I know that is first,outer, inner, last but are there other ways to solve using the method?

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- Dec 29th 2008, 03:54 PMRussianFoil Method
I know that is first,outer, inner, last but are there other ways to solve using the method?

- Dec 29th 2008, 07:17 PMMush
Ermmm what do you mean exactly?

If you have a expression , then you could re-write it as , and calculate from there.

So for example:

Essentially, it's the same as foil method. Essential, all ways of expanding the expression are very close to foil method. Can't really go wrong :P. - Dec 30th 2008, 04:35 AMHallsofIvy
To multiply together two expressions, you multiply

**every**term of the first by**every**term of the last. "FOIL" is just a way of being sure you get all of them and only apply to "two terms times two terms".

For example: to multiply (a+ b)(c+ d) I can use "FOIL": "First" terms, ab; "Outside terms", ad; "Inside" terms, bc; "Last" terms, bd. Adding all of those: ab+ ad+ bc+ bd.

I could have got the same result by using the rule: Multiply the first term in the first expression, a, by each of the terms in the second expression: a(c+ c)= ac+ ad. Then multiply the second term in the first expression by each of the terms in the second expression: b(c+ d)= bc+ bd. Adding those: ac+ ad+ bc+ bd, exactly as above.

If I have 3 terms in, say, the first expression, like (a+ b+ c)(u+ v), I can't use "FOIL" exactly but I can think:

Multiply the first term in the first expression, a, by every term in the second expression- a(u+ v)= au+ av.

Multiply the second term in the first expression, b, by every term in the second expression- b(u+ v)= bu+ bv.

Multiply the third term in the first expression, c, by every term in the second expression- c(u+ v)= cu+ cv.

Now add all of those: (a+ b+ c)(u+ v)= au+ av+ bu+ bv+ cu+ cv. - Dec 31st 2008, 12:41 PMRussian