below. The festival includes a theatre, food court, crafts area,

and kids play area. (Dimensions are in blue and orange) (4 points)

• Write a polynomial that represents the area of the entire festival.

• Write a polynomial that represents the combined area of theatre and kids play area.

• Suppose x = 8 feet and y = 12feet. Find the area of the entire festival, the combined area of

the theatre and kids play area and the combined area of the food court and crafts

area.

10x

10x

Theatre

Food Court

Kids Play Crafts Area

Area

Y

y

The length of the festival below is: the length of the theatre (y) + the length of the food court(10x), so

the entire length is (10x + y)

The width of the festival is the width of the theatre(10x) + the width of the kids play area(y), so the

entire width is (10x+y)

The total area is: (10x+y)(10x+y) or (10x+y)2

A = 100x2 +20xy +y2

The length of the theatre and kids play area is y and the width is (10+y).

A = y(10x+y)

A = 10xy + y2

In order to find the area of the entire festival,

we will substitute 8 for x and 12 for y into the

expression: A = 100x2 + 20xy + y2

A = 100(8)2 + 20(8)(12) + (12)2

A = 8464 ft2 Area of entire festival

In order to find the area of the theatre and kids

area, we will substitute 8 for x and 12 for y into

the expression: A = 10xy + y2

A = 10(8)(12) + (12)2

A = 1104 ft2 Area of theatre and kids play

In order to find the area of the food court and

crafts area, we can subtract the area of the

theatre and kids play area (1104 ft2) from the

total area (8464 ft2) because this area is the

only area remaining.

8464 ft2 – 1104 ft2 = 7360 ft2

The area of the food court and crafts area

is: 7360 ft2

You can also justify your answer by substituting 8 for

x and 12 for y using the drawing.

Total area = (y+10x) (10x+y) or (10x+y)2

Total Area = (10(8)+12)2

Total Area = 8464 ft2

Check food court + crafts area.

A = (10x)(10x+y)

A = 10(8) (10(8)+12) = 7360 ft2 This matches!!!


Algebra 2Algebra 2Algebra 2Algebra 2


Lesson 8: Dividing Polynomials

�� + �� − ��

��

Example 1: Dividing a Polynomial by a Monomial

Example 2 – Using the Division Algorithm to Divide a Polynomial by a Polynomial

(x3 +6x2 +11x + 6) ÷ (x+3)




Example 3 – More Practice with the Division Algorithm

(3x3- 10x2 +13x-10) ÷ (x-2)

Example 4 – Quotient with a Remainder

(4x2 – 2x +5)(x-2)-1




Example 5 – Quotient with a Remainder

(2x3 – 2x2 + 4x-6)(3 - x)-1




Lesson 8: Dividing Polynomials Practice Problems

Part 1: Simplify.

1. ��

� 2.

��

��

3. (x3 +x2 – 5x +3)÷ (x+3) 4. (y4 – 4y3 +7y2 +5y +2)÷ (y-2)

5. (a2+3a+6)(a+1)-1 6. ��

��

7. (a3+4a2+4a - 2)(a-1)-1 8. . (2x3 +4x2 – 2x -6)÷ (x-3)

9. ��

�� 10. (5x3 +2x2 – 3x -1)÷ (x-1)

11. (4a2-2a+4)(a-2)-1 12. (x3 -4x2 + 2x -2)÷ (1-x)

13. ��

�� 14. (3x3 -2x2 +5x -3)÷ (x-2)

15. ��

�� 16. (2a2+6a-3)(2-a)-1




Simplify:

1. (2x3 +6x2 -4x +2) ÷ (x+1)

2. (2a3 – 2a2 +6a-2)(a-4)-1



Unit 5: PolynomialsUnit 5: PolynomialsUnit 5: PolynomialsUnit 5: Polynomials

Lesson 8: Synthetic Division

Example 1

(x3 -6x2 +11x - 6) ÷ (x-3)

What is Synthetic Division?

