factoring polynomials

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Factoring Polynomials
A presentation for the
greatest Algebra I kids at RJR
By Mrs. Sexton
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Step by Step
Easy Problems
Medium Problems
4my  20m  3 py 15p
8s 2  200t 2
x 2  14x  24
3
2
3x  9 x
2
4s  16
6 y2  5y  6
Hard Problems
6 x3  15x 2  9 x
8
2 x  32
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Word Problems
Division of polynomial by monomial
Find dimensions when area is given
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Rules for Factoring Polynomials
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Step by Step
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• Is there a GCF?
– Yes
• Factor as the product of the GCF and one other factor—i.e.
GCF•(the other factor). Look at the other factor and go to the
next step below with it.
– No
• Go the the next step.
• Is it a binomial?
– Yes
• Is it a difference of two squares? (a2-b2)
– Yes—Factor as (a+b)(a-b).
– No—It can’t be factored any more.
– No
• Go to the next step.
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• Is it a trinomial?
– Yes
• Do you recognize it as a pattern for a perfect square
trinomial? (a2+2ab+b2) or (a2-2ab+b2)
– Yes—Factor as (a+b)2 or (a-b)2
– No—Go to next step.
• Use the ac and b pattern to look for factors.
• Can you find factors of ac that add up to b?
– Yes—Rewrite the equation with those factors, group, and
factor.
– No—You can’t do anything else. If there’s no GCF, it’s a
prime polynomial.
– No
• Go to the next step.
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• Is it a four-term polynomial?
– Yes
• Are there two sets of terms that you can group
together that have a common factor?
– Yes—Group and factor.
– No—If it doesn’t have a GCF, it’s a prime polynomial.
– No
• If it doesn’t have a GCF, it’s a prime polynomial.
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NOTE:
At EVERY step along the way, you
must look at the factors that you get to
see if they can be factored any more.
Factoring completely means that no factors can be
broken down any further using any of the rules
you’ve learned.
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Practice
Factor completely.
x  14x  24
2
No.
Is there a GCF?
Is it a binomial, trinomial, or four-term polynomial?
It’s a trinomial.
Do you recognize it as a perfect square trinomial?
No. Use ac and b.
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ac
1 • 24
b
14
24
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Use your handy-dandy calculator or
your super math skills to find 12
and 2 as the factors to use.
12, 2
Rewrite the equation with those two factors in the middle.
x  14x  24
2
Group.
x  12x  2 x  24
2
out the GCF
( x  12x)  (2 x  24) Factor
from each group.
x( x  12)  2( x  12) Write the two factors.
( x  12)(x  2) Neither one of these factors can be
2
broken down any more, so you’re done.
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Factor completely
3x  9 x
3
2
Is there a GCF?
Yes. Write the GCF first and the remaining factor after it.
3x ( x  3)
2
Look at the remaining factor. (x-3)
Is it a binomial, trinomial, or four-term polynomial?
It’s a binomial. Is it a difference of two squares? (a2-b2)
No. You can’t do anything else.
3x ( x  3)
2
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is the completely factored form.
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Factor completely
4s  16
2
Is there a GCF?
Yes. Write the GCF first and the remaining factor after it.
4( s  4)
2
Look at the remaining factor. (s2-4)
Is it a binomial, trinomial, or four-term polynomial?
It’s a binomial. Is it a difference of two squares? (a2-b2)
Yes. s2 is a square (s • s) and 4 is a square (2 • 2).
Factor as (s+2)(s-2). Then write the complete
factorization.
4( s  2)(s  2)
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Factor completely
4my  20m  3 py 15p
Is there a GCF?
No. There is no single factor that goes into all four of the terms.
Is it a binomial, trinomial, or four-term polynomial?
It’s a four-term polynomial. Factor by grouping.
(4my  20m)  (3 py  15 p) Factor out the
GCF from each
group.
4m( y  5)  3 p( y  5)
Write the two factors.
(4m  3 p)( y  5)
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Factor completely
8s  200t
2
2
Is there a GCF?
Yes. Write the GCF first and the remaining factor after it.
8(s  25t )
2
2
Look at the remaining factor. (s2-25t2)
Is it a binomial, trinomial, or four-term polynomial?
It’s a binomial. Is it a difference of two squares? (a2-b2)
Yes. s2 is a square (s • s) and 25t2 is a square (5t • 5t).
Factor as (s+5t)(s-5t). Then write the complete
factorization.
8(s  5t )(s  5t )
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Factor completely
6y  5y  6
2
No.
Is there a GCF?
Is it a binomial, trinomial, or four-term polynomial?
It’s a trinomial.
Do you recognize it as a perfect square trinomial?
No. Use ac and b.
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ac
6 • -6
b
-5
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Look for factors of –36 that add up
to –5. Use your calculator or your
math skills to find 4 and -9 as the
factors to use.
-36
4, -9
Rewrite the equation with those two factors in the middle.
6y  5y  6
2
6y  9y  4y  6
2
(6 y  9 y)  (4 y  6)
2
3 y(2 y  3)  2(2 y  3)
Group.
Factor out the GCF
from each group.
Write the two factors.
(2 y  3)(3 y  2)
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Factor completely
6 x  15x  9 x
3
2
Is there a GCF?
Yes. Write the GCF first and the remaining factor after it.
3x(2 x  5x  3)
2
Look at the remaining factor.
(2x2  5x  3)
Is it a binomial, trinomial, or four-term polynomial?
It’s a trinomial.
Do you recognize it as a perfect square trinomial?
No. Use ac and b.
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ac
2 • -3
b
5
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Look for factors of -6 that add up to
5. Use your calculator or your
math skills to find 6 and -1 as the
factors to use.
-6
6, -1
Rewrite the equation with those two factors in the middle.
3x(2 x  5x  3)
2
3x[2x  6x 1x  3]
2
Group. Remember to change the –3 to a
+3 because of the minus sign in the
grouping!!
3x[(2x  6x)  (1x  3)]
3x[2 x( x  3)  1( x  3)]
3x( x  3)(2 x  1)
Factor out the GCF
from each group.
Write all three factors.
2
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Factor completely
2 x  32
8
Is there a GCF?
Yes. Write the GCF first and the remaining factor after it.
2( x 16)
8
Look at the remaining factor.
( x8  16)
Is it a binomial, trinomial, or four-term polynomial?
It’s a binomial. Is it a difference of two squares? (a2-b2)
Yes. x8 is a square (x4 • x4) and 16 is a square
(4 • 4). Factor as (x4 + 4)(x4 - 4).
So far we have 2(x4 + 4)(x4 - 4).
(Please continue—not done yet!!)
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2(x4 +4)(x4 -4)
Look at what you have. Can either of the binomials be broken
down?
(x4 +4)
Is this binomial a difference of two squares? (a2-b2)
No. It can’t be broken down. So, we have to keep this factor.
What about the other binomial?
(x4 -4)
Is this binomial a difference of two squares? (a2-b2)
Yes. x4 is a square (x2 • x2) and 4 is a square
(2 • 2). Factor as (x2 + 2)(x2 - 2).
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Put it all together.
2 x  32
8
2( x 16)
8
2(x4 +4)(x4 -4)
2(x4 +4)(x2 +2)(x2 -2)
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Not a difference of
squares. Can’t go
any farther!!
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Word Problem #1
What is the quotient when
12x3  8x 2  16x is divided by 4x?
This question is
3
2
3
2
12
x

