Polynomials Sprint: Polynomial Remainder

A degree 5 monic polynomial f(x) f(x)
leaves a remainder of 1 when divided by x1 x-1,
leaves a remainder of 2 when divided by x2 x-2,
leaves a remainder of 3 when divided by x3 x-3,
leaves a remainder of 4 when divided by x4 x-4,
leaves a remainder of 5 when divided by x5 x-5.

What is f(x) f(x) ?


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Note by Calvin Lin
5 years, 2 months ago

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Constructing a new polynomial , such that the zeroes of the polynomials are satisfying above relation we get- f(x)=(x1)(x2)(x3)(x4)(x5)+x f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)+x

Krishna Ar - 5 years, 1 month ago

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Can you explain why? Thanks

Daniel Lim - 5 years, 1 month ago

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Using Remainder Factor Theorem, we can say if f(x) f(x) is divided by (xk)(x- k) where kk is a constant, the remainder is given by f(k) f(k) . Therefore, the above information can be rewritten as f(r)=r ;r=1,2,3,4,5 f(r) = r \text{ }; r = 1,2,3,4,5 . Let us define a new function g(x) g(x) as g(x)=f(x)x g(x) = f(x) - x . Therefore, the zeroes of g(x)g(x) are 1,2,3,4 and 51,2,3,4 \text{ and } 5 . Hence g(x)g(x) can be written as g(x)=a(x1)(x2)(x3)(x4)(x5) g(x) = a(x-1)(x-2)(x-3)(x-4)(x-5) where aa is some constant. Therefore, f(x)=a(x1)(x2)(x3)(x4)(x5)+xf(x) = a(x-1)(x-2)(x-3)(x-4)(x-5) + x . Since f(x) f(x) is given to be monic, therefore, a=1 a = 1 . Hence, f(x)=(x1)(x2)(x3)(x4)(x5)+xf(x) = (x-1)(x-2)(x-3)(x-4)(x-5) + x .

Hope this helps.

Sudeep Salgia - 5 years, 1 month ago

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@Sudeep Salgia You just forgot that it's a monic polynomial so a=1a=1.

Nishant Sharma - 5 years, 1 month ago

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@Nishant Sharma Oh, thanks. Edited.

Sudeep Salgia - 5 years, 1 month ago

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@Sudeep Salgia Why are the zeroes of g(x)g(x) 1,2,3,4,51, 2, 3, 4, 5?

Daniel Lim - 5 years, 1 month ago

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@Daniel Lim If α\alpha is a zero of g(x)g(x), then g(α)=0g(\alpha ) = 0 . Therefore, f(α)α=0  f(α)=αf(\alpha ) - \alpha = 0 \text{ }\Rightarrow \text{ } f(\alpha ) = \alpha . We find that α\alpha can assume the values 1,2,3,4 and 5 1,2,3,4 \text{ and } 5 ,thus the result.

Sudeep Salgia - 5 years, 1 month ago

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@Sudeep Salgia Thanks, now I understand

Daniel Lim - 5 years, 1 month ago

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@Daniel Lim Hi! The zero of a polynomial is the value that if substituted in the variable of the polynomial- gives 0. Eg- zero of (x3)is3 (x-3) is 3 and thus we get the above relations!

Krishna Ar - 5 years, 1 month ago

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The function will by f(x)=(x1)(x2)(x3)(x4)(x5)+xf(x) = (x-1)(x-2)(x-3)(x-4)(x-5)+x

Ishan Tarunesh - 5 years, 1 month ago

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There is no such polynomial with real coefficients if you solve it. by making 5 equations in a;b;c;d;e. I would love if someone proves me wrong

Ujjwal Mani Tripathi - 5 years, 1 month ago

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Real?? Really? If you'd expand the expression given above, I;m sure the answer would have integral coefficients only. Isn;t it?

Krishna Ar - 5 years, 1 month ago

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Well you are right but please try it by solving the way I told and you would end up with the same problem

Ujjwal Mani Tripathi - 5 years, 1 month ago

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@Ujjwal Mani Tripathi I'd have grey hair by the time I'd have finished solving those equations!!

Krishna Ar - 5 years, 1 month ago

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By a, b, c, d, e do you mean the coefficients of the polynomial? If so, f(x) = x + (x-1)(x-2)(x-3)(x-4)(x-5) is a counterexample to your claim. It has integral coefficients, therefore it has real coefficients.

Colin Tang - 5 years, 1 month ago

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f(x) = x

Somnath KVS - 5 years, 1 month ago

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That is a good start, but it does not have degree 5.

Calvin Lin Staff - 5 years, 1 month ago

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It can be easily made degree 5 as follows

f(x)=(x1)(x2)(x3)(x4)(x5)+xf(x) = (x-1)(x-2)(x-3)(x-4)(x-5)+x

Md Zuhair - 1 year, 5 months ago

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