Question

How do you write $\dfrac{{{x}^{4}}}{{{\left( x-1 \right)}^{3}}}$ as a partial fraction decomposition?

Answer 1

In this problem, we have to write the given expression as a partial fraction decomposition. We can first divide the given equation, and find the partial fraction. We can see that the degree of the numerator is greater than the denominator, so we can use the long division method to find the quotient and the remainder and we can find the partial decomposition.

Complete step-by-step answer:
We know that the given fraction is,
$\dfrac{{{x}^{4}}}{{{\left( x-1 \right)}^{3}}}$
We can see that the degree of the numerator is greater than the denominator, so we can use the long division method to find the quotient and the remainder.
We can now expand the denominator, we get
${{\left( x-1 \right)}^{3}}={{x}^{3}}-3{{x}^{2}}+3x-1$
We can see that the denominator expansion is the divisor.
Now we can set up the polynomials to be divided in long division, if there is any missing terms for every exponent, add one with a value of 0, we get
${{x}^{3}}-3{{x}^{2}}+3x-1\overset{{}}{\overline{\left){{{x}^{4}}+0{{x}^{3}}+0{{x}^{2}}+0x+0}\right.}}$
Now we can divide the highest order term in the dividend ${{x}^{4}}$ by the highest order term in the divisor x, we can then multiply the quotient to the divisor, we get

& {{x}^{4}}+0{{x}^{3}}+0{{x}^{2}}+0x+0 \\\ & {{x}^{4}}-3{{x}^{3}}+3{{x}^{2}}-1x \\\ \end{aligned}}\right.}}$$ We know that the expression is to be subtracted in the dividend, so we can change the sign in $${{x}^{4}}-3{{x}^{3}}+3{{x}^{2}}-1x$$, we get $${{x}^{3}}-3{{x}^{2}}+3x-1\overset{x}{\overline{\left){\begin{aligned} & {{x}^{4}}+0{{x}^{3}}+0{{x}^{2}}+0x+0 \\\ & \underline{-{{x}^{4}}+3{{x}^{3}}-3{{x}^{2}}+1x} \\\ & \text{ }3{{x}^{3}}-3{{x}^{2}}+1x \\\ \end{aligned}}\right.}}$$ Now we can bring down the next term from the dividend to the current dividend, $${{x}^{3}}-3{{x}^{2}}+3x-1\overset{x}{\overline{\left){\begin{aligned} & {{x}^{4}}+0{{x}^{3}}+0{{x}^{2}}+0x+0 \\\ & \underline{-{{x}^{4}}+3{{x}^{3}}-3{{x}^{2}}+1x} \\\ & \text{ }3{{x}^{3}}-3{{x}^{2}}+1x+0 \\\ \end{aligned}}\right.}}$$ Now we should divide the highest order term in the dividend $${{x}^{3}}$$ by the divisor y and we can simplify it, we get $${{x}^{3}}-3{{x}^{2}}+3x-1\overset{x+3}{\overline{\left){\begin{aligned} & {{x}^{4}}+0{{x}^{3}}+0{{x}^{2}}+0x+0 \\\ & \underline{-{{x}^{4}}+3{{x}^{3}}-3{{x}^{2}}+1x} \\\ & \text{ }3{{x}^{3}}-3{{x}^{2}}+1x+0 \\\ & \text{ }-\underline{3{{x}^{3}}+9{{x}^{2}}-9x+3} \\\ & \text{ }6{{x}^{2}}-8x+3 \\\ \end{aligned}}\right.}}$$ We can now write it as, Quotient + Remainder / Divisor. $$\dfrac{{{x}^{4}}}{{{\left( x-1 \right)}^{3}}}=x+3+\dfrac{6{{x}^{2}}-8x+3}{{{\left( x-1 \right)}^{3}}}$$…… (1) We can now form the partial fraction decomposition as the denominator has degree 3, which can be written as, $$\dfrac{6{{x}^{2}}-8x+3}{{{\left( x-1 \right)}^{3}}}=\dfrac{A}{x-1}+\dfrac{B}{{{\left( x-1 \right)}^{2}}}+\dfrac{C}{{{\left( x-1 \right)}^{3}}}$$……. (2) We can now cross multiply and simplify the step, we get $$\dfrac{6{{x}^{2}}-8x+3}{{{\left( x-1 \right)}^{3}}}=\dfrac{A{{\left( x-1 \right)}^{2}}+B\left( x-1 \right)+C}{\left( x-1 \right){{\left( x-1 \right)}^{2}}{{\left( x-1 \right)}^{3}}}$$ We can now take the numerator which equates the partial fractions, we get $$\Rightarrow 6{{x}^{2}}-8x+3=A{{\left( x-1 \right)}^{2}}+B\left( x-1 \right)+C$$ We can now see that, if x = 1, the C = 1. We can now compare the coefficient of $${{x}^{2}}$$, we get $$\Rightarrow A=6$$ We can now take x = 0, then $$\Rightarrow A-B+C=3$$ We can now substitute the value of A, B, we get $$\begin{aligned} & \Rightarrow 6-B+1=3 \\\ & \Rightarrow B=4 \\\ \end{aligned}$$ We can now substitute the values in (2), we get $$\dfrac{6{{x}^{2}}-8x+3}{{{\left( x-1 \right)}^{3}}}=\dfrac{6}{x-1}+\dfrac{4}{{{\left( x-1 \right)}^{2}}}+\dfrac{1}{{{\left( x-1 \right)}^{3}}}$$ We can now write equation (1) as, $$\dfrac{{{x}^{4}}}{{{\left( x-1 \right)}^{3}}}=x+3+\dfrac{6}{x-1}+\dfrac{4}{{{\left( x-1 \right)}^{2}}}+\dfrac{1}{{{\left( x-1 \right)}^{3}}}$$. Therefore, the answer is $$\dfrac{{{x}^{4}}}{{{\left( x-1 \right)}^{3}}}=x+3+\dfrac{6}{x-1}+\dfrac{4}{{{\left( x-1 \right)}^{2}}}+\dfrac{1}{{{\left( x-1 \right)}^{3}}}$$. **Note:** Students make mistakes while finding the quotient and remainder using the long division method, we should write the last step as Quotient + Remainder / Divisor. We should concentrate while comparing the coefficient of $${{x}^{2}}$$ to find the value of A.