In order to multiply rational fractions, or rational expressions, it is important to know how to multiply fractions and how to use various factoring techniques. To do so, follow these steps:
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Factor the polynomials in the numerator and denominator of each fraction, if possible.
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Set all restrictions (the denominators must not be equal to 0).
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Multiply the fractions.
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Simplify the common factors of the resulting fraction, if possible.
Find the answer to the following multiplication:||\dfrac{4-x^2}{x-2}\times \dfrac{-x}{2x+4}||
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Factor the polynomials in the numerator and denominator of each fraction.
The numerator of the first fraction can be factored using a difference of squares and the denominator of the second fraction can be factored by removing a greatest common factor.
||\begin{align}4-x^2&=-(x^2-4)\\&=-(x-2)(x+2)\end{align}||
||2x+4 = 2(x+2)||
Now we can multiply the 2 fractions:||\dfrac{-(x-2)(x+2)}{(x-2)}\times \dfrac{-x}{2(x+2)}||
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Set all restrictions.
Each factor in the denominator must not be equal to 0.
||\dfrac{-(x-2)(x+2)}{\color{#3b87cd}{(x-2)}}\times \dfrac{-x}{2\color{#fa7921}{(x+2)}}||
||\begin{align}\color{#3b87cd}{x-2}&\neq 0\\x&\neq2 \end{align}||
||\begin{align}\color{#3a9a38}{x+2}&\neq 0\\x&\neq-2 \end{align}||
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Multiply.
||\begin{align}&\dfrac{-(x-2)(x+2)}{(x-2)}\times \dfrac{-x}{2(x+2)}\\\\=&\ \dfrac{-(-x)(x-2)(x+2)}{2(x-2)(x+2)}\end{align}||
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Simplify the common factors in the resulting fraction.
||\begin{align}&\dfrac{-(-x)\cancel{(x-2)}\cancel{(x+2)}}{2\cancel{(x-2)}\cancel{(x+2)}}\\=&\ \dfrac{-(-x)}{2}\\=&\ \dfrac{x}{2}\end{align}||
Answer: The answer of the multiplication of |\dfrac{4-x^2}{x-2}\times \dfrac{-x}{2x+4}| is |\dfrac{x}{2}| where |x\neq2| and |x\neq-2.|
Find the answer to the following multiplication:||\dfrac{x^2+3x+2}{2x^2+13x+20} \times \dfrac{x^2+7x+12}{2x^2+7x+6}||
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Factor the polynomials in the numerator and denominator of each fraction.
All 4 polynomials can be factored using the product-sum method.
||\begin{gather}x^2+3x+2\\\\
\begin{aligned}\text{Product}&=1\times 2\\&=2\\ &=\color{#3b87cd}{1}\times \color{#3b87cd}{2}\end{aligned}\quad\begin{aligned}\text{Sum} &=3\\ &=\color{#3b87cd}{1}+\color{#3b87cd}{2}\\ \phantom{=} \end{aligned}\\\\
x^2+3x+2=(x+\color{#3b87cd}{1})(x+\color{#3b87cd}{2})\end{gather}||
||\begin{gather}x^2+7x+12\\\\
\begin{aligned}\text{Product}&=1\times12\\&=12\\ &=\color{#3b87cd}{3}\times \color{#3b87cd}{4}\end{aligned}\quad\begin{aligned}\text{Sum} &=7\\ &=\color{#3b87cd}{3}+\color{#3b87cd}{4}\\ \phantom{=} \end{aligned}\\\\
