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 $f(x)$ ?

Note: Discussions have been locked till June 25 5PM PST (approximately), so that you can work on this yourself.

#Algebra #Polynomials #Remainder-FactorTheorem #Sprint

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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 - 6 years, 11 months ago

Can you explain why? Thanks

Daniel Lim - 6 years, 11 months 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.

Sudeep Salgia - 6 years, 11 months ago

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

Nishant Sharma - 6 years, 11 months ago

@Nishant Sharma – Oh, thanks. Edited.

Sudeep Salgia - 6 years, 11 months ago

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

Daniel Lim - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@Sudeep Salgia – Thanks, now I understand

Daniel Lim - 6 years, 11 months ago

@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 - 6 years, 11 months ago

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

Ishan Tarunesh - 6 years, 11 months 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

Ujjwal Mani Tripathi - 6 years, 11 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?

Krishna Ar - 6 years, 11 months ago

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 - 6 years, 11 months ago

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

Krishna Ar - 6 years, 11 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.

Colin Tang - 6 years, 11 months ago

f(x) = x

Somnath KVS - 6 years, 11 months ago

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

Calvin Lin Staff - 6 years, 11 months ago

It can be easily made degree 5 as follows

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

Md Zuhair - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$