Solve 4ax^2-20ax+24a

$$4 a x ^ { 2 } - 20 a x + 24 a$$

Factor

$$4a\left(x-3\right)\left(x-2\right)$$

Solution Steps

Factor out 4.

$$4\left(ax^{2}-5ax+6a\right)$$

Consider ax^{2}-5ax+6a. Factor out a.

$$a\left(x^{2}-5x+6\right)$$

Consider x^{2}-5x+6. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+px+qx+6. To find p and q, set up a system to be solved.

$$p+q=-5 pq=1\times 6=6$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$p=-3 q=-2$$

Rewrite x^{2}-5x+6 as \left(x^{2}-3x\right)+\left(-2x+6\right).

$$\left(x^{2}-3x\right)+\left(-2x+6\right)$$

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

$$x\left(x-3\right)-2\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-2\right)$$

Rewrite the complete factored expression.

$$4a\left(x-3\right)\left(x-2\right)$$

Evaluate

$$4a\left(x-3\right)\left(x-2\right)$$