Solve x^3-3x^2-6x+8

$${ x }^{ 3 } -3 { x }^{ 2 } -6x+8$$

Factor

$$\left(x-4\right)\left(x-1\right)\left(x+2\right)$$

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. One such root is 4. Factor the polynomial by dividing it by x-4.

$$\left(x-4\right)\left(x^{2}+x-2\right)$$

Consider x^{2}+x-2. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite x^{2}+x-2 as \left(x^{2}-x\right)+\left(2x-2\right).

$$\left(x^{2}-x\right)+\left(2x-2\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-1\right)+2\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$\left(x-4\right)\left(x-1\right)\left(x+2\right)$$

Evaluate

$$\left(x-4\right)\left(x-1\right)\left(x+2\right)$$