In order to divide rational expressions, or rational fractions, it is important to know how to divide fractions and how to use various factoring techniques. Use the following approach:
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Factor the polynomials in the numerator and denominator of each fraction when possible.
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Set all restrictions (the denominators and numerator of the divisor must not be equal to 0).
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Divide by transforming it into multiplication.
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Simplify the common factors of the resulting fraction when possible.
Why must we look for restrictions in the numerator of the divisor?
In a rational expression, restrictions must be set on all polynomials found in the denominator.||\dfrac{a}{\color{#ec0000}b}\div\dfrac{c}{\color{#ec0000}d}\ \Rightarrow\ \color{#ec0000}b\ne 0\ \text{and}\ \color{#ec0000}d\ne 0||When we convert a division to a multiplication, we multiply by the reciprocal of the divisor, so the numerator becomes the denominator. This is why we need to exclude the values for which this numerator is zero.||\dfrac{a}{\color{#ec0000}b}\times\dfrac{\color{#ec0000}d}{\color{#fa7921}c}\ \Rightarrow\ \color{#ec0000}b\ne 0,\ \color{#ec0000}d\ne 0\ \underline{\text{and}\ \color{#fa7921}c\ne0}||
Find the answer of the following division:||\dfrac{c^3-9c}{c^3} \div \dfrac{c+3}{c}||
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Factor the polynomials in the numerator and denominator of each fraction.
The numerator of the 1st fraction can be factored by removing a greatest common factor, then using a difference of squares.||\begin{align}c^3-9c&=c\,(c^2-9)\\&=c\,(c-3)(c+3)\end{align}||
Now the 2 fractions can be divided.||\dfrac{c\, (c-3) (c+3)}{c^3} \div \dfrac{c+3}{c}|| -
Set all restrictions.
The 2 denominators and the numerator of the 2nd fraction cannot be equal to |0.|||\dfrac{c\, (c-3) (c+3)}{\color{#3b87cd}{c^3}} \div \dfrac{\color{#ec0000}{c+3}}{\color{#3a9a38}c}||
||\begin{align}\color{#3b87cd}{c^3}&\neq0\\c&\neq0\end{align}||
||\begin{align}\color{#ec0000}{c+3}&\neq0\\c&\neq-3\end{align}||
||\color{#3a9a38}c\neq0||
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Divide.
||\begin{align}&\dfrac{c\, (c-3) (c+3)}{c^3} \div \dfrac{c+3}{c}\\\\=&\ \dfrac{c\, (c-3) (c+3)}{c^3} \times \dfrac{c}{c+3}\\\\=&\ \dfrac{c^2\, (c-3) (c+3)}{c^3(c+3)}\end{align}|| -
Simplify the common factors of the resulting fraction.
||\begin{align}&\dfrac{\cancel{c^2}(c-3)\cancel{(c+3)}}{\cancel{c^2}(c)\cancel{(c+3)}}\\\\=&\ \dfrac{c-3}{c}\end{align}||
Answer: The answer of the division |\dfrac{c^3-9c}{c^3} \div \dfrac{c+3}{c}| is |\dfrac{c-3}{c},| where |c\neq0| and |x\neq-3.|
Find the answer of the following division.||\dfrac{x^2+8x+16}{2x^3+8x^2-3x-12} \div \dfrac{x+4}{2}||
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Factor the polynomials in the numerator and denominator of each fraction.
The numerator of the first fraction can be factored, since it is a perfect square trinomial. The denominator of the first fraction can be factored by grouping.
||\begin{align}&x^2\boldsymbol{\color{#ec0000}{+}}8x+16\\
=\ &(\color{#3a9a38}{x})^2\boldsymbol{\color{#ec0000}{+}}2(\color{#3a9a38}{x})(\color{#3b87cd}{4})+(\color{#3b87cd}{4})^2\\
=\ &(\color{#3a9a38}{x}\boldsymbol{\color{#ec0000}{+}}\color{#3b87cd}{4})(\color{#3a9a38}{x}\boldsymbol{\color{#ec0000}{+}}\color{#3b87cd}{4})\end{align}||
||\begin{align}&\ 2x^3+8x^2-3x-12\\=&\ \color{#3a9a38}{2x^2} (\color{#3b87cd}{x+4}) \color{#3a9a38}{-3} (\color{#3b87cd}{x+4}) \\ =&\ (\color{#3b87cd}{x+4}) (\color{#3a9a38}{2x^2-3}) \end{align}||
Now the 2 fractions can be divided.||\dfrac{(x+4)(x+4)}{(x+4)(2x^2-3)} \div \dfrac{x+4}{2}||
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Set all restrictions.
The 2 denominators and the numerator of the 2nd fraction cannot be equal to |0.|||\dfrac{(x+4)(x+4)}{\color{#3b87cd}{(x+4)}\color{#ec0000}{(2x^2-3)}} \div \dfrac{\color{#3b87cd}{x+4}}{2}||
||\begin{align}\color{#3b87cd}{x+4}&\neq0\\x&\neq-4\end{align}||
||\begin{align}\color{#ec0000}{2x^2-3}&\neq0\\2x^2&\neq3\\x^2&\neq\dfrac{3}{2}\\x&\neq\pm\sqrt{\dfrac{3}{2}}\end{align}||
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Divide.
||\begin{align}&\dfrac{(x+4)(x+4)}{(x+4)(2x^2-3)} \div \dfrac{x+4}{2}\\\\=&\ \dfrac{(x+4)(x+4)}{(x+4)(2x^2-3)} \times \dfrac{2}{x+4}\\\\=&\ \dfrac{2(x+4)(x+4)}{(x+4)(2x^2-3)(x+4)} \end{align}|| -
Simplify the common factors of the resulting fraction.
||\begin{align}&\dfrac{2\cancel{(x+4)}\cancel{(x+4)}}{\cancel{(x+4)}(2x^2-3)\cancel{(x+4)}}\\\\=&\ \dfrac{2}{2x^2-3}\end{align}||
Answer: The answer of the division |\dfrac{x^2+8x+16}{2x^3+8x^2-3x-12} \div \dfrac{x+4}{2}| is |\dfrac{2}{2x^2-3},| where |x\neq-4| and |x\neq\pm\sqrt{\dfrac{3}{2}}.|