Solve =a^3-a^2-10a-8

$$= a ^ { 3 } - a ^ { 2 } - 10 a - 8$$

Factor

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

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 a-4.

$$\left(a-4\right)\left(a^{2}+3a+2\right)$$

Consider a^{2}+3a+2. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+2. To find p and q, set up a system to be solved.

$$p+q=3 pq=1\times 2=2$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. The only such pair is the system solution.

$$p=1 q=2$$

Rewrite a^{2}+3a+2 as \left(a^{2}+a\right)+\left(2a+2\right).

$$\left(a^{2}+a\right)+\left(2a+2\right)$$

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

$$a\left(a+1\right)+2\left(a+1\right)$$

Factor out common term a+1 by using distributive property.

$$\left(a+1\right)\left(a+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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