# Maths Problem

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Bhav

This isn't any homework or anything, so don't think I'm trying to get the answers off you guys.

It's a problem our teach thought we might like to try:

"A certain polynomial p(x) when divided by x - a, x - b, x - c, leaves a, b, c, respectively, as the remainder.

What is the remainder when p(x) is divided by (x - a)(x - b)(x - c)?

NB: a, b, c are distinct"

Something to puzzle over... :)

edit:

Teach admits it is a difficult question.

He also adds that he worded the question poorly...in the first part where it says "leaves a, b, c respectively" - he meant to say it leaves a, b and c, respectively, as the remainder

Solution on Page 2 in PDF

Edited by Bhav
Bhav

no one? :(

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Omar Hussain

im guessing not bhav.... hehe unlucky on this one, not everyone is as educated as you i am thinkin

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User6060

I thought long and hard, and I am wondering if this is possible.

Can I clarify the question?

Correct me if I'm wrong.

P(x) / x-a = a

P(x) / x-b = b

P(x) / x-c = c

a ≠ b

a ≠ c

b ≠ c

P(x) / (x-a)(x-b)(x-b)

Remainder= ?

Edited by Co_Co
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2xSilverKnight

x=0 :rofl: :whistle:

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Rob2687
x=0  :rofl: :whistle:

586652657[/snapback]

OMG that's genious! :p

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Borbus

What maths course are you studying? I have an A-level in Maths and I don't understand most of what you have written there, unless its just poorly written due to limitation of the board.

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Bhav

i'm doing MEI.

but this isnt an exam board question.

like i mentioned, it's just a maths quiz type thing that we can do if we like. basically the teacher looks at solutions people give in, ranks them, marks them and puts up a league table.

i've written the question in the exact same way it was given to me.

this answer to the problem doesnt count towards anything, so it's not a big deal if i dont get the answer...but i just thought it'd be an interesting puzzle to work through.

i'll post the solution later this week (once my teacher's given it to me)

@Co_Co: yes you're right.

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Bhav

eh?! really?! well i'm doing further maths too...but that'll be the end of this year + next year.

we're still doing C1-4 (Pure1, 2 and more), and S1, and M1 (yes all at once lol)

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Guest

I did MEI - Further Maths and got an A :D - Ill have a look at this in a second :) - Good ole remainder theorem - easy marks in the exam. Pure 5 this is, isnt it?

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Guest

Pfft - the new maths syllabus has been dumbed down. We had Pure 1,2,3,4,5 to do and Mechs 1,2,3,4 and Stats 1,2 and D n D 1.

Where our Pure's are the equivalent of 1.5 of ure Cores :p

*steps down from high platform*

I cant remember the remainder theorem - ill have a look at my old notes :-p

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Borbus

^Yeah I did that kind of maths course, I got an A too :D

It has been made really easy now. Apparently they don't even do calculus in C1...

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Bhav

yep there's no calculus in C1 (with MEI anyway). calculus only starts at C2.

i guess you guys are the experts on whether it's been dumbed down or not (which is why i'm not condemning what you say ;))

This is the system in our school:

Everyone does C1-C4 and FP1 (as with all exam boards - they're the only compulsory modules).

Everyone, in our school, also has to do S1 and M1. Then I think we can go in whichever direction we want...but only stats or mechs. they dont teach anything like decision maths. i think we can do differential equations though.

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FiREFLi
Pfft - the new maths syllabus has been dumbed down. We had Pure 1,2,3,4,5 to do and Mechs 1,2,3,4 and Stats 1,2 and D n D 1.

586653102[/snapback]

Try Pure 1,2,3,4,5,6 and Stats 1,2,3,4,7,8 ;)
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Bhav

while we're on this, and people are thinking about the problem...could someone give me an example of a further maths question?

i just cant imagine how complex a problem can get - it feels like we're quite involved already.

