Question

Find the coefficient of ${{x}^{4}}$ in the expansion of
${{(2-4x+3{{x}^{2}})}^{-2}}$.

Answer 1

- Hint: Assume that the roots of the quadratic expression $3{{x}^{2}}-4x+2=0$ are a and b. Hence write $\dfrac{1}{{{\left( 3{{x}^{2}}-4x+2 \right)}^{2}}}$ as $\dfrac{1}{9}{{\left( \dfrac{1}{\left( x-a \right)\left( x-b \right)} \right)}^{2}}$. Decompose $\dfrac{1}{\left( x-a \right)\left( x-b \right)}$ into partial fractions by multiplying and dividing numerator by $b-a$ and writing b-a as (x-a)-(x-b). Use the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and binomial theorem for non-integral powers and hence find the coefficient of ${{x}^{4}}$ in the expansion of the term.

Complete step-by-step solution -
Let the roots of the quadratic expression $3{{x}^{2}}-4x+2=0$ be a and b.
Hence, we have
$3{{x}^{2}}-4x+2=3\left( x-a \right)\left( x-b \right)$
Hence, the given expression(E say) becomes
$E=\dfrac{1}{9}{{\left( \dfrac{1}{\left( x-a \right)\left( x-b \right)} \right)}^{2}}$
Inside the square term multiplying numerator and denominator by a-b, we get
$E=\dfrac{1}{9{{\left( a-b \right)}^{2}}}{{\left( \dfrac{a-b}{\left( x-a \right)\left( x-b \right)} \right)}^{2}}$
Writing a-b as (x-b)-(x-a) in the numerator, we get
$E=\dfrac{1}{9{{\left( a-b \right)}^{2}}}{{\left( \dfrac{x-b-\left( x-a \right)}{\left( x-a \right)\left( x-b \right)} \right)}^{2}}$
Using $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$, we get
$E=\dfrac{1}{9{{\left( a-b \right)}^{2}}}{{\left( \dfrac{1}{x-a}-\dfrac{1}{x-b} \right)}^{2}}$
Using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get
$E=\dfrac{1}{9{{\left( a-b \right)}^{2}}}\left( \dfrac{1}{{{\left( x-a \right)}^{2}}}+\dfrac{1}{{{\left( x-b \right)}^{2}}}+\dfrac{2}{\left( x-a \right)\left( x-b \right)} \right)$
From the above procedure, we know that $\dfrac{1}{\left( x-a \right)\left( x-b \right)}=\dfrac{1}{a-b}\left( \dfrac{1}{x-a}-\dfrac{1}{x-b} \right)$
Hence, we have
$E=\dfrac{1}{9{{\left( a-b \right)}^{2}}}\left( \dfrac{1}{{{\left( x-a \right)}^{2}}}+\dfrac{1}{{{\left( x-b \right)}^{2}}}+\dfrac{2}{a-b}\left( \dfrac{1}{x-a}-\dfrac{1}{x-b} \right) \right)$
Consider the term $\dfrac{1}{{{\left( x-a \right)}^{2}}}$
We know from binomial theorem for real exponents
${{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\cdots$
Hence, we have
${{\left( x-a \right)}^{-2}}={{a}^{-2}}{{\left( 1-\dfrac{x}{a} \right)}^{-2}}=\dfrac{1}{{{a}^{2}}}\left( 1+\dfrac{2x}{a}+\dfrac{3!}{2!}{{\left( \dfrac{x}{a} \right)}^{2}}+\dfrac{4!}{3!}{{\left( \dfrac{x}{a} \right)}^{3}}+\dfrac{5!}{4!}{{\left( \dfrac{x}{a} \right)}^{4}}+O\left( {{x}^{5}} \right) \right)$
Hence the coefficient of ${{x}^{4}}$ in the expansion of $\dfrac{1}{{{\left( x-a \right)}^{2}}}$ is $\dfrac{5}{{{a}^{6}}}$
Similarly, the coefficient of ${{x}^{4}}$ in the expansion of $\dfrac{1}{{{\left( x-b \right)}^{2}}}$ is $\dfrac{5}{{{b}^{6}}}$
Consider the term $\dfrac{1}{x-a}$
We know from binomial theorem for real exponents
${{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\cdots$
Hence, we have
${{\left( x-a \right)}^{-1}}=-{{a}^{-1}}{{\left( 1-\dfrac{x}{a} \right)}^{-1}}=-\dfrac{1}{a}\left( 1+\dfrac{x}{a}+{{\left( \dfrac{x}{a} \right)}^{2}}+{{\left( \dfrac{x}{a} \right)}^{3}}+{{\left( \dfrac{x}{a} \right)}^{4}}+O\left( {{x}^{5}} \right) \right)$
Hence the coefficient of ${{x}^{4}}$ in the expansion of $\dfrac{1}{\left( x-a \right)}$ is $-\dfrac{1}{{{a}^{5}}}$
Similarly, the coefficient of ${{x}^{4}}$ in the expansion of $\dfrac{1}{x-b}$ is $-\dfrac{1}{{{b}^{5}}}$
Hence, the coefficient of ${{x}^{4}}$ in the expansion of E is $\dfrac{1}{9{{\left( a-b \right)}^{2}}}\left( \dfrac{5}{{{a}^{6}}}+\dfrac{5}{{{b}^{6}}}-\dfrac{2}{a-b}\left( -\dfrac{1}{{{a}^{5}}}+\dfrac{1}{{{b}^{5}}} \right) \right)=\dfrac{1}{9{{\left( a-b \right)}^{2}}}\left( \dfrac{5}{{{\left( ab \right)}^{6}}}\left( {{a}^{6}}+{{b}^{6}} \right)-\dfrac{2\left( {{a}^{5}}-{{b}^{5}} \right)}{\left( a-b \right){{\left( ab \right)}^{5}}} \right)$
Dividing ${{a}^{5}}-{{b}^{5}}$ by a-b

