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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:
Factor the polynomials in the numerator and denominator of each fraction when possible.
Set all restrictions (the denominators and numerator of the divisor must not be equal to 0).
Divide by transforming it into multiplication.
Simplify the common factors of the resulting fraction when possible.
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}||
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||
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}||
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}||
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}||
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}}.|