Solve x^3-6x^2-16x+96=0

$$x ^ { 3 } - 6 x ^ { 2 } - 16 x + 96 = 0$$

Solve for x

$$x=4$$

$$x=6$$

$$x=-4$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 96 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±96,±48,±32,±24,±16,±12,±8,±6,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-6x^{2}-16x+96 by x-4 to get x^{2}-2x-24. Solve the equation where the result equals to 0.

$$x^{2}-2x-24=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -2 for b, and -24 for c in the quadratic formula.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-24\right)}}{2}$$

Do the calculations.

$$x=\frac{2±10}{2}$$

Solve the equation x^{2}-2x-24=0 when ± is plus and when ± is minus.

$$x=-4 x=6$$

List all found solutions.

$$x=4 x=-4 x=6$$