Question

Find the coefficient of ${{x}^{4}}$ in the expansion of ${{\left( 1+2x \right)}^{4}}{{\left( 2-x \right)}^{5}}$

Answer 1

Hint: We start solving this question by first dividing the given expression ${{\left( 1+2x \right)}^{4}}{{\left( 2-x \right)}^{5}}$ into two parts ${{\left( 1+2x \right)}^{4}}$ and ${{\left( 2-x \right)}^{5}}$. Then we use the formula for the binomial expansion ${{\left( a+x \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}x+{}^{n}{{C}_{2}}{{a}^{n-2}}{{x}^{2}}+.........+{}^{n}{{C}_{n-1}}a{{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}}$ to find the expansion of the two parts. Then we find the coefficients obtained by multiplying the terms of first part and second part that gives ${{x}^{4}}$ and add them to find the coefficient of ${{x}^{4}}$.

Complete step-by-step answer:
First, let us go through the formula for the binomial expansion of a first-degree polynomial
${{\left( a+x \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}x+{}^{n}{{C}_{2}}{{a}^{n-2}}{{x}^{2}}+.........+{}^{n}{{C}_{n-1}}a{{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}}$
Now let us divide the given polynomial into two parts ${{\left( 1+2x \right)}^{4}}$ and ${{\left( 2-x \right)}^{5}}$.
Now let us go through the first part ${{\left( 1+2x \right)}^{4}}$. By using the above discussed formula expansion and applying it we get,
${{\left( 1+2x \right)}^{4}}={}^{4}{{C}_{0}}+{}^{4}{{C}_{1}}\left( 2x \right)+{}^{4}{{C}_{2}}{{\left( 2x \right)}^{2}}+{}^{4}{{C}_{3}}{{\left( 2x \right)}^{3}}+{}^{4}{{C}_{4}}{{\left( 2x \right)}^{4}}$
Now let us consider the formula ${}^{n}{{C}_{r}}$
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$
Using that we get
$\begin{align}
  & \Rightarrow {}^{4}{{C}_{0}}=\dfrac{4!}{0!\times 4!}=1 \\
 & \Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\times 3!}=4 \\
 & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4!}{2!\times 2!}=6 \\
 & \Rightarrow {}^{4}{{C}_{3}}=\dfrac{4!}{3!\times 1!}=4 \\
 & \Rightarrow {}^{4}{{C}_{4}}=\dfrac{4!}{4!\times 0!}=1 \\
\end{align}$
Substituting the values in the above expansion, we get
$\begin{align}
  & \Rightarrow {{\left( 1+2x \right)}^{4}}=1+4\left( 2x \right)+6\left( 4{{x}^{2}} \right)+4\left( 8{{x}^{3}} \right)+1\left( 16{{x}^{4}} \right) \\
 & \Rightarrow {{\left( 1+2x \right)}^{4}}=1+8x+24{{x}^{2}}+32{{x}^{3}}+16{{x}^{4}} \\
\end{align}$
Now let us consider the second part ${{\left( 2-x \right)}^{5}}$. By applying the binomial expansion for it we get,
${{\left( 2-x \right)}^{5}}={}^{5}{{C}_{0}}{{\left( 2 \right)}^{5}}+{}^{5}{{C}_{1}}{{\left( 2 \right)}^{4}}\left( -x \right)+{}^{5}{{C}_{2}}{{\left( 2 \right)}^{3}}{{\left( -x \right)}^{2}}+{}^{5}{{C}_{3}}{{\left( 2 \right)}^{2}}{{\left( -x \right)}^{3}}+{}^{5}{{C}_{4}}\left( 2 \right){{\left( -x \right)}^{4}}+{}^{5}{{C}_{5}}{{\left( -x \right)}^{5}}$
 Now let us consider the formula ${}^{n}{{C}_{r}}$
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$
Using that we get
