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- \( f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)+x\) – Krishna Ar · 2 years, 7 months ago

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– Daniel Lim · 2 years, 7 months ago

Can you explain why? ThanksLog in to reply

Hope this helps. – Sudeep Salgia · 2 years, 7 months ago

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– Daniel Lim · 2 years, 7 months ago

Why are the zeroes of \(g(x)\) \(1, 2, 3, 4, 5\)?Log in to reply

– Sudeep Salgia · 2 years, 7 months ago

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.Log in to reply

– Daniel Lim · 2 years, 7 months ago

Thanks, now I understandLog in to reply

– Krishna Ar · 2 years, 7 months ago

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!Log in to reply

– Nishant Sharma · 2 years, 7 months ago

You just forgot that it's a monic polynomial so \(a=1\).Log in to reply

– Sudeep Salgia · 2 years, 7 months ago

Oh, thanks. Edited.Log in to reply

f(x) = x – Somnath Kvs · 2 years, 7 months ago

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– Calvin Lin Staff · 2 years, 7 months ago

That is a good start, but it does not have degree 5.Log in to reply

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 · 2 years, 7 months ago

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– Colin Tang · 2 years, 7 months ago

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.Log in to reply

– Krishna Ar · 2 years, 7 months ago

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

– Starwar Clone · 2 years, 7 months ago

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

– Krishna Ar · 2 years, 7 months ago

I'd have grey hair by the time I'd have finished solving those equations!!Log in to reply

The function will by \(f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)+x\) – Ishan Tarunesh · 2 years, 7 months ago

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