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Factor $\text{\hspace{0.17em}}x({b}^{2}-a)+6({b}^{2}-a)\text{\hspace{0.17em}}$ by pulling out the GCF.
$({b}^{2}-a)(x+6)$
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 $\text{\hspace{0.17em}}(x+2)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}(x+3).$
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 $\text{\hspace{0.17em}}(x+2)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}(x+8).$
A trinomial of the form $\text{\hspace{0.17em}}{x}^{2}+bx+c\text{\hspace{0.17em}}$ can be written in factored form as $\text{\hspace{0.17em}}(x+p)(x+q)\text{\hspace{0.17em}}$ 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.
Factor $\text{\hspace{0.17em}}{x}^{2}+2x-15.$
We have a trinomial with leading coefficient $\text{\hspace{0.17em}}1,b=2,$ and $\text{\hspace{0.17em}}c=\mathrm{-15.}\text{\hspace{0.17em}}$ We need to find two numbers with a product of $\text{\hspace{0.17em}}\mathrm{-15}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\mathrm{-15}$ | Sum of Factors |
---|---|
$1,\mathrm{-15}$ | $\mathrm{-14}$ |
$\mathrm{-1},15$ | 14 |
$3,\mathrm{-5}$ | $\mathrm{-2}$ |
$\mathrm{-3},5$ | 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 $\text{\hspace{0.17em}}\mathrm{-3}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}5,$ write the factored form as $\text{\hspace{0.17em}}(x-3)(x+5).$
Does the order of the factors matter?
No. Multiplication is commutative, so the order of the factors does not matter.
Factor $\text{\hspace{0.17em}}{x}^{2}-7x+6.$
$(x\mathrm{-6})(x\mathrm{-1})$
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 $\text{\hspace{0.17em}}(2x+3)(x+1)\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}2x(x+1)+3(x+1).\text{\hspace{0.17em}}$ We then pull out the GCF of $\text{\hspace{0.17em}}(x+1)\text{\hspace{0.17em}}$ to find the factored expression.
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.
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 $\text{\hspace{0.17em}}c=\mathrm{-6.}\text{\hspace{0.17em}}$ First, determine $\text{\hspace{0.17em}}ac=\mathrm{-30.}\text{\hspace{0.17em}}$ We need to find two numbers with a product of $\text{\hspace{0.17em}}\mathrm{-30}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}\mathrm{-30}$ | Sum of Factors |
---|---|
$1,\mathrm{-30}$ | $\mathrm{-29}$ |
$\mathrm{-1},30$ | 29 |
$2,\mathrm{-15}$ | $\mathrm{-13}$ |
$\mathrm{-2},15$ | 13 |
$3,\mathrm{-10}$ | $\mathrm{-7}$ |
$\mathrm{-3},10$ | 7 |
So $\text{\hspace{0.17em}}p=\mathrm{-3}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q=10.$
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