$\begin{align}
  & \Rightarrow {}^{5}{{C}_{0}}=\dfrac{5!}{0!\times 5!}=1 \\
 & \Rightarrow {}^{5}{{C}_{1}}=\dfrac{5!}{1!\times 4!}=5 \\
 & \Rightarrow {}^{5}{{C}_{2}}=\dfrac{5!}{2!\times 3!}=10 \\
 & \Rightarrow {}^{5}{{C}_{3}}=\dfrac{5!}{3!\times 2!}=10 \\
 & \Rightarrow {}^{5}{{C}_{4}}=\dfrac{5!}{4!\times 1!}=5 \\
 & \Rightarrow {}^{5}{{C}_{5}}=\dfrac{5!}{5!\times 0!}=1 \\
\end{align}$
Substituting the values in the above expansion, we get
$\begin{align}
  & \Rightarrow {{\left( 2-x \right)}^{5}}={}^{5}{{C}_{0}}{{\left( 2 \right)}^{5}}+{}^{5}{{C}_{1}}{{\left( 2 \right)}^{4}}\left( -x \right)+{}^{5}{{C}_{2}}{{\left( 2 \right)}^{3}}{{\left( -x \right)}^{2}}+{}^{5}{{C}_{3}}{{\left( 2 \right)}^{2}}{{\left( -x \right)}^{3}}+{}^{5}{{C}_{4}}\left( 2 \right){{\left( -x \right)}^{4}}+{}^{5}{{C}_{5}}{{\left( -x \right)}^{5}} \\
 & \Rightarrow {{\left( 2-x \right)}^{5}}=32+5\left( 16 \right)\left( -x \right)+10\left( 8 \right){{x}^{2}}+10\left( 4 \right)\left( -{{x}^{3}} \right)+5\left( 2 \right){{x}^{4}}+{}^{5}{{C}_{5}}\left( -{{x}^{5}} \right) \\
 & \Rightarrow {{\left( 2-x \right)}^{5}}=32-80x+80{{x}^{2}}-40{{x}^{3}}+10{{x}^{4}}-{{x}^{5}} \\
\end{align}$
As we need to find the coefficient of ${{x}^{4}}$ in the expansion.
We can get ${{x}^{4}}$ by multiplying the coefficients of
${{x}^{0}}$ of first part with ${{x}^{4}}$ of second part we get $1\left( 10{{x}^{4}} \right)$
${{x}^{1}}$ of first part with ${{x}^{3}}$ of second part we get $8x\left( -40{{x}^{3}} \right)$
${{x}^{2}}$ of first part with ${{x}^{2}}$ of second part we get $24{{x}^{2}}\left( 80{{x}^{2}} \right)$
${{x}^{3}}$ of first part with ${{x}^{1}}$ of second part we get $32{{x}^{3}}\left( -80x \right)$
${{x}^{4}}$ of first part with ${{x}^{0}}$ of second part we get $16{{x}^{4}}\left( 32 \right)$
We need to find the coefficient of ${{x}^{4}}$. So, by adding all the terms we get,
$\begin{align}
  & \Rightarrow 1\left( 10{{x}^{4}} \right)+8x\left( -40{{x}^{3}} \right)+24{{x}^{2}}\left( 80{{x}^{2}} \right)+32{{x}^{3}}\left( -80x \right)+16{{x}^{4}}\left( 32 \right) \\
 & \Rightarrow 10{{x}^{4}}-320{{x}^{4}}+1920{{x}^{4}}-2560{{x}^{4}}+512{{x}^{4}} \\
 & \Rightarrow -438{{x}^{4}} \\
\end{align}$
So, we get that the coefficient of ${{x}^{4}}$ in the expansion of ${{\left( 1+2x \right)}^{4}}{{\left( 2-x \right)}^{5}}$ is -438.
Hence, the answer is -438.

Note: The common mistake that one does while solving this problem is one might multiply the coefficient of ${{x}^{4}}$ in the first part with the coefficient of ${{x}^{0}}$ in second part and multiply the coefficient of ${{x}^{0}}$ in the first part with the coefficient of ${{x}^{4}}$ in second part. Then we get
$\begin{align}
  & \Rightarrow 1\left( 10{{x}^{4}} \right)+16{{x}^{4}}\left( 32 \right) \\
 & \Rightarrow 10{{x}^{4}}+512{{x}^{4}} \\
 & \Rightarrow 522{{x}^{4}} \\
\end{align}$
Then we get the coefficient of ${{x}^{4}}$ as 522. But we need to consider all the possible choices of getting ${{x}^{4}}$ in the expansion.