Remainder of a Polynomial when divided by a polynomial with nonreal roots

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What will be the remainder when $x^2015+x^2016$ is divided by $x^2+x+1$ without using the fact that $x^2+x+1$ has roots as non real cube roots of unity.

edited Jul 25 at 5:43

tinlyx

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asked Jul 25 at 4:50

Dhiraj Kumar Kushwaha

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up vote
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What will be the remainder when $x^2015+x^2016$ is divided by $x^2+x+1$ without using the fact that $x^2+x+1$ has roots as non real cube roots of unity.

edited Jul 25 at 5:43

tinlyx

90811118

asked Jul 25 at 4:50

Dhiraj Kumar Kushwaha

111

add a commentÂ |Â

up vote
2
down vote

favorite

What will be the remainder when $x^2015+x^2016$ is divided by $x^2+x+1$ without using the fact that $x^2+x+1$ has roots as non real cube roots of unity.

edited Jul 25 at 5:43

tinlyx

90811118

asked Jul 25 at 4:50

Dhiraj Kumar Kushwaha

111

What will be the remainder when $x^2015+x^2016$ is divided by $x^2+x+1$ without using the fact that $x^2+x+1$ has roots as non real cube roots of unity.

edited Jul 25 at 5:43

tinlyx

90811118

asked Jul 25 at 4:50

Dhiraj Kumar Kushwaha

111

edited Jul 25 at 5:43

tinlyx

90811118

edited Jul 25 at 5:43

tinlyx

90811118

edited Jul 25 at 5:43

tinlyx

90811118

asked Jul 25 at 4:50

Dhiraj Kumar Kushwaha

111

asked Jul 25 at 4:50

Dhiraj Kumar Kushwaha

111

asked Jul 25 at 4:50

Dhiraj Kumar Kushwaha

111

add a commentÂ |Â

3 Answers
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$$x^3-1=(x-1)(x^2+x+1)=?$$

$$x^2015=(x^3)^671x^2=?$$

$$x^2016=(x^3)^672=?$$

answered Jul 25 at 5:00

lab bhattacharjee

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Hint: Â using that $a^3-1=(a-1)(a^2+a+1)$ and so $,x^3k-1=(x-1)(x^2+x+1)(ldots),$:

$$
big(x^2016colorred-1big)+big(x^2015colorred-x^2big)colorred+1+x^2 \
=left(left(x^3right)^672-1right)+x^2left(left(x^3right)^671-1right)+big(x^2colorblue+ x + 1big)colorblue- x
$$

answered Jul 25 at 6:16

dxiv

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Let $K$ be the base field, and consider the more general problem of finding the remainder $r(x)$ of the division of a polynomial $p(x)in K[x]$ by $(x-a)(x-b)$ ($ane b$) in terms of $p,a$ and $b$.

We know that, dividing $p(x)$ by $x-a$, we obtain
$$p(x)=q(x) (x-a)+p(a),qquad q(x)in K[x].tag1$$
Now dividing $q(x)$ by $x-b$ yields similarly $;q(x)=q_1(x)(x-b)+q(b)$, so
$$p(x)=q_1(x)(x-a)(x-b)+underbraceq(b)(x-a)+p(a)_textremainder.$$

Now from $(1)$ we deduce $;q(b)=dfracp(b)-p(a)b-a$, so
beginalign
r(x)&=fracp(b)-p(a)b-a(x-a)+p(a)=fracbigl(p(b)-p(a)bigr)(x-a)+(b-a)p(a)b-a\[1.5ex]
&=fracp(b)(x-a)-p(a)(x-b)b-a=frac1b-a,beginvmatrixx-a &p(a)\x-b&p(b)endvmatrix.
endalign

answered Jul 25 at 9:20

Bernard

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3 Answers
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active

