Solve x^3+2x^2-23x-60=0

$$x ^ { 3 } + 2 x ^ { 2 } - 23 x - 60 = 0$$

Solve for x

$$x=-3$$

$$x=5$$

$$x=-4$$

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 -60 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±60,±30,±20,±15,±12,±10,±6,±5,±4,±3,±2,±1$$

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}+2x^{2}-23x-60 by x+3 to get x^{2}-x-20. Solve the equation where the result equals to 0.

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

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, -1 for b, and -20 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-20\right)}}{2}$$

Do the calculations.

$$x=\frac{1±9}{2}$$

Solve the equation x^{2}-x-20=0 when ± is plus and when ± is minus.

$$x=-4 x=5$$

List all found solutions.

$$x=-3 x=-4 x=5$$