Solve a^3-4a^2-21a

$$a ^ { 3 } - 4 a ^ { 2 } - 21 a$$

Factor

$$a\left(a-7\right)\left(a+3\right)$$

Solution Steps

Factor out a.

$$a\left(a^{2}-4a-21\right)$$

Consider a^{2}-4a-21. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-21. To find p and q, set up a system to be solved.

$$p+q=-4 pq=1\left(-21\right)=-21$$

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 -21.

$$1,-21 3,-7$$

Calculate the sum for each pair.

$$1-21=-20 3-7=-4$$

The solution is the pair that gives sum -4.

$$p=-7 q=3$$

Rewrite a^{2}-4a-21 as \left(a^{2}-7a\right)+\left(3a-21\right).

$$\left(a^{2}-7a\right)+\left(3a-21\right)$$

Factor out a in the first and 3 in the second group.

$$a\left(a-7\right)+3\left(a-7\right)$$

Factor out common term a-7 by using distributive property.

$$\left(a-7\right)\left(a+3\right)$$

Rewrite the complete factored expression.

$$a\left(a-7\right)\left(a+3\right)$$

Evaluate

$$a\left(a-7\right)\left(a+3\right)$$