Question

Factors of $p^4-81$ are

Answer 1

Factorize $p^4-81$.

We can rewrite $p^4-81$ as ${\left(p^2\right)}^2-9^2$

This is in the form of $a^2-b^2$ where $a=p^2$and $b=9$

We know that ${\mathit{a}}^{\mathbf{2}}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}\mathbf{(}\mathit{a}\mathbf{-}\mathit{b}\mathbf{)}$

Substituting the values for $a$ and $b$, we see that the factors of $p^4-81$ are $(p^2+9)$ and $(p^2-9)$

Further, factors of $(p^2-9)$ are $(p-3)(p+3)$

Therefore, the factors of $p^4-81$ are $(p^2+9)$ and $(p-3)(p+3)$