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Mathematics 18 Online
OpenStudy (anonymous):

Factor the following: a(x + 2) + 3(x + 2)

OpenStudy (mimi_x3):

(a+3)(x+2)

myininaya (myininaya):

both terms have x+2 in common factor that out (x+2)(a+3)

OpenStudy (anonymous):

you pull out the (x+2) , then you just plug in what's left.

OpenStudy (anonymous):

a(b+c) + d(b+c) is always equal to (a+d)(b+c) therefore \[a(x+2)+3(x+2) =(a+3)(x+2)\]

OpenStudy (anonymous):

remember the logics if u have learn them then no problem u would have

OpenStudy (anonymous):

(x+2)(a+3) my bad...there shouldn't have been a plus in there

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