# How to Find the Roots of a Polynomial Function

Most grade 12 math students are required to solve various polynomial functions. This tutorial explains how to solve for the roots of a polynomial function using the factor theorem. According to this theorem the factors of a polynomial are determined when $P(x)=0$. In order to find appropriate values to test the factors of a polynomial’s constant are used.

**Practice:**

Use the factor theorem to solve for the roots of the cubic function $f(x) = x^3 + 3x^2 – 6x – 8$.

**Solution:**

First, list all the possible factors of $(-8)$ and test each factor $($ 1, -1, 2, -2, 4, -4, 8, -8 $)$.

$P(1) = (1)^3 + 3(1)^2 – 6(1) – 8$

$P(1) = 1 + 3 – 6 – 8 = -10$

Since the $P(1) = -10$, $(x-1)$ is not a factor.

$P(-1) = (-1)^3 + 3(-1)^2 – 6(-1) – 8$

$P(-1) = -1 + 3 + 6 – 8 = 0$

Since the $P(-1) = 0$, we have found our first factor which is $(x+1)$.

$P(2) = (2)^3 + 3(2)^2 – 6(2) – 8$

$P(2) = 8 + 12 – 12 – 8 = 0$

Since the $P(2) = 0$, we have found our second factor which is $(x-2)$.

$P(-2) = (-2)^3 + 3(-2)^2 – 6(-2) – 8$

$P(-2) = -8 + 12 + 12 – 8 = -8$

Since $P(-2) = -8$, $(x+2)$ is not a factor.

$P(4) = (4)^3 + 3(4)^2 – 6(4) – 8$

$P(4) = 64 + 48 – 24 – 8 = 100$

Since the $P(4) = 100$, $(x-4)$ is not a factor.

$P(-4) = (-4)^3 + 3(-4)^2 – 6(-4) – 8$

$P(4) = -64 + 48 – 24 – 8 = 0$

Since the $P(-4) = 0$, we have found our third and final factor which is $(x+4)$. The roots of this cubic function are $(-1,0)$, $(2,0)$, and $(-4,0)$.

**Note: There are other ways of obtaining the zeros of this cubic function. Typically after finding the first root synthetic division and factoring is used to solve for the remaining roots.**