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(x +y)^5
Complete the polynomial operation

User Betagreg
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1 Answer

2 votes

Answer:

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

User Huron
by
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