Hello. Were you trying to use the Remainder Theorem? Please show the beginnings of your attempt. Thanks!Remainer Thm
Tried verifying but the numbers got very large … am I missing insight
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)am I missing insight that provides a short cut?
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]
x+1 is clearly a factor of p(x) but not p(x^6)Hello. Were you trying to use the Remainder Theorem? Please show the beginnings of your attempt. Thanks!
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can't see how 6x is relevant as dealing with x^6p(6x) = ( (6x)^6-1)/(x-1)
p(x) = (x^6-1)/(x-1)
So p(6x)/p(x) = ?
[math]\dfrac{y^{m+1}-1}{y-1} = y^m+y^{m-1}+\cdots+y+1[/math]Let y=x^6x+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
1+x+x^2 is clearly a factor of p(x) as well.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
Just ignore this post!p(6x) = ( (6x)^6-1)/(x-1)
p(x) = (x^6-1)/(x-1)
So p(6x)/p(x) = ?
I've worked out how to answer this using something similar to the remainder theorem...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.
p(6x) = ( (6x)^6-1)/(x-1)
p(x) = (x^6-1)/(x-1)
So p(6x)/p(x) = ?
I think we must start here.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?
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.I think we must start here.
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[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]
Yes. I thought I could get to a simplification my way. I was wrong.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
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?
keep getting in to a loop so no......
P(x^6)*(x - 1) = (x^6 - 1)*Q(x) + R(x)*(x - 1)