# What are all the rational zeros of x^3-7x-6?

May 21, 2018

Zeros are $x = - 1 , x = - 2 \mathmr{and} x = 3$

#### Explanation:

f(x)=x^3-7 x - 6 ;  By inspection $f \left(- 1\right) = 0$ ,so

$\left(x + 1\right)$ will be a factor.

${x}^{3} - 7 x - 6 = {x}^{3} + {x}^{2} - {x}^{2} - x - 6 x - 6$

$= {x}^{2} \left(x + 1\right) - x \left(x + 1\right) - 6 \left(x + 1\right)$

$= \left(x + 1\right) \left({x}^{2} - x - 6\right) = \left(x + 1\right) \left({x}^{2} - 3 x + 2 x - 6\right)$

$= \left(x + 1\right) \left\{x \left(x - 3\right) + 2 \left(x - 3\right)\right\}$

$\therefore f \left(x\right) = \left(x + 1\right) \left(x - 3\right) \left(x + 2\right) \therefore f \left(x\right)$ will be zero

for $x = - 1 , x = - 2 \mathmr{and} x = 3$

Hence zeros are $x = - 1 , x = - 2 \mathmr{and} x = 3$ [Ans]