Solve ax^2-4a+bx^2-4b

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

Factor

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

Solution Steps

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

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

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

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{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).

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

Rewrite the complete factored expression.

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

Evaluate

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