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anonymous
 one year ago
x(m+5)+7(m+5)
answer: (m+5)(x+7)
I dont get it..
anonymous
 one year ago
x(m+5)+7(m+5) answer: (m+5)(x+7) I dont get it..

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my answer is Xm+5x+7m+35

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1wait what what's the original equation ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2is it factor by grouping? or combine like terms? what's the full question? It looks like a double sword question right now

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1x(m+5)+7(m+5) is this is the question then you don't need to distribute

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do I get to the answer?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2x(m+5)+7(m+5) I would just pull out the m+5 (m+5)(x+7) the problem was done correctly :)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439609551521:dw parentheses are common so take it our or in other words just write on time

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2first term x(m+5) + second term 7(m+5) we notice that both terms have an m+5 in common so we take that out (m+5)(x+7)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh.. so (m+5) cancel out right?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm a\color{red}{(c+b)}+d\color{reD}{(c+b)}\] c+b is common in both terms so take it out \[(c+b)\] then u hve left with (a+d) now look at the question try to write that

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2^ what this user said

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1yep right and the common parentheses

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but we already take out (m+5)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1yeah but it still there

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2i'ts not just x+7 \[\huge\rm a\color{red}{(c+b)}+d\color{reD}{(c+b)}\] \[\huge\rm x\color{red}{(m+5)}+7\color{reD}{(m+5)}\] BOth terms have m+5 in common so yank them out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, I dont get it when you said "take it out", what do you mean?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2there is a m+5 in common right?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1by taking out i mean\[\huge\rm [x(m+5)+7(m+5)]\] \[(m+5)[x+7]\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439609915447:dw i mean take out m+5 from x(m+5) and take out m+5 from 7(m+5)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so we factor m+5 \[\huge\rm x\color{red}{(m+5)}+7\color{reD}{(m+5)} \] becomes \[\huge \rm \color{reD}{(m+5)} (x+7) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, but we there is two (m+5) wont it be (m+5)^2?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2we can also distribute the m+5 back as well (m+5)x+(m+5)(7) and we're back to what we had earlier m+5 is what both terms have in common, so we factor them out (m+5)(x+7) no it won't be (m+5)^2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2if we had a problem like this \[\huge\rm x\color{red}{(m+5)^3}+7\color{reD}{(m+5)^2}\] then the only thing common in these terms is the \[\huge \color{red}{(m+5)^2}\]

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0@dark_knight321 dude atleast give a medal to the person who helps you out
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