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Math Help - Factoring

  1. #1
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    Factoring

    I believe i know the answer this this one, but i just want to make sure.
    Could someone tell me the answer to the following:

    Factor the difference of two sqaures,
    25 - (z+5)
    Also, this one, i am having a hard time finding the answer to:
    Completely factor the expression,
    5(3-4x) - 8(3-4x)(5x-1)

    Thanks
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by WilliamR View Post
    I believe i know the answer this this one, but i just want to make sure.
    Could someone tell me the answer to the following:

    Factor the difference of two sqaures,
    25 - (z+5)
    Also, this one, i am having a hard time finding the answer to:
    Completely factor the expression,
    5(3-4x) - 8(3-4x)(5x-1)

    Thanks
    In the first, let z+5=a. This makes the problem look a little less distracting. We can always replace the z+5 after the end. So... How would you factor...

    25-a^2

    It would look like this

    (5+a)(5-a)

    now if I put z+5 back in...

    [5+(z+5)][5-(z+5)]

    And you can simplify further.

    In the second problem, try putting a=