# C3 Remainder theorem. how do you know what the quotient and remainder form will be?

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right, I understand when i get a question like x^3 + 2x^2+4x +3 divide that by (x+3) i get a quotient multiplied by divisor + a remainder.

This puts it into form like (AX^2+BX+C)(X+3) +D

I then solve it to find a b, c, d. I can solve this fine.

However, depending on the question the form of the quotient and the remaindeer will be different, e..g the remainder might be CX + D, etc.

How can I tell what form it will be in?

a rough method I just made up was look at the original expression and look at the divisor, if the divisor is (x-2) and the expression is x^3+3x^2-6x-3. I know that my quotient will only have things in it, which when multipled by the x from the x-2 make the original expression. e.g. the quotient will be (Ax^2+Bx-C) as I know that if i multipy everything by x I get basically the orioginal experession back except for the -3 which I treat as the remainder and thereofr just D.

Does that make sense as a method, is it ok?

This puts it into form like (AX^2+BX+C)(X+3) +D

I then solve it to find a b, c, d. I can solve this fine.

However, depending on the question the form of the quotient and the remaindeer will be different, e..g the remainder might be CX + D, etc.

How can I tell what form it will be in?

a rough method I just made up was look at the original expression and look at the divisor, if the divisor is (x-2) and the expression is x^3+3x^2-6x-3. I know that my quotient will only have things in it, which when multipled by the x from the x-2 make the original expression. e.g. the quotient will be (Ax^2+Bx-C) as I know that if i multipy everything by x I get basically the orioginal experession back except for the -3 which I treat as the remainder and thereofr just D.

Does that make sense as a method, is it ok?

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#2

(Original post by

right, I understand when i get a question like x^3 + 2x^2+4x +3 divide that by (x+3) i get a quotient multiplied by divisor + a remainder.

This puts it into form like (AX^2+BX+C)(X+3) +D

I then solve it to find a b, c, d. I can solve this fine.

However, depending on the question the form of the quotient and the remaindeer will be different, e..g the remainder might be CX + D, etc.

How can I tell what form it will be in?

a rough method I just made up was look at the original expression and look at the divisor, if the divisor is (x-2) and the expression is x^3+3x^2-6x-3. I know that my quotient will only have things in it, which when multipled by the x from the x-2 make the original expression. e.g. the quotient will be (Ax^2+Bx-C) as I know that if i multipy everything by x I get basically the orioginal experession back except for the -3 which I treat as the remainder and thereofr just D.

Does that make sense as a method, is it ok?

**f45**)right, I understand when i get a question like x^3 + 2x^2+4x +3 divide that by (x+3) i get a quotient multiplied by divisor + a remainder.

This puts it into form like (AX^2+BX+C)(X+3) +D

I then solve it to find a b, c, d. I can solve this fine.

However, depending on the question the form of the quotient and the remaindeer will be different, e..g the remainder might be CX + D, etc.

How can I tell what form it will be in?

a rough method I just made up was look at the original expression and look at the divisor, if the divisor is (x-2) and the expression is x^3+3x^2-6x-3. I know that my quotient will only have things in it, which when multipled by the x from the x-2 make the original expression. e.g. the quotient will be (Ax^2+Bx-C) as I know that if i multipy everything by x I get basically the orioginal experession back except for the -3 which I treat as the remainder and thereofr just D.

Does that make sense as a method, is it ok?

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#3

(Original post by

This is Core 3? Which chapter?

**beatleboar**)This is Core 3? Which chapter?

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#6

(Original post by

This is all in Chapter 1 (Edexcel).

**olipal**)This is all in Chapter 1 (Edexcel).

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#7

pretty sure remainder depends on the divisor, doubt there is a general form for it.

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#8

(Original post by

I do AQA so this could be why.

**beatleboar**)I do AQA so this could be why.

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#9

(Original post by

pretty sure this is C1

**boromir9111**)pretty sure this is C1

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#10

(Original post by

I'm pretty sure I did it in C1. OCR.

**Toneh**)I'm pretty sure I did it in C1. OCR.

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#12

(Original post by

anyone?

p.s. cheers for the 9 replies of "i do aqa" well done.

**f45**)anyone?

p.s. cheers for the 9 replies of "i do aqa" well done.

Highest power of remainder = 1 lower than highest power of divisor.

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#13

**f45**)

right, I understand when i get a question like x^3 + 2x^2+4x +3 divide that by (x+3) i get a quotient multiplied by divisor + a remainder.

This puts it into form like (AX^2+BX+C)(X+3) +D

I then solve it to find a b, c, d. I can solve this fine.

However, depending on the question the form of the quotient and the remaindeer will be different, e..g the remainder might be CX + D, etc.

How can I tell what form it will be in?

a rough method I just made up was look at the original expression and look at the divisor, if the divisor is (x-2) and the expression is x^3+3x^2-6x-3. I know that my quotient will only have things in it, which when multipled by the x from the x-2 make the original expression. e.g. the quotient will be (Ax^2+Bx-C) as I know that if i multipy everything by x I get basically the orioginal experession back except for the -3 which I treat as the remainder and thereofr just D.

Does that make sense as a method, is it ok?

If you are dividing by a linear function (e.g. ax + b) then the remainder will

**never**be of the Form CX+D ....... Think about it !!! CX+D can be divided by ax+b !!!!.

If you are only dividing Polynomials by linear functions then your Remainder will always be a constant.

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#14

(Original post by

anyone?

p.s. cheers for the 9 replies of "i do aqa" well done.

**f45**)anyone?

p.s. cheers for the 9 replies of "i do aqa" well done.

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#15

(Original post by

Cheers for the acknowledgement of my very helpful post.

**AnthonyH91**)Cheers for the acknowledgement of my very helpful post.

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#16

(Original post by

A quick check of his profile will tell you he hasn't been on since eight minutes before you made that post yesterday. Calm down.

**generalebriety**)A quick check of his profile will tell you he hasn't been on since eight minutes before you made that post yesterday. Calm down.

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#18

(Original post by

Highest power of quotient = (Highest power of dividend) - (Highest power of Divisor)

Highest power of remainder = 1 lower than highest power of divisor.

**member188279**)Highest power of quotient = (Highest power of dividend) - (Highest power of Divisor)

Highest power of remainder = 1 lower than highest power of divisor.

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