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What is the result when you multiply (x+3)(x+2)?
x2 x x x
xx
1 1 11 1 1
x + 3
x
+
2
Using algebra tiles, we have
The resulting trinomial is x2 + 5x + 6.
Notice that 2 + 3 = 5 which is the coefficient of the middle term. 2 x 3 = 6 which is the value of the constant. The coefficient of x2 is 1 and 1 x 6 = 6 which is again the value of the constant.
Using the distributive property, we have:
( )( ) ( ) ( )x x x x x
x x x
x x
3 2 2 3 2
2 3 6
5 6
2
2
The resulting trinomial is x2 + 5x + 6.
How can we factor trinomials such as x2 + 7x + 12?
One method would be to again use algebra tiles.
Start with the x2.
x2
Add the twelve tiles with a value of 1.
1 1 1 1 1 11 1 1 1 1 1
Try to complete the rectangle using the 7 tiles labeled x.
x x x x x x
x
How can we factor trinomials such as x2 + 7x + 12?
One method would be to again use algebra tiles.
x2
1 1 1 1 1 11 1 1 1 1 1
x x x x x x
x
Note that we have used 7 tiles with “x”, but are still short one “x”. Thus, we must rearrange the tiles with a value of 1.
How can we factor trinomials such as x2 + 7x + 12?
One method would be to again use algebra tiles.
x2
1 1 1 1
11
1 1 1 1
11
x xxx
x
x
x
We now have a rectangular array that is (x+4) by (x+3) units.
x + 4
x
+ 3
Therefore, x2 + 7x + 12 = (x + 4)(x + 3).
While the use of algebra tiles helps us to visualize these concepts, there are some drawbacks to this method, especially when it comes to working with larger numbers and the time it takes for trial and error.
Thus, we need to have a method that is fast and efficient and works for factoring trinomials.
In our previous example, we said that x2 + 7x + 12 = (x + 4)(x + 3).
Step 1: Find the factors of the coefficient of the term with x2.
1 x 1
Step 2: Find the factors of the constant.
1 x 12 2 x 6 3 x 4
Step 3: The trinomial ax2 + bx + c = (mx + p)(nx + q) where
a = m n, c = p q, and b = mq + np
While the use of algebra tiles helps us to visualize these concepts, there are some drawbacks to this method, especially when it comes to working with larger numbers.
Thus, we need to have a method that is fast and efficient and works for factoring trinomials.
In our previous example, we said that x2 + 7x + 12 = (x + 4)(x + 3).
Step 3: The trinomial ax2 + bx + c = (mx + p)(nx + q) where
a = m n, c = p q, and b = mq + np
Step 4: Write trial factors and check the middle term.
(x + 1)(x + 12) x + 12x = 13 x No
(x + 2)(x + 6) 2x + 6x = 8x No
Step 4: Write trial factors and check the middle term.
(x + 1)(x + 12) x + 12x = 13 x No
(x + 2)(x + 6) 2x + 6x = 8x No
(x + 3)(x + 4) 3x + 4x = 7x Yes
This method works for trinomials which can be factored. However, it also involves trial and error and may be somewhat time consuming.
Trinomials are written as ax2 + bx + c. However, a, b, and c may be positive or negative. Thus a trinomial may actually appear as:
ax2 + bx + c
ax2 - bx + c
ax2 - bx - c
ax2 + bx - c
AIM: How do we factor trinomials?
Do Now: Answer the following:1. How many factors are in a binomial?2. How many factors are in a trinomial?3. How many factors are in the product
of two binomials?
Reviewing Multiplying Binomials…
Simplify the following:(x – 2)(x +3)
Predict the number of terms in the product.
Today’s Big Idea!
We spent last week multiplying binomials (FOILing) to get trinomials.
Today (and probably tomorrow) we are going to factor trinomials to get two sets of binomials.
Say WHAT?!
We’re taking these types of expressions…
x2 – 2x + 5 5y2 – 7y + 2 3x2 – 2x - 8
And breaking them into these types of expressions…
(x + 2)(x + 4) (2y – 3)(y + 2) (4x – 5)(2x – 7)
First, we have to remember… The standard form of a
quadratic equation: ax2 + bx + c
Case 1: If a = 1, b is positive, and c is positive, find two numbers whose product is c and whose sum is b.
Example x2 + 10x + 16
a = 1, b = 10, c = 16The factors of 16 are 1 and 16, 2 and 8, 4 and 4.2 + 8 is 10.
x x x x x2 210 16 2 10 16
x x x2 2 8 16 ( ) ( )
x x x2 8 2 ( ) ( )
x x8 2 ( )( )
Write 10x as the sum of the two factors.Use parentheses to group terms with common factors.
