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# How do you express $\dfrac{1}{{{x}^{4}}-16}$in partial fractions?

Last updated date: 09th Aug 2024
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Hint: From the question given, we have to express the given $\dfrac{1}{{{x}^{4}}-16}$ in partial fractions. To solve the given, we have to find the factors of denominator and we have to use the basic formulae of algebra like $\Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. After using the basic formulae of algebra to the given question, we have to simplify further to get the final accurate and exact answer.

Complete step-by-step solution:
From the question, we have been given that,
$\Rightarrow \dfrac{1}{{{x}^{4}}-16}$
We have to find the factors of denominator
By using the difference of square identity,
$\Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
We can write ${{x}^{4}}$ as ${{\left( {{x}^{2}} \right)}^{2}}$ and we can $16$ as ${{4}^{2}}$
Now, we have to substitute the above conversions in the given question.
By substituting the above conversions in the given question, we get
$\Rightarrow \left( {{\left( {{x}^{2}} \right)}^{2}}-{{4}^{2}} \right)$
Now, by difference of square identity, we get,
$\Rightarrow \left( {{\left( {{x}^{2}} \right)}^{2}}-{{4}^{2}} \right)=\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)$
To make things easier we have to break down this into partial fractions in two stages,
$\Rightarrow \dfrac{1}{{{x}^{4}}-16}=\dfrac{1}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)}$
Now, we have to simplify further
$\Rightarrow \dfrac{1}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{1}{8\left( {{x}^{2}}-4 \right)}-\dfrac{1}{8\left( {{x}^{2}}+4 \right)}$
Now, further we have to find the factors of ${{x}^{2}}-4$
As we did earlier by using difference of square identity,
$\Rightarrow {{x}^{2}}-4=\left( x-2 \right)\left( x+2 \right)$
To make things easier we have to break down this into partial fractions
$\Rightarrow \dfrac{1}{\left( {{x}^{2}}-4 \right)}=\dfrac{1}{\left( x-2 \right)\left( x+2 \right)}$
Now, we have to simplify further
$\Rightarrow \dfrac{1}{\left( {{x}^{2}}-4 \right)}=\dfrac{1}{4\left( x-2 \right)}-\dfrac{1}{4\left( x+2 \right)}$
Now substitute the above conversion, we get
$\Rightarrow \dfrac{1}{\left( {{x}^{4}}-16 \right)}=\dfrac{1}{8}\left( \dfrac{1}{4\left( x-2 \right)}-\dfrac{1}{4\left( x+2 \right)} \right)-\dfrac{1}{8\left( {{x}^{2}}+4 \right)}$
Now, further simplifying we get,
$\Rightarrow \dfrac{1}{\left( {{x}^{4}}-16 \right)}=\dfrac{1}{32\left( x-2 \right)}-\dfrac{1}{32\left( x+2 \right)}-\dfrac{1}{8\left( {{x}^{2}}+4 \right)}$
Now, putting all together we get,
$\Rightarrow \dfrac{1}{\left( {{x}^{4}}-16 \right)}=\dfrac{1}{32\left( x-2 \right)}-\dfrac{1}{32\left( x+2 \right)}-\dfrac{1}{8\left( {{x}^{2}}+4 \right)}$
Therefore, the partial fraction of given $\dfrac{1}{{{x}^{4}}-16}$ is
$\Rightarrow \dfrac{1}{\left( {{x}^{4}}-16 \right)}=\dfrac{1}{32\left( x-2 \right)}-\dfrac{1}{32\left( x+2 \right)}-\dfrac{1}{8\left( {{x}^{2}}+4 \right)}$

Note: Students should be well aware of the basic formulae of algebra and also be well aware of the general identities of the algebra. Students should be very careful while doing the calculation part for the given question. Students should know what formula is to be used to solve the given question. Students should also know how they have to simplify the given question in an easier way.