Solve 4a^3+4a^2-8a

$$4 a ^ { 3 } + 4 a ^ { 2 } - 8 a$$

Factor

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

Solution Steps

Factor out 4.

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

Consider a^{3}+a^{2}-2a. Factor out a.

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

Consider a^{2}+a-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.

$$p+q=1 pq=1\left(-2\right)=-2$$

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

$$p=-1 q=2$$

Rewrite a^{2}+a-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.

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

Evaluate

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