In some cases, a satisfactory result is obtained by factoring the numerator and the denominator and simplifying if necessary. When factoring is too difficult, we often divide using long division.
To do so, refer to the following procedure.
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Write the terms of the dividend and divisor in descending order according to the degree of their terms.
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Divide the first terms of the dividend and the divisor together.
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Place the result of the division above the dividend.
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Multiply it by all the divisor’s terms.
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Subtract the new algebraic expression obtained from the dividend.
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Lower the remaining terms of the dividend to the same level as the result of the subtraction.
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Repeat steps 2 to 6 until the dividend’s degree is smaller than the divisor degree.
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If necessary, write the remainder appropriately.
Consider the following polynomials: |(2x^2 + 2x^3y+ 4x^2y^2 + 4xy)| and |(x + 2y).|
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Order the polynomials
By ordering the polynomials, the following division is obtained.
||\begin{align} \begin{aligned}\ \\ \\ \color{#ec0000}x+2\color{#3a9a38}y \\ \ \end{aligned}\ \begin{aligned}\\ \\ \Big\vert\!\!\! \\ \ \end{aligned} \begin{aligned}\ &\phantom{3x+1}\\ \hline &2\color{#ec0000}{x^3}\color{#3a9a38}y+4\color{#ec0000}{x^2}\color{#3a9a38}{y^2}+2\color{#ec0000}{x^2}+4\color{#ec0000}{x}\color{#3a9a38}y\end{aligned} \end{align}||
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Divide the first terms of the dividend and the divisor ||\dfrac{2x^3y}{x}=2x^2y||
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Write the result above the dividend
||\begin{align} \begin{aligned}\ \\ \\ \color{#ec0000}x+2\color{#3a9a38}y \\ \ \end{aligned}\ \begin{aligned}\\ \\ \Big\vert\!\!\! \\ \ \end{aligned} \begin{aligned}\ &2x^2y\\ \hline &2\color{#ec0000}{x^3}\color{#3a9a38}y+4\color{#ec0000}{x^2}\color{#3a9a38}{y^2}+2\color{#ec0000}{x^2}+4\color{#ec0000}{x}\color{#3a9a38}y\end{aligned} \end{align}||
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Multiply the result by each of the divisor’s terms ||2x^2y(x+2y)=2x^3y+4x^2y^2||
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Subtract the new algebraic expression obtained from the dividend
||\begin{align} \begin{aligned}\ \\ \\ \color{#ec0000}x+2\color{#3a9a38}y \\ \\ \ \end{aligned}\ \begin{aligned}\\ \Big\vert\!\!\! \\ \ \end{aligned} \begin{aligned}\ &2x^2y\\ \hline &2\color{#ec0000}{x^3}\color{#3a9a38}y+4\color{#ec0000}{x^2}\color{#3a9a38}{y^2}+2\color{#ec0000}{x^2}+4\color{#ec0000}{x}\color{#3a9a38}y\\ -(&2x^3y+4x^2y^2)\end{aligned} \end{align}||
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Lower the remaining terms of the dividend to the same level as the result of the subtraction
||\begin{align} \begin{aligned}\ \\ \\[-4px] \color{#ec0000}x+2\color{#3a9a38}y \\ \\ \\ \ \end{aligned}\ \begin{aligned}\\[4px] \Big\vert\!\!\!\\ \\ \\ \ \end{aligned} \begin{aligned}\ &2x^2y\\ \hline &\cancel{2\color{#ec0000}{x^3}\color{#3a9a38}y}+\cancel{4\color{#ec0000}{x^2}\color{#3a9a38}{y^2}}+2\color{#ec0000}{x^2}+4\color{#ec0000}{x}\color{#3a9a38}y\\ -(&\cancel{2x^3y}+\cancel{4x^2y^2})\ \ \ \ \downarrow\qquad\; \downarrow \\ \hline&\phantom{2x^3y+4x^2y^2+2)}2\color{#ec0000}{x^2}+4\color{#ec0000}x\color{#3a9a38}y\end{aligned} \end{align}||
