# Calculadora de división larga de polinomios

## Realiza la división larga de polinomios paso a paso

La calculadora realizará la división larga de polinomios, mostrando los pasos.

Divide (dividend):

By (divisor):

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### Solution

Your input: 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):

$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x}-1}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}\phantom{x^{2}}&\phantom{+8 x}&\phantom{+8}&\phantom{+1}\end{array}&\\x-1&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}x^{3}&+7 x^{2}&+0 x&+1\end{array}}&\\\phantom{\color{Magenta}{x}-1}&\begin{array}{rrrr}\end{array}&\begin{array}{c}\end{array}\end{array}$

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: $\left(x^{3}+7 x^{2}+1\right)-\left(x^{3}- x^{2}\right)=8 x^{2}+1$.

$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x}-1}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}\color{Crimson}{x^{2}}&\phantom{+8 x}&\phantom{+8}&\phantom{+1}\end{array}&\\\color{Magenta}{x}-1&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}\color{Crimson}{x^{3}}&+7 x^{2}&+0 x&+1\end{array}}&\frac{\color{Crimson}{x^{3}}}{\color{Magenta}{x}}=\color{Crimson}{x^{2}}\\\phantom{\color{Magenta}{x}-1}&\begin{array}{rrrr}-\phantom{x^{3}}&\phantom{+7 x^{2}}&\phantom{+0 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}x^{3}&- x^{2}\\\hline\phantom{\enclose{longdiv}{}}&8 x^{2}&+0 x&+1\end{array}&\begin{array}{c}\phantom{x^{3}+7 x^{2}+0 x+1}\\\color{Crimson}{x^{2}}\left(\color{Magenta}{x}-1\right)=x^{3}- x^{2}\\\phantom{8 x^{2}+0 x+1}\end{array}\end{array}$

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: $\left(8 x^{2}+1\right)-\left(8 x^{2}- 8 x\right)=8 x+1$.

$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x}-1}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}x^{2}&\color{Purple}{+8 x}&\phantom{+8}&\phantom{+1}\end{array}&\\\color{Magenta}{x}-1&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}x^{3}&+7 x^{2}&+0 x&+1\end{array}}&\\\phantom{\color{Magenta}{x}-1}&\begin{array}{rrrr}-\phantom{x^{3}}&\phantom{+7 x^{2}}&\phantom{+0 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}x^{3}&- x^{2}\\\hline\phantom{\enclose{longdiv}{}}&\color{Purple}{8 x^{2}}&+0 x&+1\\&-\phantom{8 x^{2}}&\phantom{+0 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}&8 x^{2}&- 8 x\\\hline\phantom{\enclose{longdiv}{}}&&8 x&+1\end{array}&\begin{array}{c}\phantom{x^{3}+7 x^{2}+0 x+1}\\\phantom{\color{Crimson}{x^{2}}\left(\color{Magenta}{x}-1\right)=x^{3}- x^{2}}\\\frac{\color{Purple}{8 x^{2}}}{\color{Magenta}{x}}=\color{Purple}{8 x}\\\phantom{8 x^{2}+0 x+1}\\\color{Purple}{8 x}\left(\color{Magenta}{x}-1\right)=8 x^{2}- 8 x\\\phantom{8 x+1}\end{array}\end{array}$

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: $\left(8 x+1\right)-\left(8 x-8\right)=9$.

$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x}-1}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}x^{2}&+8 x&\color{Brown}{+8}&\phantom{+1}\end{array}&\\\color{Magenta}{x}-1&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}x^{3}&+7 x^{2}&+0 x&+1\end{array}}&\\\phantom{\color{Magenta}{x}-1}&\begin{array}{rrrr}-\phantom{x^{3}}&\phantom{+7 x^{2}}&\phantom{+0 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}x^{3}&- x^{2}\\\hline\phantom{\enclose{longdiv}{}}&8 x^{2}&+0 x&+1\\&-\phantom{8 x^{2}}&\phantom{+0 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}&8 x^{2}&- 8 x\\\hline\phantom{\enclose{longdiv}{}}&&\color{Brown}{8 x}&+1\\&&-\phantom{8 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}&&8 x&-8\\\hline\phantom{\enclose{longdiv}{}}&&&\color{Chocolate}{9}\end{array}&\begin{array}{c}\phantom{x^{3}+7 x^{2}+0 x+1}\\\phantom{\color{Crimson}{x^{2}}\left(\color{Magenta}{x}-1\right)=x^{3}- x^{2}}\\\phantom{8 x^{2}+0 x+1}\\\phantom{8 x^{2}+0 x+1}\\\phantom{\color{Purple}{8 x}\left(\color{Magenta}{x}-1\right)=8 x^{2}- 8 x}\\\frac{\color{Brown}{8 x}}{\color{Magenta}{x}}=\color{Brown}{8}\\\phantom{8 x+1}\\\color{Brown}{8}\left(\color{Magenta}{x}-1\right)=8 x-8\\\phantom{9}\end{array}\end{array}$

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

The resulting table is shown once more:

$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x}-1}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}\color{Crimson}{x^{2}}&\color{Purple}{+8 x}&\color{Brown}{+8}&\phantom{+1}\end{array}&Hints\\\color{Magenta}{x}-1&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}\color{Crimson}{x^{3}}&+7 x^{2}&+0 x&+1\end{array}}&\frac{\color{Crimson}{x^{3}}}{\color{Magenta}{x}}=\color{Crimson}{x^{2}}\\\phantom{\color{Magenta}{x}-1}&\begin{array}{rrrr}-\phantom{x^{3}}&\phantom{+7 x^{2}}&\phantom{+0 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}x^{3}&- x^{2}\\\hline\phantom{\enclose{longdiv}{}}&\color{Purple}{8 x^{2}}&+0 x&+1\\&-\phantom{8 x^{2}}&\phantom{+0 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}&8 x^{2}&- 8 x\\\hline\phantom{\enclose{longdiv}{}}&&\color{Brown}{8 x}&+1\\&&-\phantom{8 x}&\phantom{+1}\\\phantom{\enclose{longdiv}{}}&&8 x&-8\\\hline\phantom{\enclose{longdiv}{}}&&&\color{Chocolate}{9}\end{array}&\begin{array}{c}\phantom{x^{3}+7 x^{2}+0 x+1}\\\color{Crimson}{x^{2}}\left(\color{Magenta}{x}-1\right)=x^{3}- x^{2}\\\frac{\color{Purple}{8 x^{2}}}{\color{Magenta}{x}}=\color{Purple}{8 x}\\\phantom{8 x^{2}+0 x+1}\\\color{Purple}{8 x}\left(\color{Magenta}{x}-1\right)=8 x^{2}- 8 x\\\frac{\color{Brown}{8 x}}{\color{Magenta}{x}}=\color{Brown}{8}\\\phantom{8 x+1}\\\color{Brown}{8}\left(\color{Magenta}{x}-1\right)=8 x-8\\\phantom{9}\end{array}\end{array}$

Therefore, $\frac{x^{3} + 7 x^{2} + 1}{x - 1}=x^{2} + 8 x + 8+\frac{9}{x - 1}$

Answer: $\frac{x^{3} + 7 x^{2} + 1}{x - 1}=x^{2} + 8 x + 8+\frac{9}{x - 1}$