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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## Comments

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

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Can you explain why? Thanks

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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.

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The function will by \(f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)+x\)

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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

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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?

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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

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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.

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f(x) = x

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

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It can be easily made degree 5 as follows

\(f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)+x\)

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