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m1532
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dividing-an-algebraic-expression-by-a-binomial
Grades
Secondary IV
Topic
Mathematics
Tags
diviseur
dividende
division
polynômes
termes
mathématique
division par un polynôme
division par un binôme
division d'expressions algébriques
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Corps

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.

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To do so, refer to the following procedure.

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  1. Write the terms of the dividend and divisor in descending order according to the degree of their terms.

  2. Divide the first terms of the dividend and the divisor together.

  3. Place the result of the division above the dividend.

  4. Multiply it by all the divisor’s terms.

  5. Subtract the new algebraic expression obtained from the dividend.

  6. Lower the remaining terms of the dividend to the same level as the result of the subtraction.

  7. Repeat steps 2 to 6 until the dividend’s degree is smaller than the divisor degree.

  8. If necessary, write the remainder appropriately.

Content
Title (level 3)
Dividing a Polynomial by a Binomial Without a Remainder
Title slug (identifier)
dividing-polynomial-without-remainder
Corps

Consider the following polynomials: |(2x^2 + 2x^3y+ 4x^2y^2 + 4xy)| and |(x + 2y).|

  1. 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}||

  1. Divide the first terms of the dividend and the divisor ||\dfrac{2x^3y}{x}=2x^2y||

  2. 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}||

  1. Multiply the result by each of the divisor’s terms ||2x^2y(x+2y)=2x^3y+4x^2y^2||

  2. 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}||

  1. 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}||

  1. 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.|

Corps

As with integers, the result of dividing may give a remainder.

Content
Title (level 3)
Dividing a Polynomial by a Binomial With a Remainder
Title slug (identifier)
dividing-polynomial-with-remainder
Corps

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}|

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Corps

Dividing two polynomials can also be completed by factoring the polynomials to eliminate any common factors.

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