# 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 $\text{\hspace{0.17em}}q=10.$

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