# Given polynomial f(x)=x^3-10x^2+19x+30 and a factor x-6 how do you find all other factors?

Jun 9, 2017

$\therefore {x}^{3} - 10 {x}^{2} + 19 x + 30 = \left(x - 6\right) \left(x - 5\right) \left(x + 1\right)$

#### Explanation:

if $\left(x - 6\right)$ is a factor we have

${x}^{3} - 10 {x}^{2} + 19 x + 30 = \left(x - 6\right) \left({x}^{2} + b x + c\right)$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} = {x}^{3} + b {x}^{2} + c x$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} = \textcolor{w h i t e}{\times} - 6 {x}^{2} - 6 b x - 6 c$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times x} = {x}^{3} + \textcolor{b l u e}{\left(b - 6\right)} {x}^{2} + \textcolor{red}{\left(c - 6 b\right)} x - 6 c$

we now compare coefficients and solve

(Coefficients of x^2)

$L H S = - 10$

$R H S = b - 6$

$\textcolor{b l u e}{b - 6 = - 10 \implies b = - 10 + 6}$

$b = - 4$

Coefficients of x

$L H S = 19$

$R H S = c - 6 b$

$\textcolor{red}{- 6 \times - 4 + c = 19}$

$\implies 24 + c = 19$

$c = - 5$

check constant term

$L H S = 30$

$R H S = - 6 \times - 5 = 30$

$\therefore {x}^{3} - 10 {x}^{2} + 19 x + 30 = \left(x - 6\right) \left({x}^{2} - 4 x - 5\right)$

now see if the quadratic factorises

factors $- 5$ that sum to $- 4$

$- 5 \text{ }$&$\text{ } = + 1$

we have therefore

$\therefore {x}^{3} - 10 {x}^{2} + 19 x + 30 = \left(x - 6\right) \left(x - 5\right) \left(x + 1\right)$