Question

using long division

1. $(p^{3}-10p^{2}+28p - 24)div(p - 6)$

Explanation:

Step1: Divide the first - term

Divide the first - term of the dividend $p^{3}-10p^{2}+28p - 24$ (which is $p^{3}$) by the first - term of the divisor $p - 6$ (which is $p$). So, $\frac{p^{3}}{p}=p^{2}$.

Step2: Multiply and subtract

Multiply $p - 6$ by $p^{2}$ to get $p^{3}-6p^{2}$, and subtract it from the dividend: $(p^{3}-10p^{2}+28p - 24)-(p^{3}-6p^{2})=-4p^{2}+28p - 24$.

Step3: Divide the new first - term

Divide the first - term of the new dividend $-4p^{2}+28p - 24$ (which is $-4p^{2}$) by the first - term of the divisor $p$ (which is $p$). So, $\frac{-4p^{2}}{p}=-4p$.

Step4: Multiply and subtract again

Multiply $p - 6$ by $-4p$ to get $-4p^{2}+24p$, and subtract it from the new dividend: $(-4p^{2}+28p - 24)-(-4p^{2}+24p)=4p - 24$.

Step5: Divide the new first - term

Divide the first - term of the new dividend $4p - 24$ (which is $4p$) by the first - term of the divisor $p$ (which is $p$). So, $\frac{4p}{p}=4$.

Step6: Multiply and subtract one last time

Multiply $p - 6$ by $4$ to get $4p - 24$, and subtract it from the new dividend: $(4p - 24)-(4p - 24)=0$.

Answer:

$p^{2}-4p + 4$