Question

divide. if there is a remainder, include it as a simplified fraction. (3z² - 24) ÷ (z + 3)

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(3z^2 - 24\) by the leading term of the divisor \(z + 3\). The leading term of the dividend is \(3z^2\) and the leading term of the divisor is \(z\), so \(\frac{3z^2}{z}=3z\).
Multiply the divisor \(z + 3\) by \(3z\) to get \(3z^2+9z\).
Subtract this from the dividend: \((3z^2 - 24)-(3z^2 + 9z)=3z^2-24 - 3z^2-9z=-9z - 24\).

Step2: Divide the new leading term

Now, divide the leading term of \(-9z - 24\) (which is \(-9z\)) by the leading term of the divisor \(z\), so \(\frac{-9z}{z}=-9\).
Multiply the divisor \(z + 3\) by \(-9\) to get \(-9z-27\).
Subtract this from \(-9z - 24\): \((-9z - 24)-(-9z - 27)=-9z-24 + 9z + 27 = 3\).

Step3: Write the result

The quotient is \(3z-9\) and the remainder is \(3\). So, \(\frac{3z^2 - 24}{z + 3}=3z-9+\frac{3}{z + 3}\).

Answer:

\(3z - 9+\dfrac{3}{z + 3}\)