# Algebra Mania!

Hello, I propose this problem to the Brilliant community. Hope you enjoy it! This problem was one of the questions in Olympiads.

What is the remainder obtained of the long division $$\large \dfrac{x^{81}+x^{49}+x^{25}+x^{9}+x}{x^3-x}$$? Note by Puneet Pinku
4 years, 9 months ago

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## Comments

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One way to solve this is to simplify by the common factor x first, and then to use algebraic long division in a faster way ( " +...+ " after recognising the repetitive parts):

$\frac {x^{80}+x^{48}+x^{24}+x^8+1}{x^2 - 1} = x^{78}+x^{76}+...+x^{48}+2x^{46}+2x^{44}+...+2x^{24}+3x^{22}+3x^{20}+...+3x^8+4x^6+4x^4+4x^2+4+ \frac { \boxed {5} }{x^2-1}$

- 4 years, 9 months ago

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How to find that 1 will be the coefficient for this much time or 4 will be there for only few numbers and lastly 5 will come... I mean can you explain the pattern a bit more clearly...

- 4 years, 9 months ago

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For the algebraic (or polynomial) long division method in general, you can find many notes, videos etc. on the Internet (e.g. https://brilliant.org/wiki/polynomial-division/ or https://revisionmaths.com/advanced-level-maths-revision/pure-maths/algebra/algebraic-long-division ).

Just follow the method in the case of this division and you will see. (The coefficient increases at some points, because you will have the same powers of x from your remainder (at the previous step) and you also have an original term there (e.g. $x^{48 }+ x^{48} = 2x^{46}$

- 4 years, 9 months ago

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Did you perform the whole long division or somehow you analyzed and figured out the coefficients??? I recently found a new method to solve it..... I will be posting it as question.....

- 4 years, 9 months ago

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I started the whole, but jumped to the key points (where the remainders "got company" from the original polynomial) after recognising the pattern. With further analysis, the process can be shortened even further.

- 4 years, 9 months ago

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We see that after dividing by $x$, we have the expression

$\frac{x^{80}+x^{48}+x^{24}+x^8+1}{(x+1)(x-1)}$

Now consider

$\frac{x^{80}+x^{48}+x^{24}+x^8+1}{(x+1)(x-1)}-\frac{5}{(x+1)(x-1)}$

$=\frac{x^{80}+x^{48}+x^{24}+x^8-4}{(x+1)(x-1)}$

We see that $x-1$ and $x+1$ are factors of $x^{80}+x^{48}+x^{24}+x^8-4$ by the factor theorem as 1 and -1 are roots of this polynomial. Hence we can write

$\frac{x^{80}+x^{48}+x^{24}+x^8-4}{(x+1)(x-1)}=\frac{x^{80}+x^{48}+x^{24}+x^8+1}{(x+1)(x-1)}-\frac{5}{(x+1)(x-1)}=p(x)+\frac{0}{(x+1)(x-1)}$

For some polynomial $p(x)$

Therefore

$\frac{x^{80}+x^{48}+x^{24}+x^8+1}{(x+1)(x-1)}=p(x)+\frac{5}{(x+1)(x-1)}$.

- 4 years, 9 months ago

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Can you just point out the mistake in my solution:

Let the remainder be r(x)=(Ax^2+Bx+C).

let P(x) be the polynomial on the numerator.

P(x)=(x^3-x)g(x)+r(x)

setting x=0,

P(0)=r(0)=C

or,C=0...................(1)

setting x=1,

P(1)=r(1)=A+B

or,A+B=5................(2)

setting x=-1,

P(-1)=r(-1)=A-B

or,A-B=-5..................(3)

Solving (1),(2),and (3), we get A=C=0,and B=5.

So, remainder=5x (ans)

- 4 years, 4 months ago

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The problem is, that x cannot be 0. That would make the denominator of the fraction zero (division by zero).

- 4 years, 4 months ago

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