C3 Remainder theorem. how do you know what the quotient and remainder form will be?
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?
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?
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 is Core 3? Which chapter?
pretty sure this is C1
olipal
This is all in Chapter 1 (Edexcel).
I do AQA so this could be why.
pretty sure remainder depends on the divisor, doubt there is a general form for it.
beatleboar
I do AQA so this could be why.
Haha not too sure what's in the AQA spec.. To be honest, I sort of missed out this rule anyways as I imagine it won't really come up in the actual exam. If you can do basic algebraic division (using long division, remainder theorem etc), then you should be fine. Saying that, I will probably give it a quick read the night before the exam just so I have a slight clue what it's going on about were it to pop up!
Toneh
I'm pretty sure I did it in C1. OCR.
Yeah, i did it in C1 the remainder theorem, OP wants a general form for remainder depending on the divisor which isn't necessary because this type of algebra (well it isn't really) requires you to carry out the steps in order to get the marks, if you can do that then you're sorted
f45
anyone?
p.s. cheers for the 9 replies of "i do aqa" well done.
p.s. cheers for the 9 replies of "i do aqa" well done.
Highest power of quotient = (Highest power of dividend) - (Highest power of Divisor)
Highest power of remainder = 1 lower than highest power of divisor.
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?
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.
f45
anyone?
p.s. cheers for the 9 replies of "i do aqa" well done.
p.s. cheers for the 9 replies of "i do aqa" well done.
Cheers for the acknowledgement of my very helpful post.
AnthonyH91
Cheers for the acknowledgement of my very helpful post.
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.
Oh ok. Thanks.
Original post by member188279
Highest power of quotient = (Highest power of dividend) - (Highest power of Divisor)
Highest power of remainder = 1 lower than highest power of divisor.
Highest power of remainder = 1 lower than highest power of divisor.
Thanks!
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