Solve a^2x-4x-2a^2+8

$$a ^ { 2 } x - 4 x - 2 a ^ { 2 } + 8$$

Factor

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

Solution Steps

Do the grouping a^{2}x-4x-2a^{2}+8=\left(a^{2}x-4x\right)+\left(-2a^{2}+8\right), and factor out x in the first and -2 in the second group.

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

Factor out common term a^{2}-4 by using distributive property.

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

Consider a^{2}-4. Rewrite a^{2}-4 as a^{2}-2^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Rewrite the complete factored expression.

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

Evaluate

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