Question

How do you find the binomial expansion $ {\left( {3x - 2} \right)^4} $ ?

Answer 1

Hint: In order to determine the expansion of the above function using binomial theorem, compare it with the form \[{\left( {x + y} \right)^n}\]to find our the value of x and y and use the binomial theorem $ {\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}.{x^{n - r}}.{y^r}} $ by putting the value of x and y and expand the summation .Simplifying the equation you will get your required result.

Complete step-by-step answer:
We are given a function $ {\left( {3x - 2} \right)^4} $ , and we have to expand this function with the help of binomial theorem.
Let’s first understand why we use binomial theorem. The reason is pretty simple, because the power of the function is large. When the power of any function increases the expansion becomes really very tough and lengthy.
In such situation, binomial theorem come into play which provide a simple formula to expand any function of the form \[{\left( {x + y} \right)^n}\]with the use of $ ^n{C_r} $ .
According to Binomial theorem:
 $ {\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}.{x^{n - r}}.{y^r}} $ -(1)
where $ ^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $ , $ n \in N,x,y \in R $
If we expand the summation, we get
\[{\left( {x + y} \right)^n} = {}^n{C_0}.{x^n}{ + ^n}{C_1}.{x^{n - 1}}.{y^1} + {}^n{C_2}.{x^{n - 2}}{y^2} + {}^n{C_3}.{x^{n - 3}}{y^3} + ....... + {}^n{C_n}{y^n}\]
According to our question, we have to expand $ {\left( {3x - 2} \right)^4} $ ,so if we compare this with \[{\left( {x + y} \right)^n}\]
We get $ n = 4 $ , $ x = 3x\,and\,y = - 2 $
Putting these values in the binomial theorem (1) , we get
 $ {\left( {3x - 2} \right)^4} = \sum\limits_{r = 0}^4 {^4{C_r}{{\left( {3x} \right)}^{4 - r}}.{{\left( { - 2} \right)}^r}} $
Expanding the summation, we get
 $
  {\left( {3x - 2} \right)^4} = {}^4{C_0}.{\left( {3x} \right)^4}{ + ^4}{C_1}.{\left( {3x} \right)^{4 - 1}}.\left( { - {2^1}} \right) + {}^4{C_2}.{\left( {3x} \right)^{4 - 2}}\left( { - {2^2}} \right) + {}^4{C_3}.{\left( {3x} \right)^{4 - 3}}\left( { - {2^3}} \right) + {}^4{C_4}\left( { - {2^4}} \right) \\
   = {}^4{C_0}.{\left( {3x} \right)^4}{ + ^4}{C_1}.{\left( {3x} \right)^3}.\left( { - {2^1}} \right) + {}^4{C_2}.{\left( {3x} \right)^2}\left( { - {2^2}} \right) + {}^4{C_3}.{\left( {3x} \right)^1}\left( { - {2^3}} \right) + {}^4{C_4}\left( { - {2^4}} \right) \\
   = \left( 1 \right).{\left( {3x} \right)^4} + \left( 4 \right).{\left( {3x} \right)^3}.\left( { - 2} \right) + \left( 6 \right).{\left( {3x} \right)^2}\left( 4 \right) + 4.\left( {3x} \right)\left( { - 8} \right) + \left( {16} \right) \\
   = 81{x^4} - 216{x^3} + 216{x^2} - 98x + 16 \\
  $
 $ \therefore {\left( {3x - 2} \right)^4} = 81{x^4} - 216{x^3} + 216{x^2} - 98x + 16 $
Therefore, the binomial expansion of the function $ {\left( {3x - 2} \right)^4} $ is equal to $ = 81{x^4} - 216{x^3} + 216{x^2} - 98x + 16 $ .
So, the correct answer is “$81{x^4} - 216{x^3} + 216{x^2} - 98x + 16 $ ”.

Note: 1.Factorial: The continued product of first n natural numbers is called the “n factorial “ and denoted by $ n! $ .
2.Combinations: Each of the different selections made by taking some or all of a number of objects irrespective of their arrangement is called a combination.
The combinations number of n objects, taken r at one time is generally denoted by
 $ C(n,r)\,or{\,^n}{C_r} $
Thus, $ C(n,r)\,or{\,^n}{C_r} $ = Number of ways of selecting r objects from n objects.
 $ C(n,r)\, = {\,^n}{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $