Answer
$\left( \dfrac{11}{5}y^2+7x\right)\left( \dfrac{11}{5}y^2-7x\right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
\dfrac{121}{25}y^4-49x^2
,$ use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
\dfrac{121}{25}y^4
$ and $
49x^2
$ are both perfect squares and are separated by a minus sign. Hence, $
\dfrac{121}{25}y^4-49x^2
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
=\left(\dfrac{11}{5}y^2\right)^2-\left(7x\right)^2
\\=\left( \dfrac{11}{5}y^2+7x\right)\left( \dfrac{11}{5}y^2-7x\right)
.\end{array}
