## Precalculus (6th Edition)

$\color{blue}{(y^2+9)(y+3)(y-3)}$
Note that $y^4=(y^2)^2$ and $81=9^2$. Thus, the given binomial is equivalent to: $=(y^2)^2-9^2$ Factor using the formula $a^2-b^2=(a+b)(a-b)$ where $a=y^2$ and $b=9$ to obtain: $=(y^2+9)(y^2-9) \\=(y^2+9)(y^2-3^2)$ Factor the second binomial factor using the same formula above with $a=y$ and $b=3$ to obtain: $=\color{blue}{(y^2+9)(y+3)(y-3)}$