Question

Find the quotient of the quantity negative 6 times x to the 2nd power times y to the 8th power plus 12 times x times y to the 3rd power minus 36 times x times y to the 2nd power all over 6 times x times y to the 2nd power.

−6x2y8 + 12xy3 − 6
−xy6 + 2y − 6
xy6 + 2y − 6
−xy6 + 2xy − 6

Answer 1

@johnweldon1993 help, smartiepants. :P

Answer 2

Sorry hun still on?

Well regardless

\[\large \frac{-6x^2y^8 + 12xy^3 - 36xy^2}{6xy^2}\]

same thing as before...every term in the numerator is divisible by 6...and each one here has at least an 'x' and at least a 'y^2' so we can factor each of those out

When we factor out a 6 we have

\[\large \frac{6(-x^2y^8 + 2xy^3 - 6xy^2)}{6xy^2}\]

When we then factor out an 'x' we have

\[\large \frac{6x(-xy^8 + 2y^3 - 6y^2)}{6xy^2}\]

and finally when we factor out a 'y^2'

\[\large \frac{6xy^2(-xy^6 + 2y - 6)}{6xy^2}\]

And just like last time...we now have a 6xy^2 on both the top and bottom so they cancel and we have

\[\large \frac{\cancel{6xy^2}(-xy^6 + 2y - 6)}{\cancel{6xy^2}} \rightarrow -xy^6 + 2y - 6\]