Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Remainders

I got -1, after long division, but the answer says it is 11? How? Thanks

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Remainders

I get -1 as well, but its easy enough to check.

If you call -3/2 a root, then it should give the same remainder.

f(-3/2) = 4(-3/2)³ - 5(-3/2) + 5
          = -13.5 + 7.5 + 5
          = - 1

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Remainders

Well, hmm, for every negative remainder you must have a positive remainder, and by definition, a 'remainder' is positive:

and

e.g 8/5 has remainder 3 'remainder'(?) -2.

So where is the positive remainder?

Last edited by Identity (2007-10-28 00:57:16)

landof+

Member

Registered: 2007-03-24

Posts: 131

Re: Remainders

shouldn't it be 8 - 2 then? because there is not a full '8' to divide by there.

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Remainders

Well, hmm, for every negative remainder you must have a positive remainder, and by definition, a 'remainder' is positive...

Not when you're dealing with algebra. In that example, -1 is the remainder just because it doesn't have any x's left in it.

A different example could be, say, [x³ + x - 1]/x². It's fairly easy to see there that the quotient would work out to be x, but then that leaves a remainder of x-1.
You can't fall back on your definition there, because you can't say whether x-1 is positive or negative anyway.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Remainders

The division remainder theorem says that for every integer a and b, there exists a unique q and r such that:

a = bq + r with 0 <= r < b

Now we move into the world of polynomials.  We want our polynomials to be represented by integer values so that we can have an analogous result.  Anyone know how to represent a polynomial by an integer?  Anyone?  Anyone?  Buler?

So we take two polynomials a and b, and we say:

a = bq + r with degree(r) < degree(b)

We don't specify 0 <= degree(r) for a few reasons.  First, there is not concept of having a negative integer degree.  At least not in polynomials.  But more importantly, we say that degree(0) = -infinity.  The zero polynomial has degree negative infinity mostly from the following theorem:

deg(fg) = deg(f) + deg(g)

If we have either f or g being the 0 polynomial, it's easy to see how having deg(0) = 0 would screw the above result up.  But making it -infinity makes everything work out nicely.

Getting back on track, by representing the polynomial by it's degree, the division remainder theorem becomes pretty straightforward.  And as a result of this, -1 has the same degree as positive 1, so they both fit as being r.

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Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Remainders

Thanks!