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To prove that the polynomial is divisible by (x-1)^2, that means that x = 1 is a root of polynomial and it's first derivative.
For this reason, we'll substitute x by 1 in the expresison of polynomial:
P(1) = n*1^(n+2)-(n+1)*1^(n+1)+1
Since 1 raised to any power, yields 1, we'll get:
P(1) = n - (n+1) + 1
We'll remove the brackets:
P(1) = n - n - 1 + 1
P(1) = 0
So, x = 1 is the root of the polynomial.
Let's verify if x = 1 is the root of the 1st derivative.
P'(x) = n*(n+2)*x^(n+1) - (n+1)^2*x^n + 1
We'll substitute x by 1 in the expresison of the 1st derivative:
P'(1) = n(n+2) - (n+1)^2 + 1
We'll remove the brackets and we'll raise to square the binomial:
P'(1) = n^2 + 2n - n^2 - 2n - 1 + 1
We'll eliminate like terms and we'll get:
P'(1) = 0
Therefore, x =1 is the root of the first derivative, too.
Then, the root x=1 has the order of multiplicity of 2, and the polynomial is divisible by (x-1)^2.
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