The document outlines a lesson on factoring polynomials, including reviewing factoring perfect square trinomials and introducing methods for factoring general trinomials of the form ax^2 + bx + c, such as grouping, the box method, and trial and error. It also notes an upcoming factoring test and provides examples and notes for students to complete as class work.

1. Today:
 Warm-Up
 Recognizing & Factoring Perfect Square Trinomials
 Factoring (ax2
+ bx + c) Trinomials
 Review of all other factoring methods/test review
 Reminder: Khan Academy Due Wednesday
 Class Work

2. Complete Factoring Guide Available
@v6Math
Notes:
First Factoring Test This Friday
~ Not all factoring methods will be on
first test. The review document lists the
methods on this test.

3. Warm-Up: Factoring Practice(8)
3. 4x³ - 4x
1. (x+9)(x-8) 2. 36x² - 25 2. (6x - 5)(6x + 5)
3. 4x(x+1)(x-1) 4. 21x³y + 28x²y²
7. x3
+ 2x2
+ 3x + 6
1. x² + x - 72
4. 7x²y (3x+4y)
Factor each expression completely:
5. (x + 4)² 5. (x² + 8x + 16)
The polynomial in # 2 is called a ______ polynomial
The polynomial in # 5 is called a ______ polynomial
6. (3a – 2b)²
The polynomial in # 6 is called a ______ polynomial
6. (9a² – 12ab - 4b²)
Class Notes Section of your Notebook:
8. x² - 9x + 12 Prime7. (x + 2)+ (x2
+ 3)

4. Factoring Perfect Square Trinomials
Let's look at #'s 5 & 6 from the warm – up: 5. (x + 4)²
When multiplied, we find that (x + 4) (x + 4)
are factors of (x² + 8x + 16). This type of polynomial
is known as a Perfect Square Trinomial
1. PST's have a square in the first & third terms: The
first & third terms are always positive.
2. The factors of PST's are always either the square
of a sum (x + y)², or the square of a difference (x - y)²
3. We arrive at the trinomial by performing a
" square, double, square" on the factor. To factor
then, we do the opposite, which is a sq. root, halve,
sq. root. to arrive at the factors.

5. Recognizing & Factoring Perfect Square Trinomials
Expand (Multiply):(x + 2)2
We'll start at the end and find the trinomial first :
We use square, double, square to
arrive at: x2 + 4x + 4
Our task today, then, is to determine whether this trinomial can be
factored into either the square of a sum or the square of a difference.
sq. root(x) + half(2) + sq. root(2)x2 + 4x + 4
Working the other way, the opposite of square, double, square is:
The middle term is twice the product of
the square root of the first term and the
square root of the third term.
Method # 1: A trinomial is a PST, if:
Check: Is this a PST? x2 + 4x + 4
Method #2: A trinomial is a PST, if: b 2
2
= A*C
Using Method 2: x2 + 4x + 4
= 4 2
4 = 1*4
2
PST Confirmed

6. (3x - 4)2
Is this a perfect square trinomial?
sq. root (3x) + half (12) + sq. root(4)
Always be aware of possible perfect square trinomials
(9x2 -24x + 16)
What's the Point? If you can identify a perfect square
trinomial, it makes factoring very easy!
Recognizing & Factoring Perfect Square Trinomials
Yes, b = ac
And the factored form is?= A*C
b 2
2Use method 2:
Factor the following: (x² - 16x + 64) = (x - 8)²
(x² + 8x + 16) = (x + 4)²
(x² - 15x + 36) Is not a sp. product. Can it be factored?
Last trinomial... 9y3 + 12x2 +
4x
Is it a PST? Factored Form?

