Question

1. (6t³ + 5t² - 2t + 1) ÷ (3t + 1)

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(6t^{3}\) by the leading term of the divisor \(3t\): \(\frac{6t^{3}}{3t} = 2t^{2}\)
Multiply the divisor \((3t + 1)\) by \(2t^{2}\): \(2t^{2}(3t + 1)=6t^{3}+2t^{2}\)
Subtract this from the dividend: \((6t^{3}+5t^{2}-2t + 1)-(6t^{3}+2t^{2}) = 3t^{2}-2t + 1\)

Step2: Divide the new leading term

Divide the leading term of \(3t^{2}-2t + 1\) (\(3t^{2}\)) by \(3t\): \(\frac{3t^{2}}{3t}=t\)
Multiply the divisor \((3t + 1)\) by \(t\): \(t(3t + 1)=3t^{2}+t\)
Subtract this from \(3t^{2}-2t + 1\): \((3t^{2}-2t + 1)-(3t^{2}+t)=-3t + 1\)

Step3: Divide the new leading term

Divide the leading term of \(-3t + 1\) (\(-3t\)) by \(3t\): \(\frac{-3t}{3t}=-1\)
Multiply the divisor \((3t + 1)\) by \(-1\): \(-1(3t + 1)=-3t - 1\)
Subtract this from \(-3t + 1\): \((-3t + 1)-(-3t - 1)=2\)

Answer:

\(2t^{2}+t - 1+\frac{2}{3t + 1}\)