# Question #9dcdb

Dec 10, 2014

Let us find all zeros of $f \left(x\right) = {x}^{3} - 7 x + 6$.

By inspection, we see that

$f \left(1\right) = {\left(1\right)}^{3} - 7 \left(1\right) + 6 = 0$

So, $x = 1$ is one of the zeros, which means that $f \left(x\right)$ has the factor $\left(x - 1\right)$.

By factoring out $\left(x - 1\right)$,

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

by factoring out a bit further,

$f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x + 3\right) = 0$

Hence, the zeros of $f$ are $x = - 3 , 1 , 2$.

I hope that this was helpful.