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#1 2007-11-07 11:45:40

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Polynomials - Long Multiplication

I just made a page on Polynomials - Long Multiplication

But ... does anyone know a better method when there is more than one variable? Other than "make up the columns as you go"?

How would YOU do (x[sup]2[/sup] + 2xz + z) (x + z) ?


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#2 2007-11-07 14:10:56

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Polynomials - Long Multiplication

I've never even heard of long multiplication.  The only way I know how to do it is to multiply term by term then add common ones together:

(ax^2 + bx + c)(dx^2 + ex + f) = adx^4 + aex^3 + afx^2 + bdx^3 + bex^2 + bfx + cdx^2 + cex + cf


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2007-11-07 21:55:44

JohnnyReinB
Member
Registered: 2007-10-08
Posts: 453

Re: Polynomials - Long Multiplication

Personally, I would do it Vertically:
(x2 + 2xz + z) (x + z)=

                     x^2 + 2xz + z
                                  x + z
________________________
           x^2z +2xz^2+ z^2
x^3 +2x^2z                      + xz
____________________________
x^3 +3x^2z +2xz^2+ z^2 +xz

It would just be like Arithmetic Multiplication


"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted" wink

Nisi Quam Primum, Nequequam

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#4 2007-11-07 23:13:14

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: Polynomials - Long Multiplication

It's just another way of setting things out I guess, and if you're better at it than normal expansion then that's great.

15 * 92 = 10(92)+5(92) = 920 + 460 = 1380

But personally I like normal expansion because you can do fancy tricks.

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#5 2007-11-09 01:54:14

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 45,956

Re: Polynomials - Long Multiplication

MathsIsFun wrote:

How would YOU do (x[sup]2[/sup] + 2xz + z) (x + z) ?

I'd multiply the like variables, and add the exponents, that would make things fairly easy! Moreover, multiplying a binomial of degree 2 wouldn't take much time.


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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