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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:
Factor the polynomials in the numerator and denominator of each fraction, if possible.
Set all restrictions (the denominators must not be equal to 0).
Multiply the fractions.
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}||
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)}||
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}||
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}||
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}||
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)}||
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}||
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}||
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\}||