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Algebra 9 Online
OpenStudy (anonymous):

Can someone explain the distributive rule?

OpenStudy (skullpatrol):

What don't you understand?

OpenStudy (anonymous):

Sure! basically 1(2+2) add 1 to each of the factors so you have (1+2) + (1+2) You distribute the outsider to each of the insiders. and it's not lonely anymore c: xD haha

OpenStudy (skullpatrol):

That^ is NOT right at all

OpenStudy (anonymous):

you multiply, not add, but the same idea yes.

OpenStudy (anonymous):

so if the equation is x(a+b) then what you'd end up with, after distribution is ax+bx

OpenStudy (skullpatrol):

@Profunds77 $$4( 90 + 12) = 4*90 + 4*12$$ Note 4 distributes as a MULTIPLIER of each term of the sum 90 + 12.

OpenStudy (anonymous):

what he said

OpenStudy (anonymous):

I like using letters instead :)

OpenStudy (anonymous):

I was looking at a question answered and they mentioned the distributive rule instead of the foil method just looking for new technics that I can apply

OpenStudy (anonymous):

Thanks

OpenStudy (anonymous):

FOIL as in: First Outside Inside Last Are you taliking about that foil method?

OpenStudy (anonymous):

Yes

OpenStudy (whpalmer4):

FOIL is just a memory crutch to keep the work organized, but you're doing the distributive property. \[(a + b)(c + d) = a(c+d) + b(c + d) = ac + ad + bc + bd\] Compare with \[(a+b)(c+d) = ac + ad + bc + bd\]Same thing.

OpenStudy (whpalmer4):

Of course, if you prefer, this is equally valid: \[(a+b)(c+d) = ac + bc + ad + bd\] Whatever method gets you through it systematically and correctly....

OpenStudy (unklerhaukus):

\[A\star(B+ C)=A\star B+ A\star C\]

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