Synthetic division is a simpler process for dividing a polynomial by a binomial. This process only

works if the divisor is a binomial in the form of x – r. (The coefficient of x must be 1.)

Step 1: Write the terms of the dividend so that

the degrees are in descending order.

Step 2: Write just the coefficients in the dividend.

Then write the constant of the divisor off to the

left.

Step 3: Bring the first coefficient down.

Step 4: Multiply the first coefficient by the

constant (3). Write the product under the second

coefficient. Then add.

Step 5: Multiply the sum (-3) b the constant (3)

and write the product under the next coefficient.

Then add.

Step 6: Multiply the sum (2) by the constant 3

and write the product under the next coefficient.

Then add.

The numbers along the bottom row are the

coefficients of the quotient. Start with the

power of x that is one less than the degree of

the dividend.




Example 2

(3x4 – 3x2 +2x -1) ÷ (x+2)

Example 3

(2x3 – 2x2 +4x -6)(2x-4)-1

What is Synthetic Division?

Synthetic division is a simpler process for dividing a polynomial by a binomial. This process only

works if the divisor is a binomial in the form of x – r. (The coefficient of x must be 1.)




Example 4

(4x3 –2x2 + x+2)(2x-1)-1




Lesson 8: Synthetic Division Practice Problems

Part 1: Simplify.

1. (x3 +x2 – 5x +3)÷ (x+3) 2. (y4 – 4y3 +7y2 +5y +2)÷ (y-2)

3. (3a3+a2+3a+6)(a+1)-1 4. ��

��

5. (a3+4a2+4a - 2)(a-1)-1 6. (2x3 +4x2 – 2x -6)÷ (x-3)

7. ��

�� 8. (5x4 +2x2 – 3x -1)÷ (x-1)

9. (2a4+4a2-2a+4)(a-2)-1 10. (2x3 -4x2 + 2x -2)÷ (2x-1)

11. ��

�� 12. (3x3 -2x2 +5x -3)÷ (x-2)

13. ��

�� 14. (3a2+6a-3)(a-3)-1




Simplify:

1. (2x3 +6x2 -4x +2) ÷ (x+1)

2. (2a4 – 2a2 +6a-2)(a-4)-1




Lesson 8: Synthetic Division Practice Problems – Answer Key

Part 1: Simplify.

1. (x3 +x2 – 5x +3)÷ (x+3) 2. (y4 – 4y3 +7y2 +5y +2)÷ (y-2)

3. (3a3+a2+3a+6)(a+1)-1 4. ��

��

5. (a3+4a2+4a - 2)(a-1)-1 6. (2x3 +4x2 – 2x -6)÷ (x-3)

-3 1 1 -5 3

-3 6 -3

1 -2 1 0

Solution: x2 – 2x +1

2 1 -4 7 5 2

2 -4 6 22

1 -2 3 11 24

Solution: y3 – 2y2 +3y +11 + 24/(y-2)

3a3+a2+3a+6

a+1

-1 3 1 3 6

-3 2 -5

3 -2 5 1

Solution: 3a2 – 2a +5 + 1/(a+1)

3x3 – 3x +2

x+3 Don’t forget the place holder

-3 3 0 -3 2

-9 27 -72

3 -9 24 -70

Solution: 3x2 -9x +24 -70/(x+3)

a3+4a2+4a - 2

a-1

1 1 4 4 -2

1 5 9

1 5 9 7

Solution: x2 +5x+9 +7/(x-1)

2x3 +4x2 – 2x -6

x-3

3 2 4 -2 -6

6 30 84

2 10 28 78

Solution: 2x2 +10x +28 +78/(x-3)

Solution:




7. ��

�� 8. (5x4 +2x2 – 3x -1)÷ (x-1)

9. (2a4+4a2-2a+4)(a-2)-1 10. (2x3 -4x2 + 2x -2)÷ (2x-1)

x3 – x2 – 5x +2

x+2

-2 1 -1 -5 2

-2 6 -2

1 -3 1 0

Solution: x2 -3x+1

5x4 +2x2 – 3x -1

x-1

1 5 0 2 -3 -1

5 5 3 0

5 5 3 0 -1

Solution: 5x3 +5x2 +3x – 1/(x-1)

2a4+4a2-2a+4

a+2 Don’t forget the place holder

2 2 0 4 -2 4

4 8 24 44

2 4 12 22 48

Solution: 2a3 +4a2 +12a +22 + 48/(a+2)

First divide every term by 2 to make the

divisor have an x term with a coefficient of 1.

