Solve x^3-3x^2-9x-5=0

$$x ^ { 3 } - 3 x ^ { 2 } - 9 x - 5 = 0$$

Solve for x

$$x=-1$$

$$x=5$$

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 -5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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}-3x^{2}-9x-5 by x+1 to get x^{2}-4x-5. Solve the equation where the result equals to 0.

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

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 -5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±6}{2}$$

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

$$x=-1 x=5$$

List all found solutions.

$$x=-1 x=5$$