Question

Factor the expression completely.

$\left(a^2-1\right)b^2-4\left(a^2-1\right)$

Short answer

The solution$\left(a^2-1\right)b^2-4\left(a^2-1\right)=\left(a-1\right)\left(a+1\right)\left(b-2\right)\left(b+2\right)$

Step-by-step solution

Step 1. Apply the Factoring Formula.

If A and B are any real numbers or algebraic expressions, then

$A^2-B^2=\left(A+B\right)\left(A-B\right)$

Step 2. Analyze the given expression.

Given$\left(a^2-1\right)b^2-4\left(a^2-1\right)$

First using distributive property, we write the expression as:

$\left(a^2-1\right)b^2-4\left(a^2-1\right)=\left(a^2-1\right)\left(b^2-4\right)$

Step 3. Solve the expression.

Substitute the values in the Factoring formula to get:

$\begin{array}{l}\left(a^2-1\right)b^2-4\left(a^2-1\right)=\left(a^2-1\right)\left(b^2-4\right)\\ \Rightarrow \left(a^2-1\right)b^2-4\left(a^2-1\right)=\left(a^2-1^2\right)\left(b^2-2^2\right)\\ \Rightarrow \left(a^2-1\right)b^2-4\left(a^2-1\right)=\left(a-1\right)\left(a+1\right)\left(b-2\right)\left(b+2\right)\end{array}$