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

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

Explanation

Solution

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 (y+x)n{\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: nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
Binomial expansion:
(x+y)n=nC0xn+nC1xn−1y1+nC2xn−2y2+......+nCn−1xyn−1+nCnyn{(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⩾0n \geqslant 0, is an integer and each nCk{}^n{C_k} is a positive integer known as binomial coefficient.

Complete answer:
In the above question, we can also write it as (y+x)4=(x+y)4{\left( {y + x} \right)^4} = {\left( {x + y} \right)^4}
We need to expand (x+y)4{(x + y)^4}
Now we know that, according to binomial theorem it is possible to expand any nonnegative power of x+yx + y into a sum of the form
(x+y)n=nC0xn+nC1xn−1y1+nC2xn−2y2+......+nCn−1xyn−1+nCnyn{(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⩾0n \geqslant 0, is an integer and each nCk{}^n{C_k} is a positive integer known as binomial coefficient.
Now we can use the binomial expansion putting n=4n = 4 we get,
⇒(x+y)4=4C0x4y0+4C1x4−1y1+4C2x4−2y2++4C3x4−3y3+4C4x4−4y4\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, x0=1{x^0} = 1,
⇒(x+y)4=4C0x4+4C1x3y1+4C2x2y2++4C3x1y3+4C4y4\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, nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}, we get,
Solving we get,
⇒(x+y)4=4!0!(4−0)!x4+4!1!(4−1)!x3y1+4!2!(4−2)!x2y2+4!3!(4−3)!x1y3+4!4!(4−4)!y4\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, ⇒(x+y)4=4!0!4!x4+4!1!3!x3y1+4!2!2!x2y2+4!3!1!x1y3+4!4!0!y4\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,
⇒(x+y)4=1x4+4x3y1+6x2y2+4x1y3+1y4\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,
⇒(x+y)4=x4+4x3y1+6x2y2+4x1y3+y4\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 (y+x)4{\left( {y + x} \right)^4} we get,
⇒(x+y)4=x4+4x3y1+6x2y2+4x1y3+y4\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.