Hence, we have $\dfrac{{{a}^{5}}-{{b}^{5}}}{a-b}=\left( {{a}^{4}}+{{b}^{4}}+ab\left( {{a}^{2}}+{{b}^{2}} \right)+{{a}^{2}}{{b}^{2}} \right)$
Hence, the coefficient of ${{x}^{4}}$ in the expansion of E is
$\dfrac{1}{9{{\left( a-b \right)}^{2}}}\left( \dfrac{5}{{{\left( ab \right)}^{6}}}\left( {{a}^{6}}+{{b}^{6}} \right)-\dfrac{2\left( {{a}^{4}}+{{b}^{4}}+ab\left( {{a}^{2}}+{{b}^{2}} \right)+{{a}^{2}}{{b}^{2}} \right)}{{{\left( ab \right)}^{5}}} \right)$
Now, we have $a+b=\dfrac{4}{3},ab=\dfrac{2}{3}$
Hence, we have
${{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab=\dfrac{16}{9}-\dfrac{8}{3}=\dfrac{-8}{9}$
Also, we have
$\begin{aligned} & {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab=\dfrac{16}{9}-\dfrac{4}{3}=\dfrac{4}{9} \\\ & {{a}^{4}}+{{b}^{4}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}-2{{\left( ab \right)}^{2}}=\dfrac{16}{81}-2\times \dfrac{4}{9}=\dfrac{16}{81}-\dfrac{8}{9}=\dfrac{-56}{81} \\\ & {{a}^{6}}+{{b}^{6}}=\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{a}^{4}}+{{b}^{4}}-{{\left( ab \right)}^{2}} \right)=\dfrac{4}{9}\left( \dfrac{-56}{81}-\dfrac{4}{9} \right)=\dfrac{-368}{729} \\\ \end{aligned}$
Hence, we have
The coefficient of ${{x}^{4}}$ in the expansion of E is
$\dfrac{1}{9\left( \dfrac{-8}{9} \right)}\left( \dfrac{5}{{{\left( \dfrac{2}{3} \right)}^{6}}}\left( \dfrac{-368}{729} \right)+\dfrac{2}{{{\left( \dfrac{2}{3} \right)}^{5}}}\left( \dfrac{-56}{81}+\dfrac{2}{3}\left( \dfrac{4}{9} \right)+{{\left( \dfrac{2}{3} \right)}^{2}} \right) \right)=\dfrac{59}{16}$
Hence the coefficient of ${{x}^{4}}$ in the expansion of ${{(2-4x+3{{x}^{2}})}^{-2}}$ is $\dfrac{59}{16}$

Note: These types of questions usually involve long and cumbersome calculations, and hence a student is likely to make calculation mistakes. So we should try to minimise calculations as much as possible. In the above question, we minimise our calculation by converting to partial fractions and finding the coefficients of individual terms in summation rather than in product which made the calculations relatively easy.