# 1.5 Factoring polynomials  (Page 2/6)

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Factor $\text{\hspace{0.17em}}x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right)\text{\hspace{0.17em}}$ by pulling out the GCF.

$\left({b}^{2}-a\right)\left(x+6\right)$

## Factoring a trinomial with leading coefficient 1

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial $\text{\hspace{0.17em}}{x}^{2}+5x+6\text{\hspace{0.17em}}$ has a GCF of 1, but it can be written as the product of the factors $\text{\hspace{0.17em}}\left(x+2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(x+3\right).$

Trinomials of the form $\text{\hspace{0.17em}}{x}^{2}+bx+c\text{\hspace{0.17em}}$ can be factored by finding two numbers with a product of $c\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ The trinomial $\text{\hspace{0.17em}}{x}^{2}+10x+16,$ for example, can be factored using the numbers $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}8\text{\hspace{0.17em}}$ because the product of those numbers is $\text{\hspace{0.17em}}16\text{\hspace{0.17em}}$ and their sum is $\text{\hspace{0.17em}}10.\text{\hspace{0.17em}}$ The trinomial can be rewritten as the product of $\text{\hspace{0.17em}}\left(x+2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(x+8\right).$

## Factoring a trinomial with leading coefficient 1

A trinomial of the form $\text{\hspace{0.17em}}{x}^{2}+bx+c\text{\hspace{0.17em}}$ can be written in factored form as $\text{\hspace{0.17em}}\left(x+p\right)\left(x+q\right)\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}pq=c\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}p+q=b.$

Can every trinomial be factored as a product of binomials?

No. Some polynomials cannot be factored. These polynomials are said to be prime.

Given a trinomial in the form $\text{\hspace{0.17em}}{x}^{2}+bx+c,$ factor it.

1. List factors of $\text{\hspace{0.17em}}c.$
2. Find $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q,$ a pair of factors of $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ with a sum of $\text{\hspace{0.17em}}b.$
3. Write the factored expression $\text{\hspace{0.17em}}\left(x+p\right)\left(x+q\right).$

## Factoring a trinomial with leading coefficient 1

Factor $\text{\hspace{0.17em}}{x}^{2}+2x-15.$

We have a trinomial with leading coefficient $\text{\hspace{0.17em}}1,b=2,$ and $\text{\hspace{0.17em}}c=-15.\text{\hspace{0.17em}}$ We need to find two numbers with a product of $\text{\hspace{0.17em}}-15\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$ In [link] , we list factors until we find a pair with the desired sum.

Factors of $\text{\hspace{0.17em}}-15$ Sum of Factors
$1,-15$ $-14$
$-1,15$ 14
$3,-5$ $-2$
$-3,5$ 2

Now that we have identified $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}5,$ write the factored form as $\text{\hspace{0.17em}}\left(x-3\right)\left(x+5\right).$

Does the order of the factors matter?

No. Multiplication is commutative, so the order of the factors does not matter.

Factor $\text{\hspace{0.17em}}{x}^{2}-7x+6.$

$\left(x-6\right)\left(x-1\right)$

## Factoring by grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping    by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial $\text{\hspace{0.17em}}2{x}^{2}+5x+3\text{\hspace{0.17em}}$ can be rewritten as $\text{\hspace{0.17em}}\left(2x+3\right)\left(x+1\right)\text{\hspace{0.17em}}$ using this process. We begin by rewriting the original expression as $\text{\hspace{0.17em}}2{x}^{2}+2x+3x+3\text{\hspace{0.17em}}$ and then factor each portion of the expression to obtain $\text{\hspace{0.17em}}2x\left(x+1\right)+3\left(x+1\right).\text{\hspace{0.17em}}$ We then pull out the GCF of $\text{\hspace{0.17em}}\left(x+1\right)\text{\hspace{0.17em}}$ to find the factored expression.

## Factor by grouping

To factor a trinomial in the form $\text{\hspace{0.17em}}a{x}^{2}+bx+c\text{\hspace{0.17em}}$ by grouping, we find two numbers with a product of $\text{\hspace{0.17em}}ac\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ We use these numbers to divide the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.

Given a trinomial in the form $\text{\hspace{0.17em}}a{x}^{2}+bx+c,$ factor by grouping.
1. List factors of $\text{\hspace{0.17em}}ac.$
2. Find $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q,$ a pair of factors of $\text{\hspace{0.17em}}ac\text{\hspace{0.17em}}$ with a sum of $\text{\hspace{0.17em}}b.$
3. Rewrite the original expression as $\text{\hspace{0.17em}}a{x}^{2}+px+qx+c.$
4. Pull out the GCF of $\text{\hspace{0.17em}}a{x}^{2}+px.$
5. Pull out the GCF of $\text{\hspace{0.17em}}qx+c.$
6. Factor out the GCF of the expression.

## Factoring a trinomial by grouping

Factor $\text{\hspace{0.17em}}5{x}^{2}+7x-6\text{\hspace{0.17em}}$ by grouping.

We have a trinomial with $\text{\hspace{0.17em}}a=5,b=7,$ and $\text{\hspace{0.17em}}c=-6.\text{\hspace{0.17em}}$ First, determine $\text{\hspace{0.17em}}ac=-30.\text{\hspace{0.17em}}$ We need to find two numbers with a product of $\text{\hspace{0.17em}}-30\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}7.\text{\hspace{0.17em}}$ In [link] , we list factors until we find a pair with the desired sum.

Factors of $\text{\hspace{0.17em}}-30$ Sum of Factors
$1,-30$ $-29$
$-1,30$ 29
$2,-15$ $-13$
$-2,15$ 13
$3,-10$ $-7$
$-3,10$ 7

So $\text{\hspace{0.17em}}p=-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q=10.$

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
hii
Amit
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Dorbor
well
Biswajit
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Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
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Rubben
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Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin