# How do you factor given that f(-2)=0 and f(x)=9x^3+10x^2-17x-2?

Oct 15, 2016

Use synthetic division to get $\left(9 x + 1\right) \left(x - 1\right) \left(x + 2\right)$.

#### Explanation:

Factor given that $f \left(- 2\right) = 0$

$f \left(x\right) = 9 {x}^{3} + 10 {x}^{2} - 17 x - 2$

If $f \left(- 2\right) = 0$, one of the factors is $\left(x + 2\right)$

To find the other factors, divide the original polynomial by $\left(x + 2\right)$. This could be accomplished by long division, but synthetic division is quicker.

$- 2 | 9 \textcolor{w h i t e}{a a a} 10 \textcolor{w h i t e}{a} - 17 \textcolor{w h i t e}{a} - 2$
$\textcolor{w h i t e}{a a {a}^{2}} \downarrow - 18 \textcolor{w h i t e}{a a a} 16 \textcolor{w h i t e}{a {a}^{2} a} 2$
$\textcolor{w h i t e}{a {a}^{2} a a} \textcolor{red}{9} \textcolor{w h i t e}{a a} \textcolor{red}{- 8} \textcolor{w h i t e}{a a a} \textcolor{red}{- 1} \textcolor{w h i t e}{a a {a}^{2}} 0$

The red numbers in the quotient are the coefficients of a polynomial.

$\textcolor{red}{9} {x}^{2} - \textcolor{red}{8} x \textcolor{red}{- 1}$

Factor:
$\left(9 x + 1\right) \left(x - 1\right)$

The complete factorization is $\left(9 x + 1\right) \left(x - 1\right) \left(x + 2\right)$.

Oct 15, 2016

$f \left(x\right) = \left(x + 2\right) \left(9 x + 1\right) \left(x - 1\right)$

#### Explanation:

Given: $f \left(- 2\right) = 0$

Then $x + 2$ is a factor

When you divide by $\left(x + 2\right)$, the remaining quadratic is:

$9 {x}^{2} - 8 x - 1$

Check the discriminant:

${b}^{2} - 4 \left(a\right) \left(c\right) = {\left(- 8\right)}^{2} - 4 \left(9\right) \left(- 1\right) = 100$

${r}_{1} = \frac{8 - \sqrt{100}}{2 \left(9\right)} = - \frac{1}{9}$

${r}_{2} = \frac{8 + \sqrt{100}}{2 \left(9\right)}$ = 1

This makes the factors of the quadratic:

$\left(9 x + 1\right) \left(x - 1\right)$

Multiply by $\left(x + 2\right)$ for factors of the cubic:

$f \left(x\right) = \left(x + 2\right) \left(9 x + 1\right) \left(x - 1\right)$