Mixing
How do you find the remainder when dividing polynomials?
Clash Royale CLAN TAG#URR8PPP
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Suppose that $f=(x+1)^100+(x-3)^100$ and $g=x^2-2x-3$. What is the remainder of $fracfg$? I know there must be something connected to the roots of $g$, but how I can use that? Also I know that the remainder's degree is less or equal to 1. I don't want the response, I want to find that by myself, hints are welcomed.
polynomials
edited Sep 4 at 23:49
Xander Henderson
13.3k83250
asked Sep 4 at 18:28
Numbers
925
add a comment |Â
up vote
0
down vote
favorite
Suppose that $f=(x+1)^100+(x-3)^100$ and $g=x^2-2x-3$. What is the remainder of $fracfg$? I know there must be something connected to the roots of $g$, but how I can use that? Also I know that the remainder's degree is less or equal to 1. I don't want the response, I want to find that by myself, hints are welcomed.
polynomials
edited Sep 4 at 23:49
Xander Henderson
13.3k83250
asked Sep 4 at 18:28
Numbers
925
1
Reminder or remainder? This post is now on Hot Network Questions and should probably be cleaned up.
– Wildcard
Sep 4 at 21:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $f=(x+1)^100+(x-3)^100$ and $g=x^2-2x-3$. What is the remainder of $fracfg$? I know there must be something connected to the roots of $g$, but how I can use that? Also I know that the remainder's degree is less or equal to 1. I don't want the response, I want to find that by myself, hints are welcomed.
polynomials
edited Sep 4 at 23:49
Xander Henderson
13.3k83250
asked Sep 4 at 18:28
Numbers
925
Suppose that $f=(x+1)^100+(x-3)^100$ and $g=x^2-2x-3$. What is the remainder of $fracfg$? I know there must be something connected to the roots of $g$, but how I can use that? Also I know that the remainder's degree is less or equal to 1. I don't want the response, I want to find that by myself, hints are welcomed.
polynomials
edited Sep 4 at 23:49
Xander Henderson
13.3k83250
asked Sep 4 at 18:28
Numbers
925
edited Sep 4 at 23:49
Xander Henderson
13.3k83250
edited Sep 4 at 23:49
Xander Henderson
13.3k83250
edited Sep 4 at 23:49
Xander Henderson
13.3k83250
13.3k83250
asked Sep 4 at 18:28
Numbers
925
asked Sep 4 at 18:28
Numbers
925
asked Sep 4 at 18:28
Numbers
925
925
1
Reminder or remainder? This post is now on Hot Network Questions and should probably be cleaned up.
– Wildcard
Sep 4 at 21:24
add a comment |Â
1
Reminder or remainder? This post is now on Hot Network Questions and should probably be cleaned up.
– Wildcard
Sep 4 at 21:24
1
1
Reminder or remainder? This post is now on Hot Network Questions and should probably be cleaned up.
– Wildcard
Sep 4 at 21:24
Reminder or remainder? This post is now on Hot Network Questions and should probably be cleaned up.
– Wildcard
Sep 4 at 21:24
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
7
down vote
accepted
If you put $$f(x)=q(x)g(x)+r(x)$$ where $r(x)$ has degree $1$, and you are not interested at all in $q(x)$, what can you do to make life easier?
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
1
I can use the roots of $g$ to find $r(x)$. Oo, so easy!!
