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

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

Factor

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

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 negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

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

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

Factor out x in x^{2}-2x.

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

Factor out common term x-2 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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