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Repeat steps 2 to 6
||\begin{align} \begin{aligned}\ \\ \\[-4px] \color{#ec0000}x+2\color{#3a9a38}y \\ \\ \\ \\ \ \end{aligned}\ \begin{aligned}\\[3px] \Big\vert\!\!\! \\ \\ \\ \\ \ \end{aligned} \begin{aligned}\ &2x^2y\boldsymbol{+2x}\\ \hline &\cancel{2\color{#ec0000}{x^3}\color{#3a9a38}y}+\cancel{4\color{#ec0000}{x^2}\color{#3a9a38}{y^2}}+2\color{#ec0000}{x^2}+4\color{#ec0000}{x}\color{#3a9a38}y\\ -(&\cancel{2x^3y}+\cancel{4x^2y^2})\ \ \ \ \downarrow\qquad\; \downarrow \\ \hline&\phantom{2x^3y+4x^2y^2+2)}2\color{#ec0000}{x^2}+4\color{#ec0000}x\color{#3a9a38}y\\ &\phantom{2x^3y+4x^2y^2+2)}2x^2+4xy\end{aligned} \end{align}||
||\begin{align} \begin{aligned}\ \\[4px] \color{#ec0000}x+2\color{#3a9a38}y \\ \\ \\ \\ \\[4px] \ \end{aligned}\ \begin{aligned}\\[8px] \Big\vert\!\!\! \\ \\ \\ \\ \\\\ \ \end{aligned} \begin{aligned}\ &2x^2y+2x\\ \hline &\cancel{2\color{#ec0000}{x^3}\color{#3a9a38}y}+\cancel{4\color{#ec0000}{x^2}\color{#3a9a38}{y^2}}+2\color{#ec0000}{x^2}+4\color{#ec0000}{x}\color{#3a9a38}y\\ -(&\cancel{2x^3y}+\cancel{4x^2y^2})\ \ \ \ \downarrow\qquad\; \downarrow \\ \hline&\phantom{2x^3y+4x^2y^2+2}\cancel{2\color{#ec0000}{x^2}}+\cancel{4\color{#ec0000}x\color{#3a9a38}y}\\ &\phantom{2x^3y+4x^2y^2}\,-(\cancel{2x^2}+\cancel{4xy})\\ \hline &\phantom{2x^3y+4x^2y^2+2x^2}\ \quad\, 0 \end{aligned} \end{align}||
The final answer is equal to the quotient found: |2x^2y+2x.|
As with integers, the result of dividing may give a remainder.
Consider the following polynomials: |(3x^2 + 7x + 1)| and |(x + 2).| Here are the steps for performing this division:
||\begin{align} \begin{aligned}\ \\ \\ x+2 \\ \ \end{aligned}\ \begin{aligned}\\ \\ \Big\vert\!\!\! \\ \ \end{aligned} \begin{aligned}\ &\phantom{3x+1}\\ \hline &3x^2+7x+1\end{aligned} \end{align}|| ||\begin{align} \begin{aligned}\ \\ \\ x+2\\ \\ \ \end{aligned}\ \begin{aligned}\\[12px] \Big\vert\!\!\! \\ \\ \ \end{aligned} \begin{aligned}\ &3x\phantom{+1}\quad\\ \hline &3x^2+7x+1\\ &3x^2+6x\end{aligned} \end{align}|| ||\begin{align} \begin{aligned}\ \\ \\ x+2\\ \\ \\ \ \end{aligned}\ \begin{aligned}\\[20px] \Big\vert\!\!\! \\ \\ \\ \\ \ \end{aligned} \begin{aligned}\ &3x\phantom{+1}\quad\\ \hline &\cancel{3x^2}+7x+1\\ -(&\cancel{3x^2}+6x)\ \ \downarrow \\ \hline &\phantom{3x^2+7x}x+1\end{aligned} \end{align}|| ||\begin{align} \begin{aligned}\ \\ \\ x+2\\ \\ \\ \\[4px] \ \end{aligned}\ \begin{aligned}\\[6px] \Big\vert\!\!\! \\ \\ \\ \\ \ \end{aligned} \begin{aligned}\ &3x+1\quad\\ \hline &\cancel{3x^2}+7x+1\\ -(&\cancel{3x^2}+6x)\ \\ \hline &\phantom{3x^2+7x}x+1 \\ &\phantom{3x^2+7x}x+2 \end{aligned} \end{align}|| ||\begin{align} \begin{aligned}\ \\ \\ x+2\\ \\ \\ \\ \\[4px] \ \end{aligned}\ \begin{aligned}\\[4px] \Big\vert\!\!\! \\ \\ \\ \\ \\ \ \end{aligned} \begin{aligned}\ &3x+1\quad\\ \hline &\cancel{3x^2}+7x+1\\ -(&\cancel{3x^2}+6x)\ \\ \hline &\phantom{3x^2+7}\cancel{x}+1 \\ &\,\phantom{3x^2}-(\cancel{x}+2) \\ \hline &\phantom{3x^2+62x} -1 \end{aligned} \end{align}||
In the example above, |-1| is the remainder and it is no longer possible to divide |-1| by |x.| This is why we stopped performing the algebraic division.
The answer can be written in two ways:
|3x + 1| remainder |-1|
Or
|3x + 1 + \dfrac{-1}{x + 2} = 3x + 1 - \dfrac{1}{x + 2}|
Dividing two polynomials can also be completed by factoring the polynomials to eliminate any common factors.