. (x3 -4x2 + 2x -2)÷ (2x-1)

2 2

= ( x3 – 2x2 +x -1) ÷ (x- ½ )

1/2 1 -2 1 -1

½ -3/4 1/8

1 -3/2 ¼ -7/8

Solution: x2 -3/2x +1/4 +��

��

**The remainder was

��

�

��

�

This can be written as: ��

�÷

��

� (to make x a

fraction with the same denominator multiply both the

numerator and denominator by 2.

��

� ∙

�

�� =

�

�� -14/8 simplifies to -7/4

Therefore, the remainder is: ��

��




11. ��

�� 12. (3x3 -2x2 +5x -3)÷ (x-2)

13. ��

�� 14. (3a2+6a-3)(a-3)-1

-2 4 -2 5 8

-8 20 -50

4 -10 25 -42

Solution: 4x2 – 10x +25 + -42/(x+2)

2 3 -2 5 -3

6 8 26

3 4 13 23

Solution: 3x2 +4x+13 + 23/(x-2)

3 6 -1 4 -10

18 51 165

6 17 55 155

Solution: 6x2 +17x +55 +155/(x-3)

3a2+6a-3

a-3

3 3 6 -3

9 45

3 15 42

Solution: 3a +15 + 42/(a-3)




Simplify: (2 points each)

1. (2x3 +6x2 -4x +2) ÷ (x+1)

2. (2a4 – 2a2 +6a-2)(a-4)-1

-1 2 6 -4 2

-2 -4 8

2 4 -8 10

Solution: 2x2 +4x -8 + 10/(x+1)

2a4 – 2a2 +6a-2

a-4 Don’t forget the place holder

4 2 0 -2 6 -2

8 32 120 504

2 8 30 126 502

Solution: 2x3 +8x2 +30x +126 +502/(x-4)


Unit 8: Polynomials

Unit 8: Polynomials Chapter Test

Part 1: Identify each of the following as: Monomial, binomial, or trinomial. Then give the

degree of each.

1. 9x2 – 2

2. 3

3. 2x2 + 3x + 1

4. 9y -1

Part 2: Simplify each expression.

1. (7x3 + 2x2 – 1) + (8x2 + 2x -9) 11. (2x2 + 4x – 2)(3x2 – 7x +1)

2. (3x2 + 8x -1) – (2x2 - 6x + 4) 12. (2x+4)2 – (x-5)2

3. 2x(5x + 1) + 3(x2 – 2x + 7) 13. (3x – y)2 (x+3)

4. 8x2(2x2 – 9x + 3)

5. 7(x3 – 4x + 8) – 3x(2x2 + 2x – 6)

6. (y-3)(y +8)

7. (s +4)(s+5)

8. (x -7)2

9. (p – 4)(p+4)

10. (9x- 3y)2

Part 3: Write an equation for each problem, then solve.

1. A rectangle has a length of x2 + 2x + 3 and a width of 5x – 1.

• Write an expression that represents the perimeter of the rectangle.

• Write an expression that represents the area of the rectangle.

2. A triangle has sides: 3x + 2, 8x – 4, and 4x + 5. The perimeter of the triangle is 48 units.

Find the length of the longest side of the triangle.


