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anonymous

  • 4 years ago

Factor this expression. Check by multiplying factors. 3ab+3ac+2b^2+2bc

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  1. Mertsj
    • 4 years ago
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    Are you learning how to do these? You keep posting the same type of problem.

  2. anonymous
    • 4 years ago
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    Not at all.

  3. Mertsj
    • 4 years ago
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    Do you want to learn?

  4. anonymous
    • 4 years ago
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    I would love to.

  5. Mertsj
    • 4 years ago
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    Take the first two terms: 3ab +3ac What do they have in common?

  6. anonymous
    • 4 years ago
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    Do you have Skype so we don't spam this feed?

  7. Mertsj
    • 4 years ago
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    No I don't

  8. anonymous
    • 4 years ago
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    Ok well they share "a"

  9. Mertsj
    • 4 years ago
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    What else do they share?

  10. anonymous
    • 4 years ago
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    3

  11. Mertsj
    • 4 years ago
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    Good. so factor out 3a and tell me what you get

  12. anonymous
    • 4 years ago
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    1,3

  13. Mertsj
    • 4 years ago
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    |dw:1327549920542:dw|

  14. Mertsj
    • 4 years ago
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    Do you understand that?

  15. anonymous
    • 4 years ago
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    Yes

  16. Mertsj
    • 4 years ago
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    Now take the last two terms which are 2b^+2bc What do they have in common?

  17. anonymous
    • 4 years ago
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    2b

  18. Mertsj
    • 4 years ago
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    Factor it out

  19. anonymous
    • 4 years ago
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    |dw:1327550017696:dw|

  20. Mertsj
    • 4 years ago
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    Factor out 2b. Not just 2

  21. Mertsj
    • 4 years ago
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    You said they have 2b in common and they do. So factor out 2b

  22. anonymous
    • 4 years ago
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    |dw:1327550097535:dw|

  23. Mertsj
    • 4 years ago
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    |dw:1327550099915:dw|

  24. Mertsj
    • 4 years ago
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    I think that was my fault because I didn't get the exponent on the 2b^2

  25. anonymous
    • 4 years ago
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    Hmm.

  26. Mertsj
    • 4 years ago
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    |dw:1327550268634:dw|

  27. Mertsj
    • 4 years ago
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    So you see that now we have 2 terms instead of 4?

  28. anonymous
    • 4 years ago
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    Yes.

  29. Mertsj
    • 4 years ago
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    And what do those two terms have in common?

  30. anonymous
    • 4 years ago
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    b

  31. Mertsj
    • 4 years ago
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    |dw:1327550319609:dw|

  32. Mertsj
    • 4 years ago
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    Do you see that they have (b+c) in common?

  33. anonymous
    • 4 years ago
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    Yeah

  34. Mertsj
    • 4 years ago
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    Factor it out.

  35. anonymous
    • 4 years ago
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    Factor out 9\b+c?)\

  36. anonymous
    • 4 years ago
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    Wow, backspace is retartded

  37. Mertsj
    • 4 years ago
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    What did you say they have in common/

  38. anonymous
    • 4 years ago
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    b+c

  39. Mertsj
    • 4 years ago
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    Then factor out b+c

  40. anonymous
    • 4 years ago
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    They're alreadty factored out. Is your backspace working? Mine won't work on this website but others it will.

  41. Mertsj
    • 4 years ago
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    b+c times something equals the first term which is 3a(b+c)

  42. Mertsj
    • 4 years ago
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    Yes my backspace works.

  43. Mertsj
    • 4 years ago
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    Would you agree that (b+c)(3a )=3a(b+c)?

  44. anonymous
    • 4 years ago
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    Yes.

  45. Mertsj
    • 4 years ago
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    Now consider the second term 2b(b+c) (b+c) times something = 2b(b+c) What is it?

  46. anonymous
    • 4 years ago
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    2

  47. Mertsj
    • 4 years ago
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    2(b+c) does not equal 2b(b+c)

  48. anonymous
    • 4 years ago
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    2b

  49. Mertsj
    • 4 years ago
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    Yes. So now put it all together. Look at the drawing.

  50. Mertsj
    • 4 years ago
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    And think of (b+c) as a unit.

  51. Mertsj
    • 4 years ago
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    |dw:1327550434478:dw|

  52. anonymous
    • 4 years ago
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    |dw:1327550790530:dw|

  53. anonymous
    • 4 years ago
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    Messed up on the drawing.

  54. Mertsj
    • 4 years ago
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    Do you see that (b+c)(3a) gives the first term and (b+c)(2b) gives the second term/

  55. anonymous
    • 4 years ago
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    Yes.

  56. Mertsj
    • 4 years ago
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    And now it is factored because it is only one term

  57. Mertsj
    • 4 years ago
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    now if you have another problem like that, you should try it yourself. And if you don't have another one, you should go back and try to understand the others that you posted. It's called "factoring by grouping"

  58. anonymous
    • 4 years ago
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    ok

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