Solve ax^2+4ax-12a

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

Factor

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

Solution Steps

Factor out a.

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

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

$$p+q=4 pq=1\left(-12\right)=-12$$

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.

$$p=-2 q=6$$

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

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

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

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

Factor out common term x-2 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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