# Write the polynomial in factored form? x^3 + 2x^2 - 15x

## Select one: a. $- 3 x \left(x + 5\right) \left(x + 1\right)$ b. $x \left(x - 3\right) \left(x + 5\right)$ c. $5 x \left(x + 1\right) \left(x - 3\right)$ d. $x \left(x + 5\right) \left(x + 3\right)$

Mar 3, 2017

b. $x \left(x - 3\right) \left(x + 5\right)$

#### Explanation:

Note that the coefficient of ${x}^{3}$ is $1$, so we can eliminate a and c immediately.

Looking at the coefficient of $x$, which is negative, we can also rule out d, which is all positive.

So the only possibility is b.

Does it work?

$x \left(x - 3\right) \left(x + 5\right) = x \left({x}^{2} + \left(5 - 3\right) x + \left(- 3\right) \left(5\right)\right)$

$\textcolor{w h i t e}{x \left(x - 3\right) \left(x + 5\right)} = x \left({x}^{2} + 2 x - 15\right)$

$\textcolor{w h i t e}{x \left(x - 3\right) \left(x + 5\right)} = {x}^{3} + 2 {x}^{2} - 15 x$

$\textcolor{w h i t e}{}$
Footnote

If we were factoring this without the multiple choice answers, then we could proceed as follows:

Given:

${x}^{3} + 2 {x}^{2} - 15 x$

First note that all of the terms are divisible by $x$, so we can separate that out as a factor:

${x}^{3} + 2 {x}^{2} - 15 x = x \left({x}^{2} + 2 x - 15\right)$

Next look for a pair of factors of $15$ which differ by $2$.

The pair $5 , 3$ works, so we find:

${x}^{2} + 2 x - 15 = \left(x + 5\right) \left(x - 3\right)$

Putting it all together we have:

${x}^{3} + 2 {x}^{2} - 15 x = x \left(x + 5\right) \left(x - 3\right)$