x^2+7x+12=(x+\color{#3b87cd}{3})(x+\color{#3b87cd}{4})\end{gather}||
||\begin{gather}2x^2+13x+20\\\\
\begin{aligned}\text{Product}&=2\times 20\\&=40\\ &=\color{#3b87cd}{5}\times \color{#3b87cd}{8}\end{aligned}\qquad\begin{aligned}\text{Sum} &=13\\ &=\color{#3b87cd}{5}+\color{#3b87cd}{8}\\ \phantom{=} \end{aligned}\\\\
\begin{aligned}2x^2+13x+20&=2x^2+\color{#3b87cd}{5}x+\color{#3b87cd}{8}x+20\\&=x(2x+5)+4(2x+5)\\&=(x+4)(2x+5)\end{aligned}\end{gather}||
||\begin{gather}2x^2+7x+6\\\\
\begin{aligned}\text{Product}&=2\times 6\\&=12\\ &=\color{#3b87cd}{3}\times \color{#3b87cd}{4}\end{aligned}\qquad\begin{aligned}\text{Sum} &=7\\ &=\color{#3b87cd}{3}+\color{#3b87cd}{4}\\ \phantom{=} \end{aligned}\\\\
\begin{aligned}2x^2+7x+6&=2x^2+\color{#3b87cd}{3}x+\color{#3b87cd}{4}x+6\\&=x(2x+3)+2(2x+3)\\&=(x+2)(2x+3)\end{aligned}\end{gather}||
Now we can multiply the 2 fractions:||\dfrac{(x+1)(x+2)}{(2x+5)(x+4)} \times \dfrac{(x+3)(x+4)}{(2x+3)(x+2)}||
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Set all restrictions.
Any factor in the denominator must not be equal to 0.
||\dfrac{(x+1)(x+2)}{\color{#3b87cd}{(2x+5)}\color{#3a9a38}{(x+4)}} \times \dfrac{(x+3)(x+4)}{\color{#fa7921}{(2x+3)}\color{#ec0000}{(x+2)}}||
||\begin{align}\color{#3b87cd}{2x+5} &\neq 0\\ x&\neq -\dfrac{5}{2}\end{align}||
||\begin{align}\color{#3a9a38}{x+4} &\neq\ 0\\ x &\neq -4\end{align}||
||\begin{align}\color{#fa7921}{2x+3} &\neq 0\\ x &\neq -\dfrac{3}{2}\end{align}||
||\begin{align}\color{#ec0000}{x+2} &\neq 0\\ x &\neq -2\end{align}||
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Multiply.
||\begin{align}&\dfrac{(x+1)(x+2)}{(2x+5)(x+4)}\times \dfrac{(x+3)(x+4)}{(2x+3)(x+2)}\\\\=&\ \dfrac{(x+1)(x+2)(x+3)(x+4)}{(2x+5)(x+4)(2x+3)(x+2)}\end{align}||
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Simplify the common factors of the resulting fraction.
||\begin{align}&\dfrac{(x+1)\cancel{(x+2)}(x+3)\cancel{(x+4)}}{(2x+5)\cancel{(x+4)}(2x+3)\cancel{(x+2)}}\\\\=\ &\dfrac{(x+1)(x+3)}{(2x+5)(2x+3)}\end{align}||
Answer: The answer of the multiplication of |\dfrac{x^2+3x+2}{2x^2+13x+20} \times \dfrac{x^2+7x+12}{2x^2+7x+6}| is |\dfrac{(x+1)(x+3)}{(2x+5)(2x+3)}| where |x\neq -\dfrac{5}{2},| |x\neq -4,| |x\neq -\dfrac{3}{2}| and |x\neq -2.|
The answer to a multiplication question is sometimes written in the form of a fraction where the numerator and denominator are polynomials. In this case, the factors should be multiplied, if possible. So the answer from the previous example would be as follows.||\begin{align}\dfrac{(x+1)(x+3)}{(2x+5)(2x+3)}&=\dfrac{x^2+3x+x+3}{4x^2+6x+10x+15}\\&=\dfrac{x^2+4x+3}{4x^2+16x+15}\end{align}||
When there are many restrictions, it is possible to use the notation |\notin.| The restrictions on the previous example could be written as follows.||x\notin\left\{-4, -\dfrac{5}{2},-2,-\dfrac{3}{2}\right\}||