Solve x^3-12x=-16

$${ x }^{ 3 } -12x=-16$$

Solve for x

$$x=2$$

$$x=-4$$

Solution Steps

Add 16 to both sides.

$$x^{3}-12x+16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±16,±8,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-12x+16 by x-2 to get x^{2}+2x-8. Solve the equation where the result equals to 0.

$$x^{2}+2x-8=0$$

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 -8 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$x=\frac{-2±6}{2}$$

Solve the equation x^{2}+2x-8=0 when ± is plus and when ± is minus.

$$x=-4 x=2$$

List all found solutions.

$$x=2 x=-4$$