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

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

Factor

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

Solution Steps

Factor out a.

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

Consider 12+4a-a^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$p=6 q=-2$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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