Solve -3a^2x^2+24a^2x-48a^2

$$- 3 a ^ { 2 } x ^ { 2 } + 24 a ^ { 2 } x - 48 a ^ { 2 }$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(-a^{2}x^{2}+8a^{2}x-16a^{2}\right)$$

Consider -a^{2}x^{2}+8a^{2}x-16a^{2}. Factor out a^{2}.

$$a^{2}\left(-x^{2}+8x-16\right)$$

Consider -x^{2}+8x-16. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+px+qx-16. To find p and q, set up a system to be solved.

$$p+q=8 pq=-\left(-16\right)=16$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$p=4 q=4$$

Rewrite -x^{2}+8x-16 as \left(-x^{2}+4x\right)+\left(4x-16\right).

$$\left(-x^{2}+4x\right)+\left(4x-16\right)$$

Factor out -x in the first and 4 in the second group.

$$-x\left(x-4\right)+4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x+4\right)$$

Rewrite the complete factored expression.

$$3a^{2}\left(x-4\right)\left(-x+4\right)$$

Evaluate

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