Question

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

Answer 1

`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 `p/q` ratios that have the potential to be roots.
We have `2x^5 + 7x^3+6x^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 `(x-1)` and check:
`2x^4+2x^3+9x^2+15x+15`