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The Remainder Theorem

Guys I need help with the following question.
The remainders when f (x)=2x^3-3x^2+ax+b is divided by x+1, x-1 and x-2
respectively form an Arithmetic Progression. Find a and b

Ok so any ideas in how to find and a and b? I have substituted the x values in and obtained the remainder. But now I am stuck.
Reply 1
Avatar for PrinceyJ
PrinceyJ
12
So you should have 3 equations?

Just pick two of them and solve them simultaneously.

You'll have (something)a+b=something and (something)a+b=something else.

EDIT: I didn't see the arithmetic progression bit, what I said above is nonsense.
(edited 11 years ago)
Reply 2
Avatar for steve2005
steve2005
17
Original post by Snazarina
Guys I need help with the following question.
The remainders when f (x)=2x^3-3x^2+ax+b is divided by x+1, x-1 and x-2
respectively form an Arithmetic Progression. Find a and b

Ok so any ideas in how to find and a and b? I have substituted the x values in and obtained the remainder. But now I am stuck.


I get a = 1 but I think b could be any number. Have you posted the complete question.

Sorry I now get a = 1
(edited 11 years ago)
Original post by Snazarina
Guys I need help with the following question.
The remainders when f (x)=2x^3-3x^2+ax+b is divided by x+1, x-1 and x-2
respectively form an Arithmetic Progression. Find a and b

Ok so any ideas in how to find and a and b? I have substituted the x values in and obtained the remainder. But now I am stuck.


So, you've worked out the remainders, and these remainders form an arithmetic progression, call it a0, a1, a2.

Now you need to use the structure of an AP to work out what a is.

So using the common difference of an AP we have, a1-a0 = a2-a1

Unless I'm missing something, you can't work out what b is.

Is there more info?
Original post by steve2005
I get a = 0


I make it 1.
Reply 5
Avatar for steve2005
steve2005
17
Original post by ghostwalker
I make it 1.


Yes, I agree a=1
The second post isn't correct.

f(2) - f(1) = f(1) - f(-1)

This is because the common difference is a constant.

This will give you the value of a in the cubic.

You cannot find the value of b because it could be anything and the terms would still form an A.P.
Original post by ghostwalker
So, you've worked out the remainders, and these remainders form an arithmetic progression, call it a0, a1, a2.

Now you need to use the structure of an AP to work out what a is.

So using the common difference of an AP we have, a1-a0 = a2-a1

Unless I'm missing something, you can't work out what b is.

Is there more info?


You aren't missing anything. I have been faffing about with this for about 15 mins!
Original post by Mr M
You aren't missing anything. I have been faffing about with this for about 15 mins!


I thought perhaps an additional requirement that they also form a GP, but turns out that's impossible.
Original post by ghostwalker
I thought perhaps an additional requirement that they also form a GP, but turns out that's impossible.


I expect there was something else in the question that wasn't shared with us but we may never know.
Reply 10
Avatar for Snazarina
Snazarina
OP
2
As many of you assumed that there was a lack of information in the given question, I have therefore attached an extract of the question.
Much thanks to those who have tried solving it. I guess it is one of those that had been mistyped and therefore is impossible to solve.

Posted from TSR Mobile
uploadfromtaptalk1358660672264.jpg14.0KB
Original post by Snazarina
...


Thanks for answering my question from post #4.

Yes; error in the question, then.

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