Solve -a^3-4a^2+12a=

$$- a ^ { 3 } - 4 a ^ { 2 } + 12 a =$$

Factor

$$a\left(2-a\right)\left(a+6\right)$$

Solution Steps

Factor out a.

$$a\left(-a^{2}-4a+12\right)$$

Consider -a^{2}-4a+12. Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa+12. To find p and q, set up a system to be solved.

$$p+q=-4 pq=-12=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$p=2 q=-6$$

Rewrite -a^{2}-4a+12 as \left(-a^{2}+2a\right)+\left(-6a+12\right).

$$\left(-a^{2}+2a\right)+\left(-6a+12\right)$$

Factor out a in the first and 6 in the second group.

$$a\left(-a+2\right)+6\left(-a+2\right)$$

Factor out common term -a+2 by using distributive property.

$$\left(-a+2\right)\left(a+6\right)$$

Rewrite the complete factored expression.

$$a\left(-a+2\right)\left(a+6\right)$$

Evaluate

$$a\left(2-a\right)\left(a+6\right)$$