Distributive Property: (x+2)(x+4)
Date: 09/18/2001 at 07:36:30 From: Lydia Flecha Subject: Distributive Property I'm confused about the distributive property. Example: (x+2)(x+4). Would that be x+2 * x+4 = 2x + 6 ?
Date: 09/18/2001 at 11:23:49 From: Doctor Ian Subject: Re: Distributive Property Hi Lydia, The distributive property says that a(b + c) = ab + ac It's easiest to understand as a picture: 3 4 ----- ------- @ @ @ * * * * | @ @ @ * * * * | @ @ @ * * * * | 5 5(3 + 4) = 5(3) + 5(4) @ @ @ * * * * | @ @ @ * * * * | The size of the whole rectangle is equal ------------- to the sum of the sizes of the smaller 3 + 4 rectangles. Extending this from particular cases to the general case is pretty straightforward: a b --------- ----------- @ @ ... @ * * ... * * | @ @ ... @ * * ... * * | . . . | c c(a + b) = ca + cb . . . | @ @ ... @ * * ... * * | @ @ ... @ * * ... * * | --------------------- a + b Note that c doesn't have to be a number. It can also be an expression, which is the key to expanding something like (x+2)(x+4): (x + 2)(x + 4) = (x + 2)(x) + (x + 2)(4) \_____/ | | \_____/ | \_____/ | c a b c a c b Then we use the same trick to expand each of the resulting products: = (x + 2)(x) + (x + 2)(4) | | | | | | a b c A B C = (x)(x) + (2)(x) + (x)(4) + (2)(4) | | | | | | | | a c b c A C B C = x^2 + 2x + 4x + 8 Note that you can now use the distributive property in the other direction to simplify this: = x^2 + 2x + 4x + 8 || || ac bc = x^2 + (2 + 4)x + 8 = x^2 + 6x + 8 There is a technique called FOIL that is often taught as a trick for expanding expressions like this, but applying the distributive property twice isn't any harder than applying FOIL, and the former works in a lot of cases where the latter is useless: When FOIL Fails http://mathforum.org/dr.math/problems/ryan.03.22.01.html In fact, once you move from using numbers to using variables, the distributive property becomes one of your best friends, so it's good that you want to become more familiar with it. I hope this helps. Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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