The Repeating Factor

The key to solve this is "Factor Theorem". Which actually states a basic idea that if P(x) is a polynomial and P(n) = 0, then (x-n) is a factor of P(x). $P(x) = 2x^{6}-3x^{5}-19x^{4}+30x^{3}+8x^{2}-27x+9$

First step, we put x=1 in the given polynomial. If we get zero, this means (x-1) is a factor. $P(1) = 2 - 3 - 19 +30 +8 -27 + 9 = 0$ So it is clear that (x-1) is a factor, but we don't know exactly how many (x-1)'s are hiding in there.

Second step Now, we are going to use a bit of calculus. We will take the derivative of polynomial. $P'(x) = 12x^{5} - 15 x^{4} -76x^{3}+90x^{2}+16x-27$ Now $P'(1) = 12 - 15 - 76 +90 +16 -27 = 0$ This means that (x-1) is also a factor of the derivative of P(x). From all the above results, P(x) must have atleast 2 (x-1)'s . (I mean (x-1)^2)

Third Step Computing second derivative. $P"(x) = 60 x^{4} - 60x^{3} -228x^{2}+180x+16$ $P"(1) = 60 - 60 -228 +180 +16 = -32 ≠ 0$ This means (x-1) is not a factor of P"(x).

We can conclude that (x-1)^2 is factor of P(x) and (x-1)^3 is not. And so k=2 is the answer.

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If you wonder how does that work, let f(x) and g(x) be some polynomials such that $f(x) = (x-1)^{2}g(x)$ $f'(x) = (x-1)^{2}g'(x) + 2(x-1)g(x)$ $f"(x) = (x-1)^{2}g"(x) + 2(x-1)g'(x) + 2(x-1)g'(x) + 2g(x)$ f(1)=0, f'(1)=0 but f"(1) may be zero if and only if g(1)=0

"So in general the factor (x-n)^k dies after taking k derivatives" And it is cool.

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Noticed this today. The purpose of asking the question was nothing else but actually to let everyone know the method.

If you have any doubts regarding this please you can ask, I would like to help you out.

Thanks.