anonymous
  • anonymous
x(m+5)+7(m+5) answer: (m+5)(x+7) I dont get it..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
my answer is Xm+5x+7m+35
Nnesha
  • Nnesha
wait what what's the original equation ?
UsukiDoll
  • UsukiDoll
is it factor by grouping? or combine like terms? what's the full question? It looks like a double sword question right now

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Nnesha
  • Nnesha
x(m+5)+7(m+5) is this is the question then you don't need to distribute
Nnesha
  • Nnesha
don't distribute!
anonymous
  • anonymous
How do I get to the answer?
UsukiDoll
  • UsukiDoll
x(m+5)+7(m+5) I would just pull out the m+5 (m+5)(x+7) the problem was done correctly :)
anonymous
  • anonymous
I dont get it
Nnesha
  • Nnesha
|dw:1439609551521:dw| parentheses are common so take it our or in other words just write on time
UsukiDoll
  • UsukiDoll
first 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
  • anonymous
oh.. so (m+5) cancel out right?
Nnesha
  • Nnesha
\[\huge\rm a\color{red}{(c+b)}+d\color{reD}{(c+b)}\] 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
  • UsukiDoll
^ what this user said
anonymous
  • anonymous
So it (x+7) right?
Nnesha
  • Nnesha
yep right and the common parentheses
anonymous
  • anonymous
but we already take out (m+5)
Nnesha
  • Nnesha
yeah but it still there
UsukiDoll
  • UsukiDoll
i'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
UsukiDoll
  • UsukiDoll
factor m+5 out.
anonymous
  • anonymous
sorry, I dont get it when you said "take it out", what do you mean?
UsukiDoll
  • UsukiDoll
there is a m+5 in common right?
anonymous
  • anonymous
yes
Nnesha
  • Nnesha
by taking out i mean\[\huge\rm [x(m+5)+7(m+5)]\] \[(m+5)[x+7]\]
Nnesha
  • Nnesha
|dw:1439609915447:dw| i mean take out m+5 from x(m+5) and take out m+5 from 7(m+5)
UsukiDoll
  • UsukiDoll
so 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
  • anonymous
okay, but we there is two (m+5) wont it be (m+5)^2?
UsukiDoll
  • UsukiDoll
we 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
  • UsukiDoll
if 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 \[\huge \color{red}{(m+5)^2}\]
anonymous
  • anonymous
ok thanks!
madhu.mukherjee.946
  • madhu.mukherjee.946
@dark_knight321 dude atleast give a medal to the person who helps you out

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