Question

question
find the quotient of $8y^4 + 20y^3$ divided by $-4y^2$.

Explanation:

Step1: Divide each term by \(-4y^2\)

We have \(\frac{8y^4}{-4y^2} + \frac{20y^3}{-4y^2}\)

Step2: Simplify the first term

For \(\frac{8y^4}{-4y^2}\), divide the coefficients: \(\frac{8}{-4} = -2\), and subtract the exponents of \(y\): \(y^{4 - 2}=y^2\). So this term becomes \(-2y^2\)

Step3: Simplify the second term

For \(\frac{20y^3}{-4y^2}\), divide the coefficients: \(\frac{20}{-4} = -5\), and subtract the exponents of \(y\): \(y^{3 - 2}=y\). So this term becomes \(-5y\)

Step4: Combine the simplified terms

Combining \(-2y^2\) and \(-5y\), we get \(-2y^2 - 5y\)

Answer:

\(-2y^2 - 5y\)