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 x81+x49+x25+x9+xx3x\large \dfrac{x^{81}+x^{49}+x^{25}+x^{9}+x}{x^3-x}?

Note by Puneet Pinku
3 years, 1 month 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)

Rakhi Bhattacharyya - 2 years, 8 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).

Zee Ell - 2 years, 8 months ago

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We see that after dividing by xx, we have the expression

x80+x48+x24+x8+1(x+1)(x1)\frac{x^{80}+x^{48}+x^{24}+x^8+1}{(x+1)(x-1)}

Now consider

x80+x48+x24+x8+1(x+1)(x1)5(x+1)(x1)\frac{x^{80}+x^{48}+x^{24}+x^8+1}{(x+1)(x-1)}-\frac{5}{(x+1)(x-1)}

=x80+x48+x24+x84(x+1)(x1)=\frac{x^{80}+x^{48}+x^{24}+x^8-4}{(x+1)(x-1)}

We see that x1x-1 and x+1x+1 are factors of x80+x48+x24+x84x^{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

x80+x48+x24+x84(x+1)(x1)=x80+x48+x24+x8+1(x+1)(x1)5(x+1)(x1)=p(x)+0(x+1)(x1)\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)p(x)

Therefore

x80+x48+x24+x8+1(x+1)(x1)=p(x)+5(x+1)(x1)\frac{x^{80}+x^{48}+x^{24}+x^8+1}{(x+1)(x-1)}=p(x)+\frac{5}{(x+1)(x-1)}.

Jihoon Kang - 3 years, 1 month ago

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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):

x80+x48+x24+x8+1x21=x78+x76+...+x48+2x46+2x44+...+2x24+3x22+3x20+...+3x8+4x6+4x4+4x2+4+5x21 \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}

Zee Ell - 3 years, 1 month 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...

Puneet Pinku - 3 years, 1 month 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. x48+x48=2x46 x^{48 }+ x^{48} = 2x^{46}

Zee Ell - 3 years, 1 month ago

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@Zee Ell 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.....

Puneet Pinku - 3 years, 1 month ago

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@Puneet Pinku 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.

Zee Ell - 3 years, 1 month ago

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