Solve a^3-2a^2-24a

$$a ^ { 3 } - 2 a ^ { 2 } - 24 a$$

Factor

$$a\left(a-6\right)\left(a+4\right)$$

Solution Steps

Factor out a.

$$a\left(a^{2}-2a-24\right)$$

Consider a^{2}-2a-24. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-24. To find p and q, set up a system to be solved.

$$p+q=-2 pq=1\left(-24\right)=-24$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$p=-6 q=4$$

Rewrite a^{2}-2a-24 as \left(a^{2}-6a\right)+\left(4a-24\right).

$$\left(a^{2}-6a\right)+\left(4a-24\right)$$

Factor out a in the first and 4 in the second group.

$$a\left(a-6\right)+4\left(a-6\right)$$

Factor out common term a-6 by using distributive property.

$$\left(a-6\right)\left(a+4\right)$$

Rewrite the complete factored expression.

$$a\left(a-6\right)\left(a+4\right)$$

Evaluate

$$a\left(a-6\right)\left(a+4\right)$$