Divide $x^{3} + 7 x^{2} + 1$ by $x - 1$

Find $\frac{x^{3} + 7 x^{2} + 1}{x - 1}$ using long division.

Write the problem in the special format (missed terms are written with zero coefficients):

$\begin{array}{r|r}\hline\\x-1&x^{3}+7 x^{2}+0 x+1\end{array}$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $\frac{x^{3}}{x} = x^{2}$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $x^{2} \left(x-1\right) = x^{3}- x^{2}$.

Subtract the dividend from the obtained result: $\left(x^{3}+7 x^{2}+1\right) - \left(x^{3}- x^{2}\right) = 8 x^{2}+1$.

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $\frac{8 x^{2}}{x} = 8 x$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $8 x \left(x-1\right) = 8 x^{2}- 8 x$.

Subtract the remainder from the obtained result: $\left(8 x^{2}+1\right) - \left(8 x^{2}- 8 x\right) = 8 x+1$.

Step 3

Divide the leading term of the obtained remainder by the leading term of the divisor: $\frac{8 x}{x} = 8$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $8 \left(x-1\right) = 8 x-8$.

Subtract the remainder from the obtained result: $\left(8 x+1\right) - \left(8 x-8\right) = 9$.

Since the degree of the remainder is less than the degree of the divisor, we are done.

The resulting table is shown once more:

Therefore, $\frac{x^{3} + 7 x^{2} + 1}{x - 1} = \left(x^{2} + 8 x + 8\right) + \frac{9}{x - 1}$.

$\frac{x^{3} + 7 x^{2} + 1}{x - 1} = \left(x^{2} + 8 x + 8\right) + \frac{9}{x - 1}$A