Solve x^3+4x^2-9x-36=0

$$x ^ { 3 } + 4 x ^ { 2 } - 9 x - 36 = 0$$

Solve for x

$$x=-3$$

$$x=3$$

$$x=-4$$

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 -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+4x^{2}-9x-36 by x-3 to get x^{2}+7x+12. Solve the equation where the result equals to 0.

$$x^{2}+7x+12=0$$

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, 7 for b, and 12 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{-7±1}{2}$$

Solve the equation x^{2}+7x+12=0 when ± is plus and when ± is minus.

$$x=-4 x=-3$$

List all found solutions.

$$x=3 x=-4 x=-3$$