Solve x^3-61x-180=0

$$x ^ { 3 } - 61 x - 180 = 0$$

Solve for x

$$x=-4$$

$$x=-5$$

$$x=9$$

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 -180 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±180,±90,±60,±45,±36,±30,±20,±18,±15,±12,±10,±9,±6,±5,±4,±3,±2,±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-61x-180 by x+4 to get x^{2}-4x-45. Solve the equation where the result equals to 0.

$$x^{2}-4x-45=0$$

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, -4 for b, and -45 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-45\right)}}{2}$$

Do the calculations.

$$x=\frac{4±14}{2}$$

Solve the equation x^{2}-4x-45=0 when ± is plus and when ± is minus.

$$x=-5 x=9$$

List all found solutions.

$$x=-4 x=-5 x=9$$