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anonymous

  • one year ago

x(m+5)+7(m+5) answer: (m+5)(x+7) I dont get it..

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  1. anonymous
    • one year ago
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    my answer is Xm+5x+7m+35

  2. Nnesha
    • one year ago
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    wait what what's the original equation ?

  3. UsukiDoll
    • one year ago
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    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
    • one year ago
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    x(m+5)+7(m+5) is this is the question then you don't need to distribute

  5. Nnesha
    • one year ago
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    don't distribute!

  6. anonymous
    • one year ago
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    How do I get to the answer?

  7. UsukiDoll
    • one year ago
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    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
    • one year ago
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    I dont get it

  9. Nnesha
    • one year ago
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    |dw:1439609551521:dw| parentheses are common so take it our or in other words just write on time

  10. UsukiDoll
    • one year ago
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    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
    • one year ago
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    oh.. so (m+5) cancel out right?

  12. Nnesha
    • one year ago
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    \[\huge\rm a\color{red}{(c+b)}+d\color{reD}{(c+b)}\] 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
    • one year ago
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    ^ what this user said

  14. anonymous
    • one year ago
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    So it (x+7) right?

  15. Nnesha
    • one year ago
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    yep right and the common parentheses

  16. anonymous
    • one year ago
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    but we already take out (m+5)

  17. Nnesha
    • one year ago
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    yeah but it still there

  18. UsukiDoll
    • one year ago
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    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
    • one year ago
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    factor m+5 out.

  20. anonymous
    • one year ago
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    sorry, I dont get it when you said "take it out", what do you mean?

  21. UsukiDoll
    • one year ago
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    there is a m+5 in common right?

  22. anonymous
    • one year ago
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    yes

  23. Nnesha
    • one year ago
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    by taking out i mean\[\huge\rm [x(m+5)+7(m+5)]\] \[(m+5)[x+7]\]

  24. Nnesha
    • one year ago
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    |dw:1439609915447:dw| i mean take out m+5 from x(m+5) and take out m+5 from 7(m+5)

  25. UsukiDoll
    • one year ago
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    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
    • one year ago
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    okay, but we there is two (m+5) wont it be (m+5)^2?

  27. UsukiDoll
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    ok thanks!

  30. madhu.mukherjee.946
    • one year ago
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    @dark_knight321 dude atleast give a medal to the person who helps you out

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