Question

Divide (8x^4y^3 + 4x^3y^2 - 2x^2y - 12x^2y^4) by -2x^2y.

Answer 1

\(\dfrac{8x^4y^3 + 4x^3y^2 - 2x^2y - 12x^2y^4}{-2x^2y}\)

Just regularly divide the -2 into each of the whole numbered terms:

\(\dfrac{-4x^4y^3 - 2x^3y^2 + x^2y + 6x^2y^4}{-2x^2y}\)

Answer 2

Now divide the \(x^2\) to every \(x\) term and \(y\) to every \(y\) term.

When dividing fractions with like bases, subtract the exponents and keep the base.

\(x^2\) has an exponent of 2, and \(y\) has an exponent of 1.

So:

\(-4x^2y^2 - 2xy + 6y^3\)

Answer 3

Okay!

Answer 4

Sorry let me correct this:

\(\dfrac{8x^4y^3 + 4x^3y^2 - 2x^2y - 12x^2y^4}{-2x^2y}\)

Just regularly divide the -2 into each of the whole numbered terms:

\(\dfrac{-4x^4y^3 - 2x^3y^2 + x^2y + 6x^2y^4}{x^2y}\)

Forgot to remove the \(-2\).

Answer 5

@iGreen Thanks!!! Now, How would your answer in Part A be affected if the x2 variable in the denominator was just an x? :/