Solve x^3-x^2-16x=20

$${ x }^{ 3 } - { x }^{ 2 } -16x = 20$$

Solve for x

$$x=-2$$

$$x=5$$

Solution Steps

Subtract 20 from both sides.

$$x^{3}-x^{2}-16x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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}-x^{2}-16x-20 by x+2 to get x^{2}-3x-10. Solve the equation where the result equals to 0.

$$x^{2}-3x-10=0$$

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, -3 for b, and -10 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-10\right)}}{2}$$

Do the calculations.

$$x=\frac{3±7}{2}$$

Solve the equation x^{2}-3x-10=0 when ± is plus and when ± is minus.

$$x=-2 x=5$$

List all found solutions.

$$x=-2 x=5$$