## Precalculus (6th Edition)

$\color{blue}{(\frac{11}{5}y^2+7x)(\frac{11}{5}y^2-7x)}$
With $\frac{121}{25}y^4=(\frac{11}{5}y^2)^2$ and $49x^2=(7x)^2$, the given binomial is equivalent to: $=(\frac{11}{5}y^2)^2-(7x)^2$ Factor using the formula $a^2-b^2=(a+b)(a-b)$ with $a=\frac{11}{5}y^2$ and $b=7x$ to obtain: $=\color{blue}{(\frac{11}{5}y^2+7x)(\frac{11}{5}y^2-7x)}$