Question

In the following exercises, factor completely

$n^4-81$.

Short answer

The factors of the given expression are $(n-3)(n+3)(n^2+9)$.

Step-by-step solution

Step 1. Given and explanation 

We have $n^4-81$.

We can see that this forms a pattern of difference of two perfect squares.

We know that $a^2-b^2=(a+b)(a-b)$.

Applying this, we will factorize the given expression.

Step 2. Factorizing the expression. 

We have $n^4-81$.

This can be written as role="math" localid="1644629414096" ${\left(n^2\right)}^2-{\left(3^2\right)}^2$.

Applying the property , we get

role="math" localid="1644629401616" $$\begin{gathered} a^2-b^2=(a+b)(a-b) \\ {\left(n^2\right)}^2-{\left(3^2\right)}^2=(n^2+3^2)(n^2-3^2) \\ =(n^2+9)(n+3)(n-3) \end{gathered}$$