Polynomial Long Division Calculator

Your input: find $$$\frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15}$$$ using long division.

Write the problem in the special format:

$$$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x^{2}}+8 x+15}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}\phantom{2 x}&\phantom{-1}&\phantom{+22 x}&\phantom{-11}\end{array}&\\x^{2}+8 x+15&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}2 x^{3}&+15 x^{2}&+22 x&-11\end{array}}&\\\phantom{\color{Magenta}{x^{2}}+8 x+15}&\begin{array}{rrrr}\end{array}&\begin{array}{c}\end{array}\end{array}$$$

Step 1

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

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

Multiply it by the divisor: $$$2 x\left(x^{2}+8 x+15\right)=2 x^{3}+16 x^{2}+30 x$$$.

Subtract the dividend from the obtained result: $$$\left(2 x^{3}+15 x^{2}+22 x-11\right)-\left(2 x^{3}+16 x^{2}+30 x\right)=- x^{2}- 8 x-11$$$.

$$$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x^{2}}+8 x+15}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}\color{Chartreuse}{2 x}&\phantom{-1}&\phantom{+22 x}&\phantom{-11}\end{array}&\\\color{Magenta}{x^{2}}+8 x+15&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}\color{Chartreuse}{2 x^{3}}&+15 x^{2}&+22 x&-11\end{array}}&\frac{\color{Chartreuse}{2 x^{3}}}{\color{Magenta}{x^{2}}}=\color{Chartreuse}{2 x}\\\phantom{\color{Magenta}{x^{2}}+8 x+15}&\begin{array}{rrrr}-\phantom{2 x^{3}}&\phantom{+15 x^{2}}&\phantom{+22 x}&\phantom{-11}\\\phantom{\enclose{longdiv}{}}2 x^{3}&+16 x^{2}&+30 x\\\hline\phantom{\enclose{longdiv}{}}&- x^{2}&- 8 x&-11\end{array}&\begin{array}{c}\phantom{2 x^{3}+15 x^{2}+22 x-11}\\\color{Chartreuse}{2 x}\left(\color{Magenta}{x^{2}}+8 x+15\right)=2 x^{3}+16 x^{2}+30 x\\\phantom{- x^{2}- 8 x-11}\end{array}\end{array}$$$

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{- x^{2}}{x^{2}}=-1$$$.

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

Multiply it by the divisor: $$$-1\left(x^{2}+8 x+15\right)=- x^{2}- 8 x-15$$$.

Subtract the remainder from the obtained result: $$$\left(- x^{2}- 8 x-11\right)-\left(- x^{2}- 8 x-15\right)=4$$$.

$$$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x^{2}}+8 x+15}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}2 x&\color{Purple}{-1}&\phantom{+22 x}&\phantom{-11}\end{array}&\\\color{Magenta}{x^{2}}+8 x+15&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}2 x^{3}&+15 x^{2}&+22 x&-11\end{array}}&\\\phantom{\color{Magenta}{x^{2}}+8 x+15}&\begin{array}{rrrr}-\phantom{2 x^{3}}&\phantom{+15 x^{2}}&\phantom{+22 x}&\phantom{-11}\\\phantom{\enclose{longdiv}{}}2 x^{3}&+16 x^{2}&+30 x\\\hline\phantom{\enclose{longdiv}{}}&\color{Purple}{- x^{2}}&- 8 x&-11\\&-\phantom{- x^{2}}&\phantom{- 8 x}&\phantom{-11}\\\phantom{\enclose{longdiv}{}}&- x^{2}&- 8 x&-15\\\hline\phantom{\enclose{longdiv}{}}&&&\color{OrangeRed}{4}\end{array}&\begin{array}{c}\phantom{2 x^{3}+15 x^{2}+22 x-11}\\\phantom{\color{Chartreuse}{2 x}\left(\color{Magenta}{x^{2}}+8 x+15\right)=2 x^{3}+16 x^{2}+30 x}\\\frac{\color{Purple}{- x^{2}}}{\color{Magenta}{x^{2}}}=\color{Purple}{-1}\\\phantom{- x^{2}- 8 x-11}\\\color{Purple}{-1}\left(\color{Magenta}{x^{2}}+8 x+15\right)=- x^{2}- 8 x-15\\\phantom{4}\end{array}\end{array}$$$

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:

$$$\require{enclose}\begin{array}{rlc}\phantom{\color{Magenta}{x^{2}}+8 x+15}&\phantom{\enclose{longdiv}{}-}\begin{array}{rrrr}\color{Chartreuse}{2 x}&\color{Purple}{-1}&\phantom{+22 x}&\phantom{-11}\end{array}&Hints\\\color{Magenta}{x^{2}}+8 x+15&\phantom{-}\enclose{longdiv}{\begin{array}{cccc}\color{Chartreuse}{2 x^{3}}&+15 x^{2}&+22 x&-11\end{array}}&\frac{\color{Chartreuse}{2 x^{3}}}{\color{Magenta}{x^{2}}}=\color{Chartreuse}{2 x}\\\phantom{\color{Magenta}{x^{2}}+8 x+15}&\begin{array}{rrrr}-\phantom{2 x^{3}}&\phantom{+15 x^{2}}&\phantom{+22 x}&\phantom{-11}\\\phantom{\enclose{longdiv}{}}2 x^{3}&+16 x^{2}&+30 x\\\hline\phantom{\enclose{longdiv}{}}&\color{Purple}{- x^{2}}&- 8 x&-11\\&-\phantom{- x^{2}}&\phantom{- 8 x}&\phantom{-11}\\\phantom{\enclose{longdiv}{}}&- x^{2}&- 8 x&-15\\\hline\phantom{\enclose{longdiv}{}}&&&\color{OrangeRed}{4}\end{array}&\begin{array}{c}\phantom{2 x^{3}+15 x^{2}+22 x-11}\\\color{Chartreuse}{2 x}\left(\color{Magenta}{x^{2}}+8 x+15\right)=2 x^{3}+16 x^{2}+30 x\\\frac{\color{Purple}{- x^{2}}}{\color{Magenta}{x^{2}}}=\color{Purple}{-1}\\\phantom{- x^{2}- 8 x-11}\\\color{Purple}{-1}\left(\color{Magenta}{x^{2}}+8 x+15\right)=- x^{2}- 8 x-15\\\phantom{4}\end{array}\end{array}$$$

Therefore, $$$\frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15}=2 x - 1+\frac{4}{x^{2} + 8 x + 15}$$$

Answer: $$$\frac{2 x^{3} + 15 x^{2} + 22 x - 11}{x^{2} + 8 x + 15}=2 x - 1+\frac{4}{x^{2} + 8 x + 15}$$$