Roots of a Cubic Polynomial

Problem 260: Find the other two roots of the polynomial

$f(x) = 6x^3 + 17x^2 - 31x - 12$

if $x = -4$ is a root.

Solution: Since we are given one of the roots, we can use root to set up a binomial term as follow:

$x - \text{root} = x - (-4) = x + 4$

One can now use this binomial and long division to find a quadratic polynomial which we know how to solve.

That is, set up the problem as a long division problem.

$x - 4 \overline{)6x^3 + 17x^2 - 31x - 12}$

The quotient of this long division is $6x^2 - 7x - 3$ . Now, all we have to do is factor the quadratic polynomial or use the Quadratic Formula and solve for $x$ .

$\begin{align*} 6x^2 - 7x - 3 = 0\end{align*}$

Using the quadratic formula, we found the second and third roots.

$x = 3/2, \ x = -1/3$

Thus, the third degree polynomial has roots at $x = -4, 3/2, -1/3$ .

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