FactorApply the distributive property.
Summary:
Factor this!y2 + 6y + 8
AIM: How do we factor trinomials?
Do Now: 5th Period: Emergency Quiz! All other periods: Factor this trinomial
y2 + 6y + 8
Reviewing…
Yesterday, we factored trinomials of like the following:ax2 + bx + c
Now, today, we’re going to factor trinomials like: ax2 – bx + c
Notice that the value of b is now negative
Today, we’re also going to use negative factors.
Example: Positive Factors of 24
Negative Factors of 24
Case 2: If a = 1, b is negative and c is positive, find two numbers whose product is c and whose sum is b.
Example x2 – 13x + 36
a = 1, b = -13, c = 36The factors of 36 are 1 and 36, 2 and 18, 3 and 12, 4 and 9,
-1 and –36, -2 and –28, -3 and –12, -4 and –9. -4 + (-9) = -13Write -13x as the sum of the two factors.Use parentheses to group terms with common factors.
FactorApply the distributive property.
x x x x x2 213 36 4 9 36
x x x2 4 9 36 ( ) ( )
x x x4 9 4 ( ) ( )( )
x x9 4 ( )( )
AIM: Factoring Polynomials
Do Now: Factor this polynomialx2 − 5x + 6
*Predict the sign of the factors (will they be positive or negative?)
HW on desk for credit!
HW Review:Check your answers:
7. (n + 4)(n + 4)8. (x + 4)(x + 5)9. (x – 7)(x – 2)10. (x + 4)(x + 8)11. (n – 1)(n – 9)12. (x + 5)(x + 3)13. (x – 4)(x – 5)14. (x – 4)(x – 4) 15. (x – 4)(x – 6)
16. (x – 2)(x – 3)17. (x + 3)(x + 4)18. (x – 7)(x – 3)19. (x + 5)(x + 9)20. (x – 1)(x – 17)21. (x – 4)(x – 7)22. (x + 11)(x + 2)23. (x + 2)(x + 14)24. (n – 4)(n – 12)
So far, we’ve gone over…
Polynomials that look like this:ax2 + bx + c
Their factors looked like this:(x + g)(x + h)
And polynomials that look like this: ax2 – bx + c
Their factors looked like this:(x – g)(x – h)
Today, we’re going to look at polynomials
That look like this…ax2 + bx – c
And this: ax2 – bx – c
Answer this question: What do you think their factors will
look like?
Case 3: If a = 1, b is positive and c is negative, find two numbers whose product is c and whose difference is b.
Example x2 + 5x - 14
a = 1, b = 5, c = -14The factors of –14 are –1 and 14, 1 and –14, -2 and 7, and 2 and –7. -2 + 7 = 5.
x x x x x2 25 14 2 7 14
x x x2 2 7 14 ( ) ( )
x x x2 7 2 ( ) ( )
x x7 2 ( )( )
Write 5x as the sum of the two factors.Use parentheses to group terms with common factors.
FactorApply the distributive property.
Case 4: If a = 1, b is negative and c is negative, find two numbers whose product is c and whose sum is b.
Example x2 – 8x - 20
a = 1, b = -8, c = -20The factors of -20 are 1 and -20, -1 and 20, 2 and -10, -2 and 10,
4 and –5, and –4 and 5. 2 + (-10) = -8.Write -8x as the sum of the two factors.Use parentheses to group terms with common factors.
FactorApply the distributive property.
x x x x x2 28 20 2 10 20
x x x2 2 10 20 ( ) ( )
x x x2 10 2 ( ) ( )
x x10 2 ( )( )
Summary:
What are the factors of x2−10x−24 ?[A] (x - 12)(x + 2) [B] (x + 12)(x - 2)[C] (x - 4)(x + 6) [D] (x - 4)(x - 6)
AIM: Polynomial Review!!
Do Now: What are the factors of x2−10x−24 ?
Predict the signs of the factors, too!
Slide 1: Test Strategies!
Ok, since we haven’t learned how to factor polynomials like this:2x2+10x – 12
I’m going to show you a trick…
Slide 2: Solve this! Factored completely, the
expression2y2+12y−54 is equivalent to[A] (y + 6)(2y - 9) [B] 2(y + 9)(y -
3)[C] (2y + 6)(y - 9) [D] 2(y - 3)(y - 9)
Slide 3: So, it’s complicated! There are two ways to solve this
1. Factor out a 2 (since it’s in front of the y2) from each term, then factor like we know how
OR2. Work backwards, using the provided
answers, and FOIL to find the answer
Once, you’ve figured it out, fill in the answer on your handout!
