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

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

Solve for x

$$x=-1$$

$$x=1$$

$$x=-4$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

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

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}+4x^{2}-x-4 by x-1 to get x^{2}+5x+4. Solve the equation where the result equals to 0.

$$x^{2}+5x+4=0$$

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, 5 for b, and 4 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{-5±3}{2}$$

Solve the equation x^{2}+5x+4=0 when ± is plus and when ± is minus.

$$x=-4 x=-1$$

List all found solutions.

$$x=1 x=-4 x=-1$$