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

- 5 years, 1 month ago

Can you explain why? Thanks

- 5 years, 1 month ago

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.

- 5 years, 1 month ago

You just forgot that it's a monic polynomial so $a=1$.

- 5 years, 1 month ago

Oh, thanks. Edited.

- 5 years, 1 month ago

Why are the zeroes of $g(x)$ $1, 2, 3, 4, 5$?

- 5 years, 1 month 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.

- 5 years, 1 month ago

Thanks, now I understand

- 5 years, 1 month 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!

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

I'd have grey hair by the time I'd have finished solving those equations!!

- 5 years, 1 month 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.

- 5 years, 1 month ago

f(x) = x

- 5 years, 1 month ago

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

Staff - 5 years, 1 month ago

It can be easily made degree 5 as follows

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

- 1 year, 5 months ago