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3 Answers
3

active

oldest

votes

up vote
0
down vote

$$x^3-1=(x-1)(x^2+x+1)=?$$

$$x^2015=(x^3)^671x^2=?$$

$$x^2016=(x^3)^672=?$$

answered Jul 25 at 5:00

lab bhattacharjee

215k14152264

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up vote
0
down vote

$$x^3-1=(x-1)(x^2+x+1)=?$$

$$x^2015=(x^3)^671x^2=?$$

$$x^2016=(x^3)^672=?$$

answered Jul 25 at 5:00

lab bhattacharjee

215k14152264

add a commentÂ |Â

up vote
0
down vote

$$x^3-1=(x-1)(x^2+x+1)=?$$

$$x^2015=(x^3)^671x^2=?$$

$$x^2016=(x^3)^672=?$$

answered Jul 25 at 5:00

lab bhattacharjee

215k14152264

$$x^3-1=(x-1)(x^2+x+1)=?$$

$$x^2015=(x^3)^671x^2=?$$

$$x^2016=(x^3)^672=?$$

answered Jul 25 at 5:00

lab bhattacharjee

215k14152264

answered Jul 25 at 5:00

lab bhattacharjee

215k14152264

answered Jul 25 at 5:00

lab bhattacharjee

215k14152264

answered Jul 25 at 5:00

lab bhattacharjee

215k14152264

add a commentÂ |Â

up vote
0
down vote

Hint: Â using that $a^3-1=(a-1)(a^2+a+1)$ and so $,x^3k-1=(x-1)(x^2+x+1)(ldots),$:

$$
big(x^2016colorred-1big)+big(x^2015colorred-x^2big)colorred+1+x^2 \
=left(left(x^3right)^672-1right)+x^2left(left(x^3right)^671-1right)+big(x^2colorblue+ x + 1big)colorblue- x
$$

answered Jul 25 at 6:16

dxiv

53.9k64796

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up vote
0
down vote

Hint: Â using that $a^3-1=(a-1)(a^2+a+1)$ and so $,x^3k-1=(x-1)(x^2+x+1)(ldots),$:

$$
big(x^2016colorred-1big)+big(x^2015colorred-x^2big)colorred+1+x^2 \
=left(left(x^3right)^672-1right)+x^2left(left(x^3right)^671-1right)+big(x^2colorblue+ x + 1big)colorblue- x
$$

answered Jul 25 at 6:16

dxiv

53.9k64796

add a commentÂ |Â

up vote
0
down vote

Hint: Â using that $a^3-1=(a-1)(a^2+a+1)$ and so $,x^3k-1=(x-1)(x^2+x+1)(ldots),$:

$$
big(x^2016colorred-1big)+big(x^2015colorred-x^2big)colorred+1+x^2 \
=left(left(x^3right)^672-1right)+x^2left(left(x^3right)^671-1right)+big(x^2colorblue+ x + 1big)colorblue- x
$$

answered Jul 25 at 6:16

dxiv

53.9k64796

Hint: Â using that $a^3-1=(a-1)(a^2+a+1)$ and so $,x^3k-1=(x-1)(x^2+x+1)(ldots),$:

$$
big(x^2016colorred-1big)+big(x^2015colorred-x^2big)colorred+1+x^2 \
=left(left(x^3right)^672-1right)+x^2left(left(x^3right)^671-1right)+big(x^2colorblue+ x + 1big)colorblue- x
$$

answered Jul 25 at 6:16

dxiv

53.9k64796

answered Jul 25 at 6:16

dxiv

53.9k64796

answered Jul 25 at 6:16

dxiv

53.9k64796

answered Jul 25 at 6:16

dxiv

53.9k64796

add a commentÂ |Â

up vote
0
down vote

Let $K$ be the base field, and consider the more general problem of finding the remainder $r(x)$ of the division of a polynomial $p(x)in K[x]$ by $(x-a)(x-b)$ ($ane b$) in terms of $p,a$ and $b$.

answered Jul 25 at 9:20

Bernard

110k635103

add a commentÂ |Â

up vote
0
down vote

Let $K$ be the base field, and consider the more general problem of finding the remainder $r(x)$ of the division of a polynomial $p(x)in K[x]$ by $(x-a)(x-b)$ ($ane b$) in terms of $p,a$ and $b$.

answered Jul 25 at 9:20

Bernard

110k635103

add a commentÂ |Â

up vote
0
down vote

Let $K$ be the base field, and consider the more general problem of finding the remainder $r(x)$ of the division of a polynomial $p(x)in K[x]$ by $(x-a)(x-b)$ ($ane b$) in terms of $p,a$ and $b$.

answered Jul 25 at 9:20

Bernard

110k635103

Let $K$ be the base field, and consider the more general problem of finding the remainder $r(x)$ of the division of a polynomial $p(x)in K[x]$ by $(x-a)(x-b)$ ($ane b$) in terms of $p,a$ and $b$.

answered Jul 25 at 9:20

Bernard

110k635103

answered Jul 25 at 9:20

Bernard

110k635103

answered Jul 25 at 9:20

Bernard

110k635103

answered Jul 25 at 9:20

Bernard

110k635103

add a commentÂ |Â

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