Solve m^3-3m^2-6m+8=0

$$m ^ { 3 } - 3 m ^ { 2 } - 6 m + 8 = 0$$

Solve for m

$$m=1$$

$$m=4$$

$$m=-2$$

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

$$±8,±4,±2,±1$$

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.

$$m=1$$

By Factor theorem, m-k is a factor of the polynomial for each root k. Divide m^{3}-3m^{2}-6m+8 by m-1 to get m^{2}-2m-8. Solve the equation where the result equals to 0.

$$m^{2}-2m-8=0$$

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, -2 for b, and -8 for c in the quadratic formula.

$$m=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$m=\frac{2±6}{2}$$

Solve the equation m^{2}-2m-8=0 when ± is plus and when ± is minus.

$$m=-2 m=4$$

List all found solutions.

$$m=1 m=-2 m=4$$