8
x

16
x
12
x
8
x
16x
asking you to find



4x
4x
4x 4x
the OTHER
FACTOR after you
take out the greatest
2
3x
 2x  4
common factor of 4x.
Simplify each term.
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3x  2 x  4
2
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Word Problem #2
A rectangular garden plot has an area
represented by the expression
18x 2  3x  28
Find the dimensions of the garden plot.
This is a factoring problem. You need to
find the two factors that multiply together
to give you 18x 2  3x  28
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18x  3x  28
2
No.
Is there a GCF?
Is it a binomial, trinomial, or four-term polynomial?
It’s a trinomial.
Do you recognize it as a perfect square trinomial?
No. Use ac and b.
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ac
18 • -28
b
-3
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Look for factors of –504 that add
up to –3. Use your calculator or
your math skills to find 21 and -24
as the factors to use.
-504
21, -24
Rewrite the equation with those two factors in the middle.
18x  3x  28
2
18x  24x  21x  28
2
(18x  24x)  (21x  28)
2
6 x(3x  4)  7(3x  4)
(3x  4)(6 x  7)
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Group.
Factor out the GCF
from each group.
Write the two factors.
Length is 3x - 4 and width is 6x + 7
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To find factors of the ac term, use the following steps in your
calculator:
•Press the Y= button.
•In Y1=, type the ac value / X.
•In Y2=, type X + VARS, arrow to Y_VARS, Enter, Enter
•Go to Table and look for the b in column Y2. When you find it, use the
values in the X column and the Y1 column as your two factors to put in the
equation. IF YOU CAN’T find the b value in the Y2 column, the trinomial
can’t be factored.
•NOTE: Remember that you might need to scroll up the screen
to find negative numbers that give you the correct value in the Y2
column.
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