What are the possible integral zeros of P(x)=x^3+3x^2-6x-8?

Jan 5, 2018

$x = - 4 , x = - 1 \text{ and } x = 2$

Explanation:

$\text{note that }$

$f \left(2\right) = 8 + 12 - 12 - 8 = 0$

$\Rightarrow \left(x - 2\right) \text{ is a factor}$

$\textcolor{red}{{x}^{2}} \left(x - 2\right) \textcolor{m a \ge n t a}{+ 2 {x}^{2}} + 3 {x}^{2} - 6 x - 8$

$= \textcolor{red}{{x}^{2}} \left(x - 2\right) \textcolor{red}{+ 5 x} \left(x - 2\right) \textcolor{m a \ge n t a}{+ 10 x} - 6 x - 8$

$= \textcolor{red}{{x}^{2}} \left(x - 2\right) \textcolor{red}{+ 5 x} \left(x - 2\right) \textcolor{red}{+ 4} \left(x - 2\right) \cancel{\textcolor{m a \ge n t a}{+ 8}} \cancel{- 8}$

$\Rightarrow {x}^{3} + 3 {x}^{2} - 6 x - 8$

$= \left(x - 2\right) \left({x}^{2} + 5 x + 4\right)$

$= \left(x - 2\right) \left(x + 1\right) \left(x + 4\right) = 0$

$\Rightarrow x = - 4 , x = - 1 \text{ and "x=2larrcolor(blue)"are the zeros}$