# The Remainder Theorem Watch

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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.

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.

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#2

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.

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.

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#3

(Original post by

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.

**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.

Sorry I now get a = 1

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#4

**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.

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?

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#7

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.

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.

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#8

(Original post by

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?

**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?

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#9

(Original post by

You aren't missing anything. I have been faffing about with this for about 15 mins!

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

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#10

(Original post by

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

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

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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

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

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#12

(Original post by

...

**Snazarina**)...

Yes; error in the question, then.

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