# How do you use the rational roots theorem to find all possible zeros of  p(x)= 2x^5+7x^3+6x^2-15?

Mar 1, 2016

p/q: ⇒ ±1/2, ±1, ±2

#### Explanation:

If f(x)=a_nx^n+a_(n−1)x^(n−1)+…+a_1x+a_0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then $p$ is a factor of the constant term ${a}_{0}$ and $q$ is a factor of the leading coefficient ${a}_{n}$.
Given this all what we need to do here is identify the factors of ${a}_{0}$ and ${a}_{n}$ and build the $\frac{p}{q}$ ratios that have the potential to be roots.
We have $2 {x}^{5} + 7 {x}^{3} + 6 {x}^{2} - 15$

so p⇒ ±1,±2 and q⇒±1,±2
and the ratio:
p/q: ⇒ ±1/2, ±1, ±2

This are the potential roots, but remember they are not guaranteed... In this case you can try (x-1), use long division and see if you can divide without a remainder...
You were not asked to do that but you can divide $\left(x - 1\right)$ and check:
$2 {x}^{4} + 2 {x}^{3} + 9 {x}^{2} + 15 x + 15$