# How do you factor given that f(-5)=0 and f(x)=x^3-x^2-21x+45?

Oct 10, 2016

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

#### Explanation:

$f \left(x\right) = {x}^{3} - {x}^{2} - 21 x + 45$ and $f \left(- 5\right) = 0$

$f \left(- 5\right) = 0$ indicates that one of the factors is $\left(x + 5\right)$ and the zero is $- 5$.

Use synthetic division with the given zero of $- 5$.

-5color(white)(a)|1color(white)(aa)-1color(white)(a)-21color(white)(a aa^1)45
$\textcolor{w h i t e}{a a a {a}^{1}} \downarrow \textcolor{w h i t e}{a} - 5 \textcolor{w h i t e}{a {a}^{1} a} 30 \textcolor{w h i t e}{a} - 45$
$\textcolor{w h i t e}{a a a a {a}^{1}} 1 \textcolor{w h i t e}{{a}^{1}} - 6 \textcolor{w h i t e}{a a a a} 9 \textcolor{w h i t e}{a a a a a} 0$

Use the result of synthetic division as the coefficients of a new polynomial, i.e $\left({x}^{3} - {x}^{2} - 21 x + 4\right) \div \left(x + 5\right) = \left({x}^{2} - 6 x + 9\right)$.

$1 {x}^{2} - 6 x + 9$

$\left(x - 3\right) \left(x - 3\right) \textcolor{w h i t e}{a a a}$Factor

The complete factorization is

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