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

1. anonymous

2. Nnesha

wait what what's the original equation ?

3. UsukiDoll

is it factor by grouping? or combine like terms? what's the full question? It looks like a double sword question right now

4. Nnesha

x(m+5)+7(m+5) is this is the question then you don't need to distribute

5. Nnesha

don't distribute!

6. anonymous

How do I get to the answer?

7. UsukiDoll

x(m+5)+7(m+5) I would just pull out the m+5 (m+5)(x+7) the problem was done correctly :)

8. anonymous

I dont get it

9. Nnesha

|dw:1439609551521:dw| parentheses are common so take it our or in other words just write on time

10. 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)

11. anonymous

oh.. so (m+5) cancel out right?

12. 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

13. UsukiDoll

^ what this user said

14. anonymous

So it (x+7) right?

15. Nnesha

yep right and the common parentheses

16. anonymous

but we already take out (m+5)

17. Nnesha

yeah but it still there

18. 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

19. UsukiDoll

factor m+5 out.

20. anonymous

sorry, I dont get it when you said "take it out", what do you mean?

21. UsukiDoll

there is a m+5 in common right?

22. anonymous

yes

23. Nnesha

by taking out i mean$\huge\rm [x(m+5)+7(m+5)]$ $(m+5)[x+7]$

24. Nnesha

|dw:1439609915447:dw| i mean take out m+5 from x(m+5) and take out m+5 from 7(m+5)

25. UsukiDoll

so we factor m+5 $\huge\rm x\color{red}{(m+5)}+7\color{reD}{(m+5)}$ becomes $\huge \rm \color{reD}{(m+5)} (x+7)$

26. anonymous

okay, but we there is two (m+5) wont it be (m+5)^2?

27. 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

28. 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 $\huge \color{red}{(m+5)^2}$

29. anonymous

ok thanks!