Slide 4: Topics Covered on the Test
Adding and Subtracting Polynomials
Multiplying and Dividing Polynomials
Factoring binomials (including GCF)
Factoring polynomials
Say WHAT?!
Slide 6: So here are some examples in case you’ve forgotten what that stuff means! Adding Polynomials:
1. The sum of 8x2 −x+4 and x − 5 is…
Slide #7: Subtracting Polynomials
When 3a2 −2a+5 is subtracted from a2 +a−1, the result is…
Slide #8: Factoring Binomials
If one factor of 56x4y3 − 42x2y6 is 14x 2y 3,what is the other factor?
Slide #9: Factoring Polynomials
t2 – 4t – 21
Now, for the fun part!
Get into groups of 3 or 4… On a transparency, you are going to
pick one type of problem and solve it
You will then present that type of problem to the class
Ms. G will assign groups if necessary
Summary:
Student presentations on overhead projector.
Case 5: If a 1, find the ac. If c is positive, find two factors of acwhose sum is b.
Example 6x2 + 13x + 5
a = 6, b = 13, c = 5, ac=30The factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6,
-1 and –30, -2 and –15, -3 and –10, -5 and –6. 3 + 10 = 13.Write 13x as the sum of the two factors.Use parentheses to group terms with common factors.
FactorApply the distributive property.
6 13 5 6 3 10 52 2x x x x x
6 3 10 52x x x ( ) ( )
3 2 5 5 2 1x x x ( ) ( )
3 5 2 1x x ( )( )
Case 6: If a 1, find the ac. If c is negative, find two factors of acwhose difference is b.
Example 8x2 + 2x - 15
a = 8, b = 2, c = -15, ac= 120
The factors of -120 are 1,120,2,60,3,40,4,30,5,26,6,20,8,15,10,12.
12 – 10 = 2.Write 2x as the sum of the two factors.Use parentheses to group terms with common factors.
FactorApply the distributive property.
8 2 15 8 12 10 152 2x x x x x
8 12 10 152x x x ( ) ( )
4 2 3 5 2 3x x x ( ) ( )
4 5 2 3x x ( )( )
Factor each trinomial if possible.
1) t2 – 4t – 21
2) x2 + 12x + 32
3) x2 –10x + 24
4) x2 + 3x – 18
5) 2x2 – x – 21
6) 3x2 + 11x + 10
7) x2 –2x + 35
t2 – 4t – 21
a = 1, b = -4, c = -21
The factors of –21 are –1,21, 1,-21, -3,7, 3,-7.
3 + (-7) = -4.
t t t t t
t t t
t t t
t t
2 2
2
4 21 3 7 21
3 7 21
3 7 3
7 3
( ) ( )
( ) ( )
( )( )
x2 + 12x + 32
a = 1, b = 12, c = 32
The factors of 32 are 1,32, 2,16, 4,8.
4 + 8 = 32
x x x x x
x x x
x x x
x x
2 2
2
12 32 4 8 32
4 8 32
4 8 4
8 4
( ) ( )
( ) ( )
( )( )
x2 - 10x + 24
a = 1, b = -10, c = 24
The factors of 24 are 1,24, 2,12, 3,8, 4,6, -1,-24, -2,-12, -3,-8, -4,-6
-4 + (-6) = -10
x x x x x
x x x
x x x
x x
2 2
2
10 24 4 6 24
4 6 24
4 6 4
6 4
( ) ( )
( ) ( )
( )( )
x2 + 3x - 18
a = 1, b = 3, c = -18
The factors of -18 are 1,-18, -1,18, 2,-9, -2,9, 3,-6, -3,6
-3 + 6 = 3
x x x x x
x x x
x x x
x x
2 2
2
3 18 3 6 18
3 6 18
3 6 3
6 3
( ) ( )
( ) ( )
( )( )
2x2 - x - 21
a = 2, b = -1, c = -21, ac=42
The factors of 42 are 1,42, 2,21, 3,14, 6,7.
6 – 7 = -1
2 21 2 6 7 21
2 6 7 21
2 3 7 3
2 7 3
2 2
2
x x x x x
x x x
x x x
x x
( ) ( )
( ) ( )
( )( )
3x2 + 8x + 5
a = 3, b = 8, c = 5, ac=15
The factors of 15 are 1,15, 3,5.
3 + 5 = 8
3 8 5 3 3 5 5
3 3 5 5
3 1 5 1
3 5 1
2 2
2
x x x x x
x x x
x x x
x x
( ) ( )
( ) ( )
( )( )
x2 - 2x + 35
a = 1, b = -2, c = 35,
The factors of 35 are 1,35, -1,-35, 5,7, and –5,-7
None of these pairs of factors gives a sum of –2.
Therefore, this trinomial can’t be factored by this method.
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