7. When Factoring:
1. Look for the GCF
2. Look for special cases.
a. difference of two
squares
b. perfect square
trinomial
3. If a trinomial is not a
perfect square, look for
two different binomial
factors.
 8t4 – 32t3 + 40t
= 8t(t3 - 4t + 5)
 4x2 – 9y2
=(2x)2 – (3y)2
 x2 + 8x + 16
= x2 + 8x + 42 = (x +
4)2
 x2 + 11x – 10
= (x + 10)(x – 1)

8. Factoring 2nd degree trinomials
with a leading coefficient > 1
Factoring (ax2
+ bx + c) Trinomials

9. Factoring Trinomials
 Use any of the following methods to take the
guesswork out of factoring trinomials.
 It would be a good idea to write the steps down
once, as they are easy to forget when away
from class
 You can use these steps for any ax2
+ bx + c
polynomial, and for any polynomial you are
having difficulty factoring.

10. Method #1: Grouping

11. Step
1
Multiply the leading coefficient and the constant
term
Method #1: Grouping

12. Step 2
Find the two factors of 24 that add to
the coefficient of the middle term.
Notice the 'plus, plus' signs in the
original trinomial.
Factors of 24:
1 24
2 12
3 8
4 6
Our two factors are 4 &
6

13. Step 3
Re-write the original trinomial
and replace 10x with 6x + 4x.
3x2
+ 6x + 4x + 8
Step
4 Factor by Grouping

14. Step
5 Factor out the GCF of each pair of
terms
After doing so, you will have...
Step 6 Factor out the common binomial,
check that no further factoring is
possible, and the complete
factorization is..

15. Practice: Factor by GroupingFactor: 6x3 + 14x2 +
8x In this case, step 1 is...
and we are left with.. 2x(3x2 + 7x + 4)
1 12
2 6
3 4
Multiply the leading coefficient and the constant
termFactors of 12: Our two factors are 3 &
4
Re-write the original trinomial
and replace 7x with 3x + 4x.
2x(3x2
+3x + 4x +
4)Factor by Grouping
2x(3x2
+3x) + (4x + 4)
=
2x • 3x(x +1) + 4(x + 1)
=
2x (x +1)(3x +
4)

16. Method #2: The Box Method

17. Using the Box Method to factor (ax2
+ bx + c)
Trinomials
9x3 + 12x2 + 4x
As usual, we
are looking
for factors
that add to
'b', and
multiply to
'ac'
Is there a GCF
to Factor?
x(9x2 + 12x + 4)
3x,3x
9x, x
4,1:
2,2
3x 3x
1.Draw binomials with correct
signs
4
1
x( + )( +
)2. Multiply
diagonally to
mentally check, or
fill in the binomial.
3x 4 3x 13x 2 3x 2
x( 3x+2)(3x +2)or,
The correct factorization is:
x( 3x + 2)2
Factors
of 'a'
2
2

18. Practice: Factor using the Box Method
10x2 + 21x -
10
10,-1: -
10,1
5,-2: -5,2
10x,x
5x,2x
Factors
of 'a'
Draw binomials with correct
signs( + )( +
)
( 2x+5)(5x- 2)

19. Method #3: Trial & Error

20. Factor the polynomial 25x2 + 20x + 4.
Possible factors of 25x2 are {x, 25x} or {5x, 5x}.
Possible factors of 4 are {1, 4} or {2, 2}.
We need to try each pair of factors until we find a
combination that works, or exhaust all of our possible
pairs of factors.
Keep in mind that, because some of our pairs are not
identical factors, we may have to switch some pairs of
factors and make 2 attempts before we can definitely
decide a particular pair of factors will not work.
Method #3: Trial & Error

21. We are looking for a combination that gives the sums to
the middle term and are factors of the last term
{x, 25x} {1, 4} (x + 1)(25x + 4) 4x 25x 29x
(x + 4)(25x + 1) x 100x 101x
{x, 25x} {2, 2} (x + 2)(25x + 2) 2x 50x 52x
{5x, 5x} {2, 2} (5x + 2)(5x + 2) 10x 10x 20x
Method #3: Trial & Error

22. Class Work: 3.10