Question

find the quotient of $-16p^4 + 28p^3$ divided by $-4p^2$.

Explanation:

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

We have to divide \(-16p^4 + 28p^3\) by \(-4p^2\), so we can split this into two separate divisions: \(\frac{-16p^4}{-4p^2}+\frac{28p^3}{-4p^2}\)

Step2: Simplify the first term

For \(\frac{-16p^4}{-4p^2}\), we divide the coefficients and subtract the exponents of \(p\). The coefficient: \(\frac{-16}{-4} = 4\), and for the variable part, using the rule \(a^m\div a^n=a^{m - n}\), we have \(p^{4-2}=p^2\). So the first term simplifies to \(4p^2\)

Step3: Simplify the second term

For \(\frac{28p^3}{-4p^2}\), divide the coefficients: \(\frac{28}{-4}=-7\), and for the variable part, \(p^{3 - 2}=p\). So the second term simplifies to \(-7p\)

Step4: Combine the terms

Combining the two simplified terms, we get \(4p^2-7p\)

Answer:

\(4p^2 - 7p\)