– Numbers
Sep 4 at 18:35
add a comment |Â
up vote
6
down vote
Hint
We have that $$f(x)=(x^2-2x-3)q(x)+ax+b.$$ Since $$x^2-2x-3=(x-3)(x+1)$$ we get
$$f(-1)=-a+b$$ and $$f(3)=3a+b.$$
answered Sep 4 at 18:33
mfl
25.2k12141
add a comment |Â
up vote
2
down vote
Recall that
$$newcommandremoperatornamerem
rem!big(p(x)s(x),q(x)s(x)big)=s(x)rem!big(p(x),q(x)big)
$$
and that
$$
rem!big(p(x),x-abig)=p(a)
$$
Now, since $x^2-2x-3=(x-3)(x+1)$, what we are looking for is
$$
beginalign
&remleft((x+1)^100,(x-3)(x+1)right)+remleft((x-3)^100,(x-3)(x+1)right)\
%&=(x+1)remleft((x+1)^99,x-3right)+(x-3)remleft((x-3)^99,x+1right)\
%&=(x+1)(3+1)^99+(x-3)(-1-3)^99\[3pt]
%&=4^99-3(-4)^99\[3pt]
%&=4^100
endalign
$$
answered Sep 4 at 19:15
robjohn♦
259k26298613
add a comment |Â
up vote
0
down vote
Complicated solution with binomial theorem:
If you write $x-1=t$ then $q(x) = t^2-4$ and $$p(x) = 2Big[100choose 0t^100+...+
2^96100choose 96t^4 +2^98100choose 98t^2 + 2^100Big]$$
Now let $s=t^2$ then $q(x) = s-4= q_1(s)$ and
$$p(x) = 2Big[100choose 0s^50+...+
2^96100choose 96s^2 +2^98100choose 98s + 2^100Big]= p_1(s)$$
So the remanider when $p_1(s)$ is divided by $s-4$ is $$p_1(4)=2Big[100choose 02^100+...+
2^100100choose 96 +2^100100choose 98 + 2^100Big]$$
$$ = 2^101Big[100choose 0+...+
100choose 96 +100choose 98 + 1Big] $$
$$ = 2^100Big((1+1)^100+(1-1)^100Big) =boxed2^200$$
answered Sep 4 at 18:55
greedoid
28k93776
Please be nice???? It take 10 minutes of my life to write this down and then downvoted. Who is not nice? @robjohn
– greedoid
Sep 4 at 19:38
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
If you put $$f(x)=q(x)g(x)+r(x)$$ where $r(x)$ has degree $1$, and you are not interested at all in $q(x)$, what can you do to make life easier?
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
1
I can use the roots of $g$ to find $r(x)$. Oo, so easy!!
– Numbers
Sep 4 at 18:35
add a comment |Â
up vote
7
down vote
accepted
If you put $$f(x)=q(x)g(x)+r(x)$$ where $r(x)$ has degree $1$, and you are not interested at all in $q(x)$, what can you do to make life easier?
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
1
I can use the roots of $g$ to find $r(x)$. Oo, so easy!!
– Numbers
Sep 4 at 18:35
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
If you put $$f(x)=q(x)g(x)+r(x)$$ where $r(x)$ has degree $1$, and you are not interested at all in $q(x)$, what can you do to make life easier?
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
If you put $$f(x)=q(x)g(x)+r(x)$$ where $r(x)$ has degree $1$, and you are not interested at all in $q(x)$, what can you do to make life easier?
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
answered Sep 4 at 18:33
Mark Bennet
77.3k773172
77.3k773172
1
I can use the roots of $g$ to find $r(x)$. Oo, so easy!!
– Numbers
Sep 4 at 18:35
add a comment |Â
1
I can use the roots of $g$ to find $r(x)$. Oo, so easy!!
– Numbers
Sep 4 at 18:35
1
1
I can use the roots of $g$ to find $r(x)$. Oo, so easy!!
– Numbers
Sep 4 at 18:35
I can use the roots of $g$ to find $r(x)$. Oo, so easy!!
– Numbers
Sep 4 at 18:35
add a comment |Â
up vote
6
down vote
Hint
We have that $$f(x)=(x^2-2x-3)q(x)+ax+b.$$ Since $$x^2-2x-3=(x-3)(x+1)$$ we get
$$f(-1)=-a+b$$ and $$f(3)=3a+b.$$
answered Sep 4 at 18:33
mfl
25.2k12141
add a comment |Â
up vote
6
down vote
Hint
We have that $$f(x)=(x^2-2x-3)q(x)+ax+b.$$ Since $$x^2-2x-3=(x-3)(x+1)$$ we get
$$f(-1)=-a+b$$ and $$f(3)=3a+b.$$
answered Sep 4 at 18:33
mfl
25.2k12141
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Hint
We have that $$f(x)=(x^2-2x-3)q(x)+ax+b.$$ Since $$x^2-2x-3=(x-3)(x+1)$$ we get
$$f(-1)=-a+b$$ and $$f(3)=3a+b.$$
answered Sep 4 at 18:33
mfl
25.2k12141
Hint
We have that $$f(x)=(x^2-2x-3)q(x)+ax+b.$$ Since $$x^2-2x-3=(x-3)(x+1)$$ we get
$$f(-1)=-a+b$$ and $$f(3)=3a+b.$$
answered Sep 4 at 18:33
mfl
25.2k12141
answered Sep 4 at 18:33
mfl
25.2k12141
answered Sep 4 at 18:33
mfl
25.2k12141
answered Sep 4 at 18:33
mfl
25.2k12141
25.2k12141
add a comment |Â
add a comment |Â
up vote
2
down vote
Recall that
$$newcommandremoperatornamerem
rem!big(p(x)s(x),q(x)s(x)big)=s(x)rem!big(p(x),q(x)big)
$$
and that
$$
rem!big(p(x),x-abig)=p(a)
$$
Now, since $x^2-2x-3=(x-3)(x+1)$, what we are looking for is
$$
beginalign
&remleft((x+1)^100,(x-3)(x+1)right)+remleft((x-3)^100,(x-3)(x+1)right)\
%&=(x+1)remleft((x+1)^99,x-3right)+(x-3)remleft((x-3)^99,x+1right)\
%&=(x+1)(3+1)^99+(x-3)(-1-3)^99\[3pt]
%&=4^99-3(-4)^99\[3pt]
%&=4^100
endalign
$$
answered Sep 4 at 19:15
robjohn♦
259k26298613
add a comment |Â
up vote
2
down vote
Recall that
$$newcommandremoperatornamerem
rem!big(p(x)s(x),q(x)s(x)big)=s(x)rem!big(p(x),q(x)big)
$$
and that
$$
rem!big(p(x),x-abig)=p(a)
$$
Now, since $x^2-2x-3=(x-3)(x+1)$, what we are looking for is
$$
beginalign
&remleft((x+1)^100,(x-3)(x+1)right)+remleft((x-3)^100,(x-3)(x+1)right)\
%&=(x+1)remleft((x+1)^99,x-3right)+(x-3)remleft((x-3)^99,x+1right)\
%&=(x+1)(3+1)^99+(x-3)(-1-3)^99\[3pt]
%&=4^99-3(-4)^99\[3pt]
%&=4^100
endalign
$$
answered Sep 4 at 19:15
robjohn♦
259k26298613
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Recall that
$$newcommandremoperatornamerem
rem!big(p(x)s(x),q(x)s(x)big)=s(x)rem!big(p(x),q(x)big)
$$
and that
$$
rem!big(p(x),x-abig)=p(a)
$$
Now, since $x^2-2x-3=(x-3)(x+1)$, what we are looking for is
$$
beginalign
&remleft((x+1)^100,(x-3)(x+1)right)+remleft((x-3)^100,(x-3)(x+1)right)\
%&=(x+1)remleft((x+1)^99,x-3right)+(x-3)remleft((x-3)^99,x+1right)\
%&=(x+1)(3+1)^99+(x-3)(-1-3)^99\[3pt]
%&=4^99-3(-4)^99\[3pt]
%&=4^100
endalign
$$
answered Sep 4 at 19:15
robjohn♦
259k26298613
Recall that
$$newcommandremoperatornamerem
rem!big(p(x)s(x),q(x)s(x)big)=s(x)rem!big(p(x),q(x)big)
$$
and that
$$
rem!big(p(x),x-abig)=p(a)
$$
Now, since $x^2-2x-3=(x-3)(x+1)$, what we are looking for is
$$
beginalign
&remleft((x+1)^100,(x-3)(x+1)right)+remleft((x-3)^100,(x-3)(x+1)right)\
%&=(x+1)remleft((x+1)^99,x-3right)+(x-3)remleft((x-3)^99,x+1right)\
%&=(x+1)(3+1)^99+(x-3)(-1-3)^99\[3pt]
%&=4^99-3(-4)^99\[3pt]
%&=4^100
endalign
$$
answered Sep 4 at 19:15
robjohn♦
259k26298613
edited Sep 4 at 19:31
answered Sep 4 at 19:15
robjohn♦
259k26298613
answered Sep 4 at 19:15
robjohn♦
259k26298613
answered Sep 4 at 19:15
robjohn♦
259k26298613
259k26298613
add a comment |Â
add a comment |Â
up vote
0
down vote
Complicated solution with binomial theorem:
If you write $x-1=t$ then $q(x) = t^2-4$ and $$p(x) = 2Big[100choose 0t^100+...+
2^96100choose 96t^4 +2^98100choose 98t^2 + 2^100Big]$$
Now let $s=t^2$ then $q(x) = s-4= q_1(s)$ and
$$p(x) = 2Big[100choose 0s^50+...+
2^96100choose 96s^2 +2^98100choose 98s + 2^100Big]= p_1(s)$$
So the remanider when $p_1(s)$ is divided by $s-4$ is $$p_1(4)=2Big[100choose 02^100+...+
2^100100choose 96 +2^100100choose 98 + 2^100Big]$$
