Factorizing expressions

Question

I am having trouble solving this problem $81f^2- \dfrac{9}{e^2}$. How do you begin when solving this problem? Do you move $f^2$ by replacing the $9$ and vice versa and does the minus change to plus?

Answer 1

You should notice that $81=9^2$ and $9=3^2$

You can then write the expression as $(9f)^2-(\dfrac 3e)^2$

It looks like $a^2-b^2$, isn't it? You should be able to finish from here.

EDIT: even if $81$ and $9$ were no squares, the method works, you just pu the square roots of the numbers...

Answer 2

This expression is what is often termed a "difference of squares": one perfect square minus another—symbolically, $a^2-b^2$. It can be factored as $(a+b)(a-b)$.

In this case, we have $81f^2$, the square of $9f$, minus $9/e^2$, the square of $3/e$. The expression can therefore be factored, using the framework above, as $(9f-3/e)(9f+3/e)$.