# How do you use the rational roots theorem to find all possible zeros of f(x)=x^3+x^2-8x-6 ?

Mar 20, 2016

The Rational Root Theorem states: that the set $\left(\pm 1 , \pm 2 , \pm 3 , \pm 6\right)$ constitute the set of all possible zero roots to $f \left(x\right)$

#### Explanation:

Given $f \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \cdots + {a}_{1} x 6 + {a}_{0} = {\Sigma}_{i = 0}^{n} {a}_{i} {x}^{i}$
let $p$ and $q$ be $\forall p : {a}_{0} | p$ and $q : {a}_{n} | q$,
then f(x) potential roots are ${x}_{i} : {x}_{i} \in \left({p}_{l} / {q}_{k}\right)$
where ${p}_{l}$ and ${q}_{k}$ are the ${l}_{t h} \mathmr{and} {k}_{t h}$ factors

${a}_{0} = 6 \implies \pm \left(1 , 2 , 3 , 6\right)$
${a}_{3} = 1$
Thus the Rational Root Theorem states:
that the set $\left(\pm 1 , \pm 2 , \pm 3 , \pm 6\right)$ constitute the set of all possible zero roots to $f \left(x\right)$