3 (4 5)x x 212 15x x

4 (6 2)x x 224 8x x

6 ( 1)x x 26 6x x

(2 1)(2 1)x x 24 1x

23 (2 4 5)x x x 3 26 12 15x x x

(6 5)(6 5)x x 236 25x

M3U4D1 Warm-UPDistribute each problem:

NEW SEATS!

M3U4D1 Evaluating and Operating with

Polynomials

OBJ: To review adding, subtracting, multiplying,

and factoring polynomials

How do I evaluate polynomial functions?

You have 3 minutes to complete the top of

handout page 1.

Discuss

How do I operate with polynomial functions?

Let’s review…

The sum f + g

xgxfxgf This just says that to find the sum of two functions, add them together. You should simplify by finding like terms.

1432 32 xxgxxf

1432 32 xxgf

424 23 xx

Combine like terms & put in descending order

The difference f - g

xgxfxgf To find the difference between two functions, subtract the first from the second. CAUTION: Make sure you distribute the – to each term of the second function. You should simplify by combining like terms.

1432 32 xxgxxf

1432 32 xxgf

1432 32 xx

Distribute negative

224 23 xx

The product f • g

xgxfxgf To find the product of two functions, put parenthesis around them and multiply each term from the first function to each term of the second function.

1432 32 xxgxxf

1432 32 xxgf31228 325 xxx

FOIL

Good idea to put in

descending order

32128 235 xxx

The quotient f /g

xgxf

xg

f

To find the quotient of two functions, put the first one over the second.

1432 32 xxgxxf

14

323

2

x

x

g

f Nothing more you could do here. (If you can reduce these you should. More later…)

Operations with

Polynomials

Now you try the four operations on the bottom of page 1 of your handout.

Factoring Review:#1: GCF Method

GCF Method is just

distributing backwards!!

Review: What is the GCF of

25a2 and 15a?5a

Let’s go one step further…1) FACTOR 25a2 + 15a.

Find the GCF and divide each term25a2 + 15a = 5a( ___ + ___ )

Check your answer by distributing.

225

5

a

a

15

5

a

a

5a 3

2) Factor 18x2 - 12x3.Find the GCF

6x2

Divide each term by the GCF18x2 - 12x3 = 6x2( ___ - ___ )

Check your answer by distributing.

2

2

18

6

x

x

3

2

12

6

x

x

3 2x

3) Factor 28a2b + 56abc2.

GCF = 28abDivide each term by the GCF

28a2b + 56abc2 = 28ab ( ___+ ___)

Check your answer by distributing.28ab(a + 2c2)

228

28

a b

ab

256

28

abc

ab

a 2c2

4) Factor 20x2 - 24xy

1. x(20 – 24y)2. 2x(10x – 12y)3. 4(5x2 – 6xy)4. 4x(5x – 6y)

5) Factor 28a2 + 21b - 35b2c2

GCF = 7Divide each term by the GCF

28a2 + 21b - 35b2c2 = 7 ( ___ + ___ - ____ )

Check your answer by distributing.7(4a2 + 3b – 5b2c2)

228

7

a 21

7

b

4a2 5b2c2

2 235

7

b c

3b

Factor 16xy2 - 24y2z + 40y2

1. 2y2(8x – 12z + 20)2. 4y2(4x – 6z + 10)3. 8y2(2x - 3z + 5)4. 8xy2z(2 – 3 + 5)

Factor out the GCF for each polynomial:Factor out means you need the GCF times the

remaining parts.

a) 2x + 4yb) 5a – 5bc) 18x – 6yd) 2m + 6mne) 5x2y – 10xy

2(x + 2y)

6(3x – y)

5(a – b)

5xy(x - 2)

2m(1 + 3n)

Greatest Common Factorsaka GCF’s

How can you check?

Ex 1

•15x2 – 5x•GCF = 5x•5x(3x - 1)

Ex 2

•8x2 – x•GCF = x•x(8x - 1)

Ex 3

•8x2y4+ 2x3y5 - 12x4y3

•GCF = 2x2y3

•2x2y3 (4y + xy2 – 6x2)

#2: X-box Factoringaka Diamond Method

#2: X-box Factoringaka Diamond Method

X- Box

3 -9

Product

Sum

X-box FactoringX-box Factoring• This is a guaranteed method for

factoring quadratic equations—no guessing necessary!

• We will review how to factor quadratic equations using the x-box method

• Background knowledge needed:

– Basic x-solve problems

– General form of a quadratic equation

Factor the x-box wayExample: Factor 3x2 -13x -10

-13

(3)(-10)=

-30

-15 2

-10

-15x

2x

3x2

x -5

3x

+2

3x2 -13x -10 = (x-5)(3x+2)

Factor the x-box way

Middleb=m+

nSum

Product

ac=mnm n

First and Last

Coefficients

y = ax2 + bx + c

Last term

1st Term

Factorn

Factorm

Base 1 Base 2

GCF

Height

ExamplesExamplesFactor using the x-box method.1. x2 + 4x – 12

a) b)

x -12

4 6 -2

x2 6x

-2x -12

x

-2

+6

Solution: x2 + 4x – 12 = (x + 6)(x - 2)

Examples continuedExamples continued

2. x2 - 9x + 20

a) b) 20

-9

x2 -4x

-5x 20

x

x -4

-5

Solution: x2 - 9x + 20 = (x - 4)(x - 5)

-4 -5

Examples continuedExamples continued

3. 2x2 - 5x - 7

a) b) -14

-5

2x2 -7x

2x -7

x

2x -7

+1

Solution: 2x2 - 5x – 7 = (2x - 7)(x + 1)

-7 2

Examples continuedExamples continued

3. 15x2 + 7x - 2

a) b) -30

7

15x2 10x

-3x -2

5x

3x +2

-1

Solution: 15x2 + 7x – 2 = (3x + 2)(5x - 1)

10 -3

#3: Difference of Squares

•a2 – b2 = (a + b)(a - b)

What is a Perfect Square

• Any term you can take the square root evenly (No decimal)

• 25• 36• 1• x2

• y4

5

6

1

x

2y

Difference of Perfect Squares

x2 – 4=

the answer will look like this: ( )( )

take the square root of each part:( x 2)(x 2)

Make 1 a plus and 1 a minus:(x + 2)(x - 2 )

FACTORINGDifference of Perfect

Squares

EX:

x2 – 64

How:

Take the square root of each part. One gets a + and one gets a -.

Check answer by FOIL.

Solution:

(x – 8)(x + 8)

Example 1

•9x2 – 16•(3x + 4)(3x – 4)

Example 2

•x2 – 16•(x + 4)(x –4)

Ex 3

•36x2 – 25•(6x + 5)(6x – 5)

More than ONE Method• It is very possible to use more than one

factoring method in a problem• Remember:

•ALWAYS use GCF first

Example 1

•2b2x – 50x•GCF = 2x•2x(b2 – 25) •2nd term is the diff of 2 squares•2x(b + 5)(b - 5)

ClassworkM3U4D1 Factoring Review

Evens

Distribute Interims!

Due back TOMORROW with parent signature

Distribute Interims!

Due back TOMORROW with parent signature

HomeworkM3U4D1 Factoring Review

Odds

Unit 3 Geometry Test due tomorrow!!!

Show all your work to receive credit– don’t forget to check by

multiplying!