Q 189 - Intermediate Algebra

Question

In the following exercises, factor completely using the difference of squares pattern, if possible.
$162a^4{b}^2-32b^2$

Step-by-Step Solution

Answer

The complete factor of ploynomial is  $2b^2\left(3a-2\right)\left(3a+2\right)\left(9a^2+4\right)$

Step 1. Given information

Consider the given ploynomial.

$162a^4{b}^2-32b^2$

Step 2. Formula used

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

Step 3. Find the factor using difference of square pattern.

$$\begin{gathered} 162a^4{b}^2-32b^2=2b^2\left(81a^4-16\right) \\ =2b^2\left({\left(9a^2\right)}^2-4^2\right) \\ =2b^2\left(9a^2-4\right)\left(9a^2+4\right) \end{gathered}$$

Step 4. Write them as product of conjugate.

$$\begin{gathered} 2b^2\left(9a^2-4\right)\left(9a^2+4\right)=2b^2\left({\left(3a\right)}^2-2^2\right)\left(9a^2+4\right) \\ =2b^2\left(3a-2\right)\left(3a+2\right)\left(9a^2+4\right) \end{gathered}$$