Question

(x +y)^5
Complete the polynomial operation

Answer 1

Please check the explanation!

Explanation:

Given the polynomial

`\left(x+y\right)^5`

`\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _(i=0)^n\binom{n}{i}a^(\left(n-i\right))b^i`

`a=x,\:\:b=y`

`=\sum _(i=0)^5\binom{5}{i}x^(\left(5-i\right))y^i`

so expanding summation

`=(5!)/(0!\left(5-0\right)!)x^5y^0+(5!)/(1!\left(5-1\right)!)x^4y^1+(5!)/(2!\left(5-2\right)!)x^3y^2+(5!)/(3!\left(5-3\right)!)x^2y^3+(5!)/(4!\left(5-4\right)!)x^1y^4+(5!)/(5!\left(5-5\right)!)x^0y^5`

solving

`(5!)/(0!\left(5-0\right)!)x^5y^0`

`=1\cdot (5!)/(0!\left(5-0\right)!)x^5`

`=1\cdot \:1\cdot \:x^5`

`=x^5`

also solving

`=(5!)/(1!\left(5-1\right)!)x^4y`

`=(5)/(1!)x^4y`

`=(5)/(1!)x^4y`

`=(5x^4y)/(1)`

`=(5x^4y)/(1)`

`=5x^4y`

similarly, the result of the remaining terms can be solved such as

`(5!)/(2!\left(5-2\right)!)x^3y^2=10x^3y^2`

`(5!)/(3!\left(5-3\right)!)x^2y^3=10x^2y^3`

`(5!)/(4!\left(5-4\right)!)x^1y^4=5xy^4`

`(5!)/(5!\left(5-5\right)!)x^0y^5=y^5`

so substituting all the solved results in the expression

`=(5!)/(0!\left(5-0\right)!)x^5y^0+(5!)/(1!\left(5-1\right)!)x^4y^1+(5!)/(2!\left(5-2\right)!)x^3y^2+(5!)/(3!\left(5-3\right)!)x^2y^3+(5!)/(4!\left(5-4\right)!)x^1y^4+(5!)/(5!\left(5-5\right)!)x^0y^5`

`=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5`

Therefore,

`\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5`