Ahhh, thanks!
Year 13 Maths Help Thread
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 1001
 18102016 19:44

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 1002
 18102016 19:49
Also, what does this even mean...

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 1003
 18102016 20:09
In this case, dividing p(x) by (x1) gives a remainder of 2, so we need to add two in order to get a remainder of 0, therefore p(x)=(x1)q(x)2.
I'm not 100% sure on this though...Last edited by solC; 18102016 at 21:18. 
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 1004
 18102016 20:15

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 1005
 18102016 20:20
(Original post by solC)
If p(1)=0 then (x1) is a factor of p(x) from the remainder theorem, so we can rewrite p(x) as q(x)(x1) for some polynomial q(x).
In this case, dividing p(x) by (x1) gives a remainder of 2, so we need to subtract two in order to get a remainder of 0, therefore p(x)=(x1)q(x)2.
I'm not 100% sure on this though... 
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 1006
 18102016 20:21
(Original post by solC)
If p(1)=0 then (x1) is a factor of p(x) from the remainder theorem, so we can rewrite p(x) as q(x)(x1) for some polynomial q(x).
In this case, dividing p(x) by (x1) gives a remainder of 2, so we need to subtract two in order to get a remainder of 0, therefore p(x)=(x1)q(x)2.
I'm not 100% sure on this though...(Original post by RDKGames)
Well firstly we can eliminate options which prevent p(1)=2 and then investigate the remaining ones, you can quickly see that all of them depend on various conditions satisfied by q(1) apart from one which will always be true, so I'd go for that one. 
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 1007
 18102016 20:28
(Original post by RDKGames)
Can't be true. Plugging x=1 into that would leave you with 2 so p(1)=2 is not satisfied. 
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 1008
 18102016 20:38
(Original post by solC)
Ah yeah, didn't notice that. so you would have to add on 2 rather than subtract?Post rating:1 
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 1009
 18102016 20:42
(Original post by RDKGames)
Well going off your attempt it wouldn't make sense because you said "p(1)=0" which is also false so you cannot use the remainder theorem.
jamestg Ignore what I said! 
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 1010
 18102016 21:03
Erm duno wtf is goingn on above.
Consider p(x)2=r(x)
x=1 is solution to r(x) so r(x)=(x1)q(x)
so p(x)=(x1)q(x)+2
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 1011
 19102016 02:00
Last edited by NotNotBatman; 19102016 at 02:03. 
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 1012
 19102016 15:18
An employer finds that if he increase weekly wages of each worker by £30 and employs one worker less,he reduces his weekly wage bill from £8160 to £7810, taking the weekly wage of each worker as x ,find out the weekly wage of each worker
how do I do this?
I can't find a final answer to this either 
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 1013
 19102016 15:31
Post rating:1 
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 1014
 19102016 15:32
(Original post by physicsmaths)
Erm duno wtf is goingn on above.
Consider p(x)2=r(x)
x=1 is solution to r(x) so r(x)=(x1)q(x)
so p(x)=(x1)q(x)+2
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 1015
 19102016 15:46
(Original post by Zacken)
init plebs everywhere 
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 1016
 19102016 17:53
So basically the question says to find the stationary points of a curve which I've found. It then says justify which is a max and which is a min. Surely that is obvious which one is higher from the y coordinates I just worked out?

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 1017
 19102016 18:04
(Original post by medhelp)
An employer finds that if he increase weekly wages of each worker by £30 and employs one worker less,he reduces his weekly wage bill from £8160 to £7810, taking the weekly wage of each worker as x ,find out the weekly wage of each worker
how do I do this?
The first equation would be nx = 8160. The rest is left as an exercise for the reader
If only the question writer knew the difference between less and fewer ..Last edited by RogerOxon; 19102016 at 18:05. Reason: Fewer! 
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 1018
 19102016 18:16
(Original post by RogerOxon)
You have two unknowns, the number of employees (n) and their weekly wage (x). You can form two equations for the before and after wage bills. You can then solve for x.
The first equation would be nx = 8160. The rest is left as an exercise for the reader
If only the question writer knew the difference between less and fewer ..
okay, so I got the second equation as (n1) x (x+30) = 7810 and substituted in x = 8160 /n
I did a load of working out and somehow I've gotten to 3n^2  32n  816 = 0 which seems like hell to factorise so I'm assuming I've done it wrong,
:/ 
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 1019
 19102016 19:14
(Original post by medhelp)
what do you mean by the bit in bold
(Original post by medhelp)
I did a load of working out and somehow I've gotten to 3n^2  32n  816 = 0 which seems like hell to factorise so I'm assuming I've done it wrong,:/
You have:
nx = 8160 and (n1)(x+30) = 7810
Expanding the second gives: (n1)x + 30(n1) = 7810
To make the numbers simpler, I then subtracted the second from the first, giving:
x  30(n1) = 350
We're after x, so substitute for n (from nx=8160), and you should get an equation that you can solve with the quadratic formula, taking the positive root.Last edited by RogerOxon; 19102016 at 19:16. 
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 1020
 19102016 19:49
(Original post by RogerOxon)
Sorry  I'm a (partial) grammar pedant. It's fewer for countable things, not less.
That quadratic doesn't look correct to me.
You have:
nx = 8160 and (n1)(x+30) = 7810
Expanding the second gives: (n1)x + 30(n1) = 7810
To make the numbers simpler, I then subtracted the second from the first, giving:
x  30(n1) = 350
We're after x, so substitute for n (from nx=8160), and you should get an equation that you can solve with the quadratic formula, taking the positive root.
x^2  320 x + 244800 = 0
that almost looks worse than the last quadratic
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Updated: December 7, 2016
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