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kylejn

Stop saying "maths," you crazy UK'ers! It's "math," dammit! :rofl:

As for the problem, been a while since I've done anything that wasn't, like, Calc 3D or anything like that, so I can't help ya.

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Bhav

no - it's maths because it's mathematics and not mathematic ;)

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amrinders87
no - it's maths because it's mathematics and not mathematic ;)

586655117[/snapback]

Just because mathematics has a s at the end doesnt say that math should have an s.

evo0o

(a+b+c)

(:

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Borbus
while we're on this, and people are thinking about the problem...could someone give me an example of a further maths question?

586653395[/snapback]

Well, theres no such thing as "a further maths question". MEI Structured Maths is made up of a load of modules as you know. Basically six modules is one A level. Therefore a maths A level consists of 6 modules. A further maths A level consists of a further 6 modules.

I don't know what the system is this year, but last year, for AS Maths you had to do P1, P2 and either S1 or M1. For the A2 you had to do P3 and a "2" (eg M2 or S2) (follwing dependencies, eg, you can't do M2 without doing M1) and then you get a choice of what other one you want to do. Your school makes these choices usually depending on what teachers they have.

Further maths, as far as I know can be any 6 of the load of MEI courses, again following dependancies. There are loads, Pur, Mechanics and Stats go up to 6, and there are weird ones like Desicion and Discrete maths and stupid stuff like that. Of course, this is the old courses, the new ones will be different but similar.

Anyway, after that massive explanation (sorry, I have a bit of time to waste now between getting back from work and dinner :p), to answer your question. I suppose "a further maths question" will be one from P3, M3 or S4, or whatever the equivalent is now. Any question from a course that you can't do in normal maths, like those mentioned above, can be considered a further maths question. Its not really more complicated, theres just new topics. Although the questions to draw on topics from previous courses (for example, M2 draws on all of M1 plus its own stuff) so in that sense it gets more complicated. But I found that the stuff from M1 gets embedded in your head so much that its easy to remember it for M2, it just becomes natural.

I don't have any of my text books any more but at school just ask your teacher if you can take a look in a M3 or S3 book. You won't understand the questions though becasue they will be about stuff you won't have done. Hope that answers your question :)

Oh, and by the way, my Grandma says Math instead of Maths, maybe it is us that are wrong...

Hum
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Bhav

thanks BobMarley :)

and you're right about the way the further maths a level is made up.

the normal maths can be made up of C1-C4, and then any 2, and f.maths is FP1, and any 5. But MEI themselves mix and match so that you get the best possible a level maths grade with the lowest possible marks - ie they use your worst modules to get you the A, so you have the best chance of getting an A in f.maths - that's nice of them :D

I've added some stuff to the original post that my teacher told me today

He says it is a difficult question.

He also adds that he worded the question poorly...in the first part where it says "leaves a, b, c respectively" - he meant to say it leaves a, b and c, respectively, as the remainder

hope that helps.

Edited by Bhav
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kylejn
no - it's maths because it's mathematics and not mathematic ;)

586655117[/snapback]

True, but what about the abbreviation for economics, econ? It's not "econs" in the UK, is it?

Point, USA. ;)

Now with the added info about the remainder being a, b, or c, well, I still don't remember how it should be done. Sorry!

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brandon.mat

"A certain polynomial p(x) when divided by x - a, x - b, x - c, leaves a, b, c, respectively, as the remainder.

What is the remainder when p(x) is divided by (x - a)(x - b)(x - c)?

NB: a, b, c are distinct"

you must work backwords on problem like these then make equations out of it. So far you know the remainder and the equation. What i dont understand though is did he first divide it by x- a then x - b then x - c then it lieft abc? or did he divide it by a and it left a as the remainder? Becuase if so then the polynomial would be a perfect square of all 3 numbers, which is theoretically impossible, but if so in the first place then. (x - a)(x-b)(x-c) = p(x) therefore the remainder of p(x) / (x-a)(x-b)(x-c) would be 0 ...the answer of that division is 1.