Ok, so we have this on homework and never covered in clas and I dont even know where to begin
Says complete the factorization
-48x^5y^8 = -6x^3y^4 ( )
$\displaystyle \displaystyle \begin{align*} -48x^5y^8 &= -6x^3y^4 \\ 0 &= 48x^5y^8 - 6x^3y^4 \\ 0 &= 6x^3y^4\left(8x^2y^4 - 1\right) \\ 0 &= 6x^3y^4\left[\left(2\sqrt{2}\,x\,y^2\right)^2 - 1^2\right] \\ 0 &= 6x^3y^4\left(2\sqrt{2}\,x\,y^2 - 1\right)\left(2\sqrt{2}\,x\,y^2 + 1\right) \end{align*} $
Can you solve the equation now?
Assuming that is what it means (and it could really mean anything...)
$\displaystyle \displaystyle \begin{align*} -48x^5y^8 &= -6\cdot 8x^3x^2y^4y^4 \end{align*} $
What is left over when you take out $\displaystyle \displaystyle \begin{align*} -6x^3y^4 \end{align*} $ as a factor?