Calculadora de divisão polinomial longa

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

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

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

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

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

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

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:

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

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

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