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Algebra

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8 Jun 2008 15:18:26 IST

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857

Remainder

None

Let

Find the remainder when is divided by f(x)

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Comments (10)

ragesh

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9 Jun 2008 16:37:40 IST

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Re:Remainder

ragesh

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9 Jun 2008 16:38:18 IST

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put x=o

Ans is 0

Hari Shankar

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9 Jun 2008 17:46:59 IST

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Cud u kindly explain why we must put x = 0?

In any case, thats not the answer.

vasanth_mech pilani

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9 Jun 2008 18:21:43 IST

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f(x) = x⁵ -1 /x-1

f(x⁵) = x²⁵- 1/x⁵ - 1

= x²⁵ - x⁵ + x⁵ - 1/x⁵ - 1

= (x⁵ - 1)+ x⁵(x²⁰ - 1)/x⁵ - 1

= (x⁵ - 1) + x⁵(x⁵ - 1)(x¹⁰ + 1)(x⁵ + 1)/x⁵ - 1

=[ x⁵ - 1] (x¹⁰ + 1)(x⁵ + 1)²/(x⁵ - 1)

= (x¹⁰ + 1)(x⁵ + 1)²

= (x¹⁰ - 1 + 2)(x⁵ - 1 + 2)²

= [ (x⁵ - 1)(x⁵ + 1) + 2] [ F(x)*(x⁵ - 1) + 4]

= G(x)*(x⁵ - 1) + 8

= G(x) (x-1) f(x) + 8

when divided by f(x)

will leave remainder 8

am i rite??

Utkarsh Singh

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9 Jun 2008 18:31:51 IST

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@vasanth....
can u plz explain how u got this step?

"" = [ (x5 - 1)(x5 + 1) + 2] [ F(x)*(x5 - 1) + 4 ]""

Utkarsh Singh

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9 Jun 2008 18:34:54 IST

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is the answer 5??

vasanth_mech pilani

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9 Jun 2008 18:35:54 IST

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x¹⁰ - 1 = (x⁵ - 1)(x⁵ + 1)

and

(x⁵ - 1 + 2)² = (x⁵ - 1)² + 2*2*(x⁵ - 1) + 4

= (x⁵ - 1) * [(x⁵ - 1) + 4] + 4

= (x⁵ - 1)*F(x) + 4

where F(x) = (x⁵ - 1) + 4

Hari Shankar

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9 Jun 2008 19:08:23 IST

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utkarsh's answer is right. but working...

SUNDEEP ALLAMRAJU

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9 Jun 2008 21:54:38 IST

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Finally after working hard,I got the answer.

f(x⁵)=x²⁵-1/x⁵-1=(x⁵-1)+x⁵(x²⁰-1)/x⁵-1=1+[x⁵(x¹⁰+1)(x⁵+1)(x⁵-1)/(x⁵-1)]

Now,let us try to write the term in the nr of above fraction as a multiple of (x⁵-1)².

Thus,Nr=[(x⁵-1)+1][(x⁵-1)²+2x⁵][(x⁵-1)+2][x⁵-1]

=[(x⁵-1)][(x⁵-1)+1][(x⁵-1)+2][(x⁵-1)²+2(x⁵-1)+2]

On multiplying these terms and simplifying,we get,

(x⁵-1)⁵+5(x⁵-1)⁴+10(x⁵-1)³+10(x⁵-1)²+4(x⁵-1)

Substituting this in f(x⁵),cancelling (x⁵-1),we get,

f(x⁵)=1+4+(x⁵-1)[(x⁵-1)³+5(x⁵-1)²+10(x⁵-1)+10]

=5+(x-1).f(x).g(x)

where g(x)=(x⁵-1)³+5(x⁵-1)²+10(x⁵-1)+10.

Hence,the remainder is 5.

Hari Shankar

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10 Jun 2008 09:33:10 IST

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There are at least two other ways to go about it. I will present both here:

Method I:

$f(x) = x^4+x^3+x^2+x+1 = \frac{x^5-1}{x-1} \\ \\ f(x^5) = x^{20}+x^{15}+x^{10}+x^5+1 = (x^{20}-1)+(x^{15}-1)+(x^{10}-1)+(x^5-1) + 5\\ \\ \text{Each of the bracketed terms is divisible by} \ (x^5-1) \\ \\ \text{and hence by f(x), thus leaving a remainder of 5 on division by} \ f(x^5)\\ \\ $

Method II:

$f(z^5) = f(z) q(z) + r(z) \\ \\ \text{Now} \ f(z) = (z-\omega) (z-\omega^2) (z-\omega^3) (z-\omega^4)\\ \\ \text{where} \ \omega, \omega^2, \omega^3 \ \text{and} \ \omega^4 \ \text{are the non-real roots of the equation} \ z^5=1 \\ \\ \text{Now, substituting each of the roots, we obtain} \\ \\ r(\omega) = r(\omega^2) = r(\omega^3) = r(\omega^4) = 5 \\ \\ \text{However h(z) = r(z) - 5 is a polynomial of degree 3 that vanishes for 4 distinct values of z} \\ \\ \text{This means r(z) = 5 for all values of z}$

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Comments (10)

Finally after working hard,I got the answer.

f(x5)=x25-1/x5-1=(x5-1)+x5(x20-1)/x5-1=1+[x5(x10+1)(x5+1)(x5-1)/(x5-1)]

Now,let us try to write the term in the nr of above fraction as a multiple of (x5-1)2.

Thus,Nr=[(x5-1)+1][(x5-1)2+2x5][(x5-1)+2][x5-1]

=[(x5-1)][(x5-1)+1][(x5-1)+2][(x5-1)2+2(x5-1)+2]

On multiplying these terms and simplifying,we get,

(x5-1)5+5(x5-1)4+10(x5-1)3+10(x5-1)2+4(x5-1)

Substituting this in f(x5),cancelling (x5-1),we get,

f(x5)=1+4+(x5-1)[(x5-1)3+5(x5-1)2+10(x5-1)+10]

=5+(x-1).f(x).g(x)

where g(x)=(x5-1)3+5(x5-1)2+10(x5-1)+10.

Hence,the remainder is 5.

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f(x⁵)=x²⁵-1/x⁵-1=(x⁵-1)+x⁵(x²⁰-1)/x⁵-1=1+[x⁵(x¹⁰+1)(x⁵+1)(x⁵-1)/(x⁵-1)]

Now,let us try to write the term in the nr of above fraction as a multiple of (x⁵-1)².

Thus,Nr=[(x⁵-1)+1][(x⁵-1)²+2x⁵][(x⁵-1)+2][x⁵-1]

=[(x⁵-1)][(x⁵-1)+1][(x⁵-1)+2][(x⁵-1)²+2(x⁵-1)+2]

(x⁵-1)⁵+5(x⁵-1)⁴+10(x⁵-1)³+10(x⁵-1)²+4(x⁵-1)

Substituting this in f(x⁵),cancelling (x⁵-1),we get,

f(x⁵)=1+4+(x⁵-1)[(x⁵-1)³+5(x⁵-1)²+10(x⁵-1)+10]

where g(x)=(x⁵-1)³+5(x⁵-1)²+10(x⁵-1)+10.