Answer:
Please check the explanation!
Step-by-step explanation:
Given the polynomial
[tex]\left(x+y\right)^5[/tex]
[tex]\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i[/tex]
[tex]a=x,\:\:b=y[/tex]
[tex]=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i[/tex]
so expanding summation
[tex]=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5[/tex]
solving
[tex]\frac{5!}{0!\left(5-0\right)!}x^5y^0[/tex]
[tex]=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5[/tex]
[tex]=1\cdot \:1\cdot \:x^5[/tex]
[tex]=x^5[/tex]
also solving
[tex]=\frac{5!}{1!\left(5-1\right)!}x^4y[/tex]
[tex]=\frac{5}{1!}x^4y[/tex]
[tex]=\frac{5}{1!}x^4y[/tex]
[tex]=\frac{5x^4y}{1}[/tex]
[tex]=\frac{5x^4y}{1}[/tex]
[tex]=5x^4y[/tex]
similarly, the result of the remaining terms can be solved such as
[tex]\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2[/tex]
[tex]\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3[/tex]
[tex]\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4[/tex]
[tex]\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5[/tex]
so substituting all the solved results in the expression
[tex]=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5[/tex]
[tex]=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5[/tex]
Therefore,
[tex]\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5[/tex]
