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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
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.
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)
Factoring Perfect Square TrinomialsLet'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.
Recognizing & Factoring Perfect Square Trinomials
Expand (Multiply):(x + 2)2We'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 + 4Working 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 + 4Method #2: A trinomial is a PST, if: b 2
2= A*C
Using Method 2: x2 + 4x + 4 = 4 2 4 = 1*4
2PST Confirmed
(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 = acAnd 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 +
4xIs it a PST? Factored Form?
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)
Factoring 2nd degree trinomials
with a leading coefficient > 1
Factoring (ax2 + bx + c) Trinomials
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.
Method #1: Grouping
Step
1Multiply the leading coefficient and the constant
term
Method #1: Grouping
Step 2Find 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
Step 3Re-write the original trinomial
and replace 10x with 6x + 4x.
3x2 + 6x + 4x + 8Step
4 Factor by Grouping
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..
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 &
4Re-write the original trinomial
and replace 7x with 3x + 4x.
2x(3x2 +3x + 4x +
4)Factor by Grouping2x(3x2 +3x) + (4x + 4)
=
2x • 3x(x +1) + 4(x + 1)
=2x (x +1)(3x +
4)
Method #2: The Box Method
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
1x( + )( +
)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
Practice: Factor using the Box Method10x2 + 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)
Method #3: Trial & Error
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
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
Class Work: 3.10
