Solve x^3+x^2-4x-4=0

$${ x }^{ 3 } + { x }^{ 2 } -4x-4=0$$

Solve for x

$$x=-1$$

$$x=2$$

$$x=-2$$

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 -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+x^{2}-4x-4 by x+1 to get x^{2}-4. Solve the equation where the result equals to 0.

$$x^{2}-4=0$$

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, 0 for b, and -4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$x=\frac{0±4}{2}$$

Solve the equation x^{2}-4=0 when ± is plus and when ± is minus.

$$x=-2 x=2$$

List all found solutions.

$$x=-1 x=-2 x=2$$