Question

How do you expand ${\left( {y + x} \right)^4}$?

Answer 1

Hint: Here we will apply the Binomial Expansion to solve the given problem. The binomial theorem (or binomial expansion) describes the algebraic expansion of power of a binomial. According to the theorem, it is possible to expand the polynomial ${\left( {y + x} \right)^n}$. First, we have to Put the given value in the place of n and use the formula. Then solving this with the help of combination rule and factorial n and simplifying the result we will get the solution.

Formula used: Combination rule: \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Binomial expansion:
\[{(x + y)^n}{ = ^n}{C_0}{x^n}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......{ + ^n}{C_{n - 1}}x{y^{n - 1}}{ + ^n}{C_n}{y^n}\] ,
 where $n \geqslant 0$, is an integer and each \[{}^n{C_k}\] is a positive integer known as binomial coefficient.

Complete answer:
In the above question, we can also write it as \[{\left( {y + x} \right)^4} = {\left( {x + y} \right)^4}\]
We need to expand \[{(x + y)^4}\]
Now we know that, according to binomial theorem it is possible to expand any nonnegative power of $x + y$ into a sum of the form
\[{(x + y)^n}{ = ^n}{C_0}{x^n}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......{ + ^n}{C_{n - 1}}x{y^{n - 1}}{ + ^n}{C_n}{y^n}\]
Where $n \geqslant 0$, is an integer and each \[{}^n{C_k}\] is a positive integer known as binomial coefficient.
Now we can use the binomial expansion putting \[n = 4\] we get,
\[ \Rightarrow {(x + y)^4}{ = ^4}{C_0}{x^4}{y^0}{ + ^4}{C_1}{x^{4 - 1}}{y^1}{ + ^4}{C_2}{x^{4 - 2}}{y^2} + { + ^4}{C_3}{x^{4 - 3}}{y^3}{ + ^4}{C_4}{x^{4 - 4}}{y^4}\]
Using, ${x^0} = 1$,
\[ \Rightarrow {(x + y)^4}{ = ^4}{C_0}{x^4}{ + ^4}{C_1}{x^{3}}{y^1}{ + ^4}{C_2}{x^{2}}{y^2} + { + ^4}{C_3}{x^{1}}{y^3}{ + ^4}{C_4}{y^4}\]
We can use the combination, \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], we get,
Solving we get,
\[ \Rightarrow {(x + y)^4}=\dfrac{{4!}}{{0!\left( {4 - 0} \right)!}}{x^4}+\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}{x^{3}}{y^1}+\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}{x^{2}}{y^2} + \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}{x^{1}}{y^3}+\dfrac{{4!}}{{4!\left( {4 - 4} \right)!}}{y^4}\]
Again, \[\Rightarrow{(x+y)^4}=\dfrac{{4!}}{{0!4!}}{x^4}+\dfrac{{4!}}{{1!3!}}{x^{3}}{y^1}+\dfrac{{4!}}{{2!2!}}{x^{2}}{y^2} + \dfrac{{4!}}{{3!1!}}{x^{1}}{y^3}+\dfrac{{4!}}{{4!0!}}{y^4}\]
Cancelling common factors in numerator and denominator, we get,
\[\Rightarrow{(x+y)^4}=1{x^4}+4{x^{3}}{y^1}+6{x^{2}}{y^2} + 4{x^{1}}{y^3}+1{y^4}\]
Simplifying we get,
\[\Rightarrow{(x+y)^4}={x^4}+4{x^{3}}{y^1}+6{x^{2}}{y^2} + 4{x^{1}}{y^3}+{y^4}\]
Hence expanding ${\left( {y + x} \right)^4}$ we get,
\[\Rightarrow{(x+y)^4}={x^4}+4{x^{3}}{y^1}+6{x^{2}}{y^2} + 4{x^{1}}{y^3}+{y^4}\]

Note:
In mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.