Unit 8: Polynomials

Unit 8: Polynomials Chapter Test – Answer Key

Part 1: Identify each of the following as: Monomial, binomial, or trinomial. Then give the

degree of each. (1 point each)

1. 9x2 – 2 - Binomial (2 terms); Degree 2 (2 is the largest exponent)

2. 3 - Monomial (1 term); Degree 0 (Constant term with no variable – degree of 0)

3. 2x2 + 3x + 1 – Trinomial (3 terms); Degree 2 (2 is the largest exponent)

4. 9y -1- Binomial; Degree 1 (y is raised to the first power although the exponent is not written)

Part 2: Simplify each expression. (2 points each)

1. (7x3 + 2x2 – 1) + (8x2 + 2x -9)

2. (3x2 + 8x -1) – (2x2 - 6x + 4)


7x3 + 2x2 + 0x – 1

+ 8x2 + 2x - 9

7x3 +10x2 + 2x – 10

Final Answer: 7x3 + 10x2 + 2x - 10


(3x2 + 8x – 1) + (-2x2 + 6x – 4)


3x2 + 8x – 1

+ -2x2 + 6x – 4

x2 + 14x - 5

Final Answer: x2 + 14x - 5

3. 2x(5x + 1) + 3(x2 – 2x + 7)

4. 8x2(2x2 – 9x + 3)

Step 1: Distribute the 2x & the 3 throughout

the parenthesis.

(10x2 + 2x) + (3x2 – 6x + 21)


10x2 + 2x + 0

+ 3x2 - 6x + 21

13x2 – 4x + 21


Step 1: Distribute the 8x2 throughout the

parenthesis.

16x4 – 72x3 + 24x2



Unit 8: Polynomials

5. 7(x3 – 4x + 8) – 3x(2x2 + 2x – 6)

6. (y-3)(y +8)

7. (s +4)(s+5)

Step 1: Distribute the 7 throughout the 1st parenthesis, and -3x

throughout the 2nd parenthesis. (By distributing a negative

3x, you will be able to add the 2 polynomials)

(7x3 – 28x + 56) + (-6x3 – 6x2 + 18x)


7x3 + 0x2 – 28x + 56

+ -6x3 -6x2 + 18x + 0

x3 – 6x2 – 10x + 56

Final Answer: x3 – 6x2 – 10x + 56

Step 1: Use FOIL to multiply the two binomials.

y(y) + y(8) + (-3)(y) + (-3)(8)

y2 + 8y - 3y - 24

Step 2: Combine like terms

y2 + 5y – 24

Final Answer: y2 + 5y – 24

Step 1: Use FOIL to multiply the two binomials.

s(s) + s(5) + (4)(s) + (4)(5)

s2 + 5s + 4s +20

Step 2: Combine like terms

s2 + 9s + 20

Final Answer: s2 + 9s + 20


Unit 8: Polynomials

8. (x -7)2

9. (p – 4)(p+4)

10. (9x- 3y)2

11. (2x2 + 4x – 2)(3x2 – 7x +1)

Since we are squaring a binomial, we can use our special rule: Square of a Difference.

(a-b)2 = a2 – 2ab + b2

(x-7)2 = x2 – 2(x)(7) + 72

(x-7)2 = x2 – 14x + 49

Final Answer: x2 – 14x + 49

This problem utilizes our special rule: Difference of Two Squares

(a+b)(a-b) = a2 – b2

(p-4)(p+4) = p2 - 42

(p-4)(p+4) = p2 - 16

Final Answer: p2 - 16

Since we are squaring a binomial, we can use our special rule: Square of a Difference.

(a-b)2 = a2 – 2ab + b2

(9x-3y)2 =(9x)2 – 2(9x)(3y) + (3y)2

(9x-3y)2 = 81x2 – 54xy + 9y2

Final Answer: = 81x2 – 54xy + 9y2

Step 1: Use the extended distributive property to multiply.

2x2(3x2) +2x2(-7x) + 2x2 (1) + 4x(3x2) +4x(-7x) + 4x (1)+ (-2)(3x2) +(-2)(-7x) + (-2) (1)

6x4 - 14x3 + 2x2 + 12x3 - 28x2 + 4x - 6x2 + 14x - 2

Step 2: Rewrite with like terms together.

6x4 – 14x3 + 12x3 + 2x2 – 28x2 – 6x2 + 4x + 14x – 2

Step 3: Combine like terms.

Final Answer: 6x4 -2x3 -32x2 + 18x - 2


Unit 8: Polynomials

12. (2x+4)2 – (x-5)2

13. (3x – y)2 (x+3)

Step 1: Use the square of a binomial rule to expand both binomials.

(a+b)2 = a2 + 2ab+ b2

(2x+4)2 = (2x)2 +2(2x)(4)+ 42

(2x+4)2 = 4x2 + 16x + 16

(a-b)2 = a2 – 2ab +b2

(x-5)2 = x2 – 2(x)(5) + 52

(x-5)2 = x2 – 10x + 25

Step 2: Use the expanded forms to subtract.

(2x+4)2 – (x-5)2

(4x2 + 16x + 16) – (x2 – 10x + 25)


(4x2 + 16x + 16) + (-x2 + 10x - 25)


4x2 + 16x + 16

+ -x2 + 10x - 25

3x2 + 26x – 9

Final Answer: 3x2 + 26x – 9

Step 1: Expand the first binomial.

(3x-y)2 = (3x)2 – 2(3x)(y) + y2

(3x-y)2 = 9x2 – 6xy + y2

Step 2: Take this product and multiply it by (x+3) – Use the extended distributive prop.

(x+3) (9x2 – 6xy + y2)

x(9x2) +x(-6xy) +x(y2) + 3(9x2) +3(-6xy) +3(y2)

9x3 - 6x2y + xy2 + 27x2 - 18xy + 3y2

Final Answer: 9x3- 6x2y + xy2 + 27x2 - 18xy + 3y2 (There are no like terms)


Unit 8: Polynomials

Part 3: Write an equation for each problem, then solve. (2 points each)

1. A rectangle has a length of x2 + 2x + 3 and a width of 5x – 1.

• Write an expression that represents the perimeter of the rectangle.

• Write an expression that represents the area of the rectangle.

2. A triangle has sides: 3x + 2, 8x – 4, and 4x + 5. The perimeter of the triangle is 48 units.

Find the length of the longest side of the triangle.

P = 2l + 2W

P = 2(x2 + 2x + 3) + 2(5x-1) - Substitute each expression for length and width

P = (2x2 + 4x + 6) + (10x – 2) (Distribute)

P = 2x2 + 4x + 10x + 6 – 2 - Write like terms together

P = 2x2 + 14x + 4 - Final Answer

A = lw

A = (x2 + 2x + 3) (5x-1) - Substitute each expression for length and width

Or A = (5x-1) (x2 + 2x +3) - Switch terms around so that it’s easier to use the distributive prop.

A = 5x(x2) + 5x(2x) + 5x(3) + (-1)(x2) + (-1)(2x) + (-1)(3)

A = 5x3 + 10x2 + 15x – x2 – 2x – 3

A = 5x3 + 9x2 +13x – 3 (combine like terms)

Final Answer: A = 5x3 + 9x2 +13x – 3

P = S1 + S2 + S3

48= 3x+2 + 8x-4 + 4x + 5 Substitute the perimeter and lengths of sides.

48 = 3x + 8x + 4x + 2 – 4 + 5 Rewrite with like terms together

48 = 15x + 3 Simplify like terms

48 – 3 = 15x + 3 -3 Subtract 3 from both sides to begin solving for x.

45 = 15x Simplify: (48-3 = 45)

45/15 = 15x/15 Divide by 15 on both sides.

3 = x or x = 3

S1 = 3x+2 S2 = 8x-4 S3 = 4x+5

S1 = 3(3) + 2 S2 = 8(3) – 4 S3 = 4(3) + 5

S1 = 11 S2 = 20 S3 = 17

The length of the

longest side is 20

units.

This test is worth 34

points.



Documents

polynomial is: - Algebra Class E-course 8 Polynomials.pdf · A polynomial is: ... 1 Side 1 =x 2 – 3x -4 Side 2: x 2 + 2x -3 2 3 Side 3: 5x -2 Side 4: 5x -3 4 Step 1: Rewrite vertically