Solve a^2x+a^2-9b^2x-9b^2=

$$a ^ { 2 } x + a ^ { 2 } - 9 b ^ { 2 } x - 9 b ^ { 2 } =$$

Factor

$$\left(a-3b\right)\left(x+1\right)\left(a+3b\right)$$

Solution Steps

Do the grouping a^{2}x+a^{2}-9b^{2}x-9b^{2}=\left(a^{2}x+a^{2}\right)+\left(-9b^{2}x-9b^{2}\right), and factor out a^{2} in the first and -9b^{2} in the second group.

$$a^{2}\left(x+1\right)-9b^{2}\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(a^{2}-9b^{2}\right)$$

Consider a^{2}-9b^{2}. Rewrite a^{2}-9b^{2} as a^{2}-\left(3b\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a-3b\right)\left(a+3b\right)$$

Rewrite the complete factored expression.

$$\left(x+1\right)\left(a-3b\right)\left(a+3b\right)$$

Evaluate

$$\left(x+1\right)\left(a^{2}-9b^{2}\right)$$