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Polynomials Sprint: Polynomial Remainder

A degree 5 monic polynomial \( f(x) \)
leaves a remainder of 1 when divided by \( x-1\),
leaves a remainder of 2 when divided by \( x-2\),
leaves a remainder of 3 when divided by \( x-3\),
leaves a remainder of 4 when divided by \( x-4\),
leaves a remainder of 5 when divided by \( x-5\).

What is \( f(x) \)?


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Note by Calvin Lin
3 years, 1 month ago

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Constructing a new polynomial , such that the zeroes of the polynomials are satisfying above relation we get- \( f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)+x\) Krishna Ar · 3 years ago

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@Krishna Ar Can you explain why? Thanks Daniel Lim · 3 years ago

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

Hope this helps. Sudeep Salgia · 3 years ago

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@Sudeep Salgia Why are the zeroes of \(g(x)\) \(1, 2, 3, 4, 5\)? Daniel Lim · 3 years ago

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

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@Sudeep Salgia Thanks, now I understand Daniel Lim · 3 years 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 \( (x-3) is 3\) and thus we get the above relations! Krishna Ar · 3 years ago

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@Sudeep Salgia You just forgot that it's a monic polynomial so \(a=1\). Nishant Sharma · 3 years ago

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@Nishant Sharma Oh, thanks. Edited. Sudeep Salgia · 3 years ago

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f(x) = x Somnath Kvs · 3 years ago

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@Somnath Kvs That is a good start, but it does not have degree 5. Calvin Lin Staff · 3 years 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 Starwar Clone · 3 years ago

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@Starwar Clone 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 · 3 years ago

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@Starwar Clone 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 · 3 years ago

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@Krishna Ar Well you are right but please try it by solving the way I told and you would end up with the same problem Starwar Clone · 3 years ago

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@Starwar Clone I'd have grey hair by the time I'd have finished solving those equations!! Krishna Ar · 3 years ago

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The function will by \(f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)+x\) Ishan Tarunesh · 3 years ago

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