Solve x^3-2x^2-5x+6=0

$$x ^ { 3 } - 2 x ^ { 2 } - 5 x + 6 = 0$$

Solve for x

$$x=1$$

$$x=3$$

$$x=-2$$

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

$$±6,±3,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-2x^{2}-5x+6 by x-1 to get x^{2}-x-6. Solve the equation where the result equals to 0.

$$x^{2}-x-6=0$$

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 -6 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$x=\frac{1±5}{2}$$

Solve the equation x^{2}-x-6=0 when ± is plus and when ± is minus.

$$x=-2 x=3$$

List all found solutions.

$$x=1 x=-2 x=3$$