Math Is Fun Forum

Anakin

Member

Registered: 2009-10-04

Posts: 145

Remainder Theorem [Hard]

Well, here it goes.

Q: When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2. Determine the remainder when the polynomial is divided by (x+2)(x-1).

Attempt: I got as far as P(-2) = -19 and P(1) = 2.

No clue where to go after that. It's not for homework or anything so I'm not terribly worried about it but it would be greatly awesome to find guidance on how to solve the problem. Thanks

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Remainder Theorem [Hard]

Hi Anakin;

I found P(x) = 7x - 5 so I got a remainder of 7x - 5, for P(x) / ( (x+2) (x-1) ).

Last edited by bobbym (2009-10-04 18:37:47)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Remainder Theorem [Hard]

That's the correct answer!

Bobby, would you be kind enough to post your steps here for reference so when exam time comes around, I'll know what to do?

It would be greatly appreciated. Thanks.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Remainder Theorem [Hard]

Hi Anakin;

Perhaps a little unorthodox , but it is the way I try to do these problems.

I figured that P(x) is of the form a*x+b in other words a linear equation because of the the constant remainder. At least it was worth a try. Also I start with the smallest degree poly first.

So I divide (a*x + b) / ( x -1 )       Quotient = a and remainder = a + b   and

                  (a*x + b) / ( x + 2)      Quotient = a and remainder = b - 2a

Solve a+ b = 2 and b - 2a = -19 simultaneously and get a = 7 ,  b = - 5

So P(x) = 7 x - 5, now just check.

Last edited by bobbym (2009-10-04 18:55:26)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Remainder Theorem [Hard]

Alright so let me try to decipher this and you can tell me if I did this correctly or not.

y = mx+b is the standard form for a linear equation

So we divide the polynomial by a linear equation in the form (x-b) where b is equal to 1
(mx +b) / (x-1) = m with remainder m + b

(mx+b) / (x+2) = m with remainder -2m + b

We solve for the variable b in each case.
m + b = 2
b = 2-m

-2m + b = -19
b = 2m - 19

Let b = b and solve for m
2-m = 2m-19
3m = 21
m = 7

Substitute m = 7 into b = 2-m
b=2-m
b=2-7
b=-5

But these into the form P(x) = mx +b
P(x) = mx +b
P(x) = 7x -5

How's that?

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Remainder Theorem [Hard]

You said you always start with the lowest degree polynomial first. So if P(x) = mx +b did not work out, would you move to a quadratic equation next?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Remainder Theorem [Hard]

Hi Anakin;

Yes, you did it right.

You said you always start with the lowest degree polynomial first. So if P(x) = mx +b did not work out, would you move to a quadratic equation next?

No, I would have been stuck. I wouldn't be able to exactly solve for the 3 coefficients of a quadratic, because you only provided 2 function evaluations. I am pretty sure that it had to be a linear equation though.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Remainder Theorem [Hard]

Awesome, thanks a lot for the help. I don't know why you decided to divide a standard linear polynomial though. I would've NEVER thought of that.

Glad I stumbled across these forums, just for general purpose as math is awesome.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Remainder Theorem [Hard]

Hi Anakin;

Glad to help a little and welcome to the forum!

Last edited by bobbym (2009-10-04 19:27:25)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Remainder Theorem [Hard]

You really shouldn't assume that P(x) is a linear polynomial.
P(x) = x^2 + 8x - 7 is another polynomial with the required properties.

We know that when P(x) is divided by (x+2) the remainder in -19.
This means that P(x) = (x+2)Q(x) - 19 for some polynomial Q(x).

We also know that when P(x) is divided by (x-1) the remainder is 2.
As you observe this means that P(1) = 2.
So 2 = P(1) = (1+2)Q(1) - 19 so 3Q(1) = 21 and hence Q(1) = 7.
This means that Q(x) = (x-1)R(x) + 7 for some polynomial R(x).

Therefore P(x) = (x+2)[(x-1)R(x) + 7] - 19 = (x+2)(x-1)R(x) +7(x+2) - 19
so P(x) = (x+2)(x-1)R(x) + 7x - 5.

The remainder when P(x) is divided by (x+2)(x-1) is 7x - 5.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Remainder Theorem [Hard]

Hi Avon;

At least it was worth a try. Also I start with the smallest degree poly first.

Yes, there are an infinity of them, He didn't seem to want a general answer. What is wrong with looking for the smallest order poly first. That solution is the only linear solution. The linear form is the only one that is unique.  I had to make a decision as to what he probably wanted...

Last edited by bobbym (2009-10-05 11:10:47)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Remainder Theorem [Hard]

I talked to my teacher about it and he said that the reason why we use a linear equation is because the remainder is always one degree lower than the divisor.

Divisor: (x-1)(x+2) = (x^2+x-2) which is a quadratic. So we want an equation that is one degree less than a quadratic, which would be a linear one.

Is that right?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Remainder Theorem [Hard]

Yes, that is why I went linear, I know teachers. But look at Avon's fine post to learn a lot.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Remainder Theorem [Hard]

Yup, I had a good look at it.

Took me a while to understand it all but I see the logic behind it.

Thanks for the taking the time to explain all the steps guys. I appreciate it.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Remainder Theorem [Hard]

Hi bobbym

My main complaint was that you were assuming that P(x) was linear.
There is no harm in assuming that the remainder is linear, since it must have degree less that (x+2)(x-1).

Hence you could start by writing P(x) = (x+2)(x-1)Q(x) + ax + b  for some polynomial Q(x)
and then prove that a = 7 and b = -5 as you have done.

It probably is a little neater to do it that way than the way that I did it.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Remainder Theorem [Hard]

Hi Avon;

I appreciated your post and your method was quite good. Thanks !

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.