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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,581

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) ?

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

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

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" *

Nisi Quam Primum, Nequequam

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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 18,616

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 is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

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