# 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.