Remainder Thm

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SiJo

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If p(x) = 1 + x + x^2 + x^3 + x^4 + x^5, what is the remainder when p(x^6) is divided by p(x)?

Tried verifying but the numbers got very large, long division seems long - am I missing insight that provides a short cut?
 
SiJo said:
Remainer Thm

Tried verifying but the numbers got very large … am I missing insight
Click to expand...
Hello. Were you trying to use the Remainder Theorem? Please show the beginnings of your attempt. Thanks!

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SiJo said:
am I missing insight that provides a short cut?
Click to expand...
See the following page about the all one polynomial (click). Pay particular attention to the third definition (and write it out for the case when m=5)

EDIT: From that page
[math]AOP_m(x) =\sum_{i=0}^mx^i\\ =x^m+x^{m-1}+\cdots+x+1\\ =\dfrac{x^{m+1}-1}{x-1}[/math]
 
Last edited:
p(6x) = ( (6x)^6-1)/(x-1)

p(x) = (x^6-1)/(x-1)

So p(6x)/p(x) = ?
 
Cubist said:
See the following page about the all one polynomial (click). Pay particular attention to the third definition (and write it out for the case when m=5)

EDIT: From that page
[math]AOP_m(x) =\sum_{i=0}^mx^i\\ =x^m+x^{m-1}+\cdots+x+1\\ =\dfrac{x^{m+1}-1}{x-1}[/math]
Click to expand...
mmm4444bot said:
Hello. Were you trying to use the Remainder Theorem? Please show the beginnings of your attempt. Thanks!

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...
www.freemathhelp.com
[imath]\;[/imath]
Click to expand...
x+1 is clearly a factor of p(x) but not p(x^6)
the notes from wikipedia cover dividing by a linear factor but this is a quintic so not sure how to apply
 
SiJo said:
x+1 is clearly a factor of p(x) but not p(x^6)
the notes from wikipedia cover dividing by a linear factor but this is a quintic so not sure how to apply
Click to expand...
[math]\dfrac{y^{m+1}-1}{y-1} = y^m+y^{m-1}+\cdots+y+1[/math]Let y=x^6
[math]\dfrac{x^{6(m+1)}-1}{x^6-1}=x^{6m}+x^{6(m-1)}+\cdots+x^6+1[/math][math]x^{6(m+1)}-1 =(x^6-1) * (\text{a poly in x}) \,\text{ call this eqn AA}[/math]--
P(x^6) = x^30 + x^24 + x^18 + x^12 + x^6 + 1

Can you use AA to make the above look like...

P(x^6) = (x^6 - 1)*(some polynomial in x) + (something simple that can't be divided by x^6-1)

(You need to find the bold part of the above - but you can leave the "some polynomial in x" part as it is. You don't need to know that polynomial. ). Then continue to divide this by P(x)...
 
While the method proposed in post#7 definitely works, I'm concerned that there might be an easier way to solve this using the remainder theorem. The title of this thread kind of implies this is required. I can't (yet) see a way of doing that.
 
SiJo said:
x+1 is clearly a factor of p(x) but not p(x^6)
the notes from wikipedia cover dividing by a linear factor but this is a quintic so not sure how to apply
Click to expand...
1+x+x^2 is clearly a factor of p(x) as well.
 
Cubist said:
While the method proposed in post#7 definitely works, I'm concerned that there might be an easier way to solve this using the remainder theorem. The title of this thread kind of implies this is required. I can't (yet) see a way of doing that.
Click to expand...
I've worked out how to answer this using something similar to the remainder theorem...

P(x^6) = P(x)*Q(x) + R(x)
P(x^6) = (x^6 - 1)/(x - 1)*Q(x) + R(x)
P(x^6)*(x - 1) = (x^6 - 1)*Q(x) + R(x)*(x - 1)
...can you see what to do now in order to find R(x), which could be a poly of maximum degree 4?
 
Steven G said:
p(6x) = ( (6x)^6-1)/(x-1)

p(x) = (x^6-1)/(x-1)

So p(6x)/p(x) = ?
Click to expand...
Cubist said:
I've worked out how to answer this using something similar to the remainder theorem...

