Question

How do you find the coefficient of \[{x^3}{y^2}\] in the expansion of \[{(x - 3y)^5}?\]

Answer 1

Hint: In the binomial type problem, firstly we have to compare the given expression which we have to expand with the binomial theorem and try to find out the value of \[x,y\] and \[n\]. After finding the value of \[x,y\] and \[n\] by comparing, start expanding the given expression with the help of a binomial theorem.

Formula used:
Binomial theorem, \[{(p + q)^n} = {}^n{c_0}{p^n}{q^0} + {}^n{c_1}{p^{n - 1}}{q^1} + {}^n{c_2}{p^{n - 2}}{q^2} + ..... + {}^n{c_n}{p^0}{q^n}\]
\[{a^0} = 1\] and \[{x^0} = 1\]

Complete step by step solution:
Comparing given expression with binomial theorem to find out value of \[p,q\] and \[n\], we get
\[ \Rightarrow p = x\]
\[ \Rightarrow q = - 3y\] and
\[ \Rightarrow n = 5\]
First start expanding binomial theorem in the following form
\[{(p + q)^n} = {p^n}{q^0} + n{p^{n - 1}}{q^1} + \dfrac{{n(n - 1)}}{{2!}}{p^{n - 2}}{q^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{p^{n - 3}}{q^3} + \dfrac{{n(n - 1)(n - 2)(n - 3)}}{{4!}}{p^{n - 4}}{q^4} + ..... + {p^0}{q^n}\]
Now, replacing the value of \[p,q\] and \[n\]with above value that is \[p = x\],\[q = - 3y\] and \[n = 5\], we get
\[ \Rightarrow {(x + ( - 3y))^5} = {x^5}{( - 3y)^0} + 5{x^{5 - 1}}{( - 3y)^1} + \dfrac{{5(5 - 1)}}{{2!}}{x^{5 - 2}}{( - 3y)^2} + \dfrac{{5(5 - 1)(5 - 2)}}{{3!}}{x^{5 - 3}}{( - 3y)^3} + ..... + {x^0}{( - 3y)^n}\]
Reducing the terms till where we get 5 when we adding power of ‘\[p\]’ and ‘\[q\]’ as following,
\[ \Rightarrow {(x + ( - 3y))^5} = {x^5}{( - 3y)^0} + 5{x^{5 - 1}}{( - 3y)^1} + \dfrac{{5(5 - 1)}}{{2!}}{x^{5 - 2}}{( - 3y)^2}......\]
Simplifying at left-hand side and writing it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5}{( - 3y)^0} + 5{x^{5 - 1}}{( - 3y)^1} + \dfrac{{5(5 - 1)}}{{2!}}{x^{5 - 2}}{( - 3y)^2}......\]
Simplifying power of variables and write it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5}{( - 3y)^0} + 5{x^4}{( - 3y)^1} + \dfrac{{5(5 - 1)}}{{2!}}{x^3}{( - 3y)^2}......\]
Simplifying factorial part in each denominator of fraction and write it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5}{( - 3y)^0} + 5{x^4}{( - 3y)^1} + \dfrac{{5(5 - 1)}}{2}{x^3}{( - 3y)^2}......\]
Simplifying numerator part of each fraction and write it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5}{( - 3y)^0} + 5{x^4}{( - 3y)^1} + \dfrac{{5(4)}}{2}{x^3}{( - 3y)^2}......\]
Expanding numerator part of each fraction and write it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5}{( - 3y)^0} + 5{x^4}{( - 3y)^1} + \dfrac{{5 \times 4}}{2}{x^3}{( - 3y)^2}......\]
Multiplying numerator part of each fraction and write it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5}{( - 3y)^0} + 5{x^4}{( - 3y)^1} + \dfrac{{20}}{2}{x^3}{( - 3y)^2}......\]
Simplifying each fraction part by dividing and write it as following
\[ \Rightarrow {(x - 3y)^5} = {x^5}{( - 3y)^0} + 5{x^4}{( - 3y)^1} + 10{x^3}{( - 3y)^2}......\]
By using the above given formula \[{( - 3y)^0}\] becomes equal to one. So, it can be written as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5} + 5{x^4}{( - 3y)^1} + 10{x^3}{( - 3y)^2}......\]
Expanding each term of right-hand side and writing it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5} + 5 \times {x^4} \times {( - 3y)^1} + 10 \times {x^3} \times ( - 3y)( - 3y)......\]
Multiplying each term on right-hand side and writing it as following,
\[ \Rightarrow {(x - 3y)^5} = {x^5} - 15{x^4} + 90{x^3}{y^2}....\]
So, we have obtained the coefficient of \[{x^3}{y^2}\] that is equal to 90.

Note: The same number of terms need to expand from binomial theorem expansion as equal to ‘\[n\]’ here 5 when we add powers of ‘\[p\]’ and ‘\[q\]’. It means that only 5 terms need to be expanded as when we add the power of \[p = 3\] and \[q = 2\] it will give 5. It is because we will get the required term for which coefficient has to be calculated within these 5 terms.