$$ = 2^101Big[100choose 0+...+
100choose 96 +100choose 98 + 1Big] $$
$$ = 2^100Big((1+1)^100+(1-1)^100Big) =boxed2^200$$
answered Sep 4 at 18:55
greedoid
28k93776
Please be nice???? It take 10 minutes of my life to write this down and then downvoted. Who is not nice? @robjohn
– greedoid
Sep 4 at 19:38
add a comment |Â
up vote
0
down vote
Complicated solution with binomial theorem:
If you write $x-1=t$ then $q(x) = t^2-4$ and $$p(x) = 2Big[100choose 0t^100+...+
2^96100choose 96t^4 +2^98100choose 98t^2 + 2^100Big]$$
Now let $s=t^2$ then $q(x) = s-4= q_1(s)$ and
$$p(x) = 2Big[100choose 0s^50+...+
2^96100choose 96s^2 +2^98100choose 98s + 2^100Big]= p_1(s)$$
So the remanider when $p_1(s)$ is divided by $s-4$ is $$p_1(4)=2Big[100choose 02^100+...+
2^100100choose 96 +2^100100choose 98 + 2^100Big]$$
$$ = 2^101Big[100choose 0+...+
100choose 96 +100choose 98 + 1Big] $$
$$ = 2^100Big((1+1)^100+(1-1)^100Big) =boxed2^200$$
answered Sep 4 at 18:55
greedoid
28k93776
Please be nice???? It take 10 minutes of my life to write this down and then downvoted. Who is not nice? @robjohn
– greedoid
Sep 4 at 19:38
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Complicated solution with binomial theorem:
If you write $x-1=t$ then $q(x) = t^2-4$ and $$p(x) = 2Big[100choose 0t^100+...+
2^96100choose 96t^4 +2^98100choose 98t^2 + 2^100Big]$$
Now let $s=t^2$ then $q(x) = s-4= q_1(s)$ and
$$p(x) = 2Big[100choose 0s^50+...+
2^96100choose 96s^2 +2^98100choose 98s + 2^100Big]= p_1(s)$$
So the remanider when $p_1(s)$ is divided by $s-4$ is $$p_1(4)=2Big[100choose 02^100+...+
2^100100choose 96 +2^100100choose 98 + 2^100Big]$$
$$ = 2^101Big[100choose 0+...+
100choose 96 +100choose 98 + 1Big] $$
$$ = 2^100Big((1+1)^100+(1-1)^100Big) =boxed2^200$$
answered Sep 4 at 18:55
greedoid
28k93776
Complicated solution with binomial theorem:
If you write $x-1=t$ then $q(x) = t^2-4$ and $$p(x) = 2Big[100choose 0t^100+...+
2^96100choose 96t^4 +2^98100choose 98t^2 + 2^100Big]$$
Now let $s=t^2$ then $q(x) = s-4= q_1(s)$ and
$$p(x) = 2Big[100choose 0s^50+...+
2^96100choose 96s^2 +2^98100choose 98s + 2^100Big]= p_1(s)$$
So the remanider when $p_1(s)$ is divided by $s-4$ is $$p_1(4)=2Big[100choose 02^100+...+
2^100100choose 96 +2^100100choose 98 + 2^100Big]$$
$$ = 2^101Big[100choose 0+...+
100choose 96 +100choose 98 + 1Big] $$
$$ = 2^100Big((1+1)^100+(1-1)^100Big) =boxed2^200$$
answered Sep 4 at 18:55
greedoid
28k93776
edited Sep 4 at 19:01
answered Sep 4 at 18:55
greedoid
28k93776
answered Sep 4 at 18:55
greedoid
28k93776
answered Sep 4 at 18:55
greedoid
28k93776
28k93776
Please be nice???? It take 10 minutes of my life to write this down and then downvoted. Who is not nice? @robjohn
– greedoid
Sep 4 at 19:38
add a comment |Â
Please be nice???? It take 10 minutes of my life to write this down and then downvoted. Who is not nice? @robjohn
– greedoid
Sep 4 at 19:38
Please be nice???? It take 10 minutes of my life to write this down and then downvoted. Who is not nice? @robjohn
– greedoid
Sep 4 at 19:38
Please be nice???? It take 10 minutes of my life to write this down and then downvoted. Who is not nice? @robjohn
– greedoid
Sep 4 at 19:38
add a comment |Â
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1
Reminder or remainder? This post is now on Hot Network Questions and should probably be cleaned up.
– Wildcard
Sep 4 at 21:24