P(x^6) = P(x)*Q(x) + R(x)
P(x^6) = (x^6 - 1)/(x - 1)*Q(x) + R(x)
P(x^6)*(x - 1) = (x^6 - 1)*Q(x) + R(x)*(x - 1)
...can you see what to do now in order to find R(x), which could be a poly of maximum degree 4?
Click to expand...
I think we must start here.

[math]p(x) = 1 + x + x^2 + x^3 + x^4 + x^5 \implies\\ p(x^6) = 1 + x^6 + x^{12} + x^{18} + x^{24} + x^{30}, \\ p(x) = \dfrac{x^6 - 1}{x - 1} \text { if } x \ne 1 \text { and } p(x) = 6 \text { if } x = 1, \text { and}\\ p(x^6)= \dfrac{x^{36} - 1}{x^6 - 1} \text { if } x \ne 1 \text { and } p(x^6) = 6 \text { if } x = 1. [/math]
Or have I missed the boat?
 
JeffM said:
I think we must start here.
Click to expand...
I can't see where this would lead, but you might be right. Please feel free to send me a private message and I'll do my best to verify your working (without giving the full answer to SiJo, who will learn more if they can complete most of this puzzle themselves). In fact, I'll aim to send you and Steven G a private message with my own full answer so that you can see where I'm trying to lead SiJo in post#11. Sorry, I'm a bit busy today so it might take a few hours before I get around to this.

JeffM said:
[math]p(x) = 1 + x + x^2 + x^3 + x^4 + x^5 \implies\\ p(x) = \dfrac{x^6 - 1}{x - 1} \text { if } x \ne 1 \text { and } p(x) = 6 \text { if } x = 1, \text { and}\\ [/math]
Click to expand...
Normally using the remainder theorem we'd plug x=1 into x^6 - 1 rather than into (x^6-1)/(x-1) and doing this would give a remainder of 0. We'd expect this because (x^6-1)/(x-1) divides exactly producing 1+x+x^2+x^3+x^4+x^5
 
Cubist said:
I can't see where this would lead, but you might be right. Please feel free to send me a private message and I'll do my best to verify your working (without giving the full answer to SiJo, who will learn more if they can complete most of this puzzle themselves). In fact, I'll aim to send you and Steven G a private message with my own full answer so that you can see where I'm trying to lead SiJo in post#11. Sorry, I'm a bit busy today so it might take a few hours before I get around to this.


Normally using the remainder theorem we'd plug x=1 into x^6 - 1 rather than into (x^6-1)/(x-1) and doing this would give a remainder of 0. We'd expect this because (x^6-1)/(x-1) divides exactly producing 1+x+x^2+x^3+x^4+x^5
Click to expand...
Yes. I thought I could get to a simplification my way. I was wrong.
 
Cubist said:
I've worked out how to answer this using something similar to the remainder theorem...

P(x^6) = P(x)*Q(x) + R(x)
P(x^6) = (x^6 - 1)/(x - 1)*Q(x) + R(x)
P(x^6)*(x - 1) = (x^6 - 1)*Q(x) + R(x)*(x - 1)
...can you see what to do now in order to find R(x), which could be a poly of maximum degree 4?
Click to expand...
 
SiJo said:
keep getting in to a loop so no......
Click to expand...

Cubist said:
P(x^6)*(x - 1) = (x^6 - 1)*Q(x) + R(x)*(x - 1)
Click to expand...

P(x^6)*(x - 1) = (x^6 - 1)*Q(x) + R(x)*(x - 1)

What 5 values of x make the red term zero, but leave the other two terms non-zero? This leads to a solvable set of 5 equations to find all the constants in...

R(x) = a + b*x + c*x^2 + d*x^3 + e*x^4

For example (these aren't the numbers you'll get with the question in this thread), we could find a,b,c,d and e if we knew the following...
P(1^6)=8=R(1), P(2^6)=8=R(2), R(3)=8, R(4)=8, R(5)=8
a + b*1 + c*1^2 + d*1^3 + e*1^4 = 8
a + b*2 + c*2^2 + d*2^3 + e*2^4 = 8
a + b*3 + c*3^2 + d*3^3 + e*3^4 = 8
a + b*4 + c*4^2 + d*4^3 + e*4^4 = 8
a + b*5 + c*5^2 + d*5^3 + e*5^4 = 8
...obviously a=8, b=c=d=e=0

Plug the 5 values of x into P(x^6), and note that x is raised to power 6, and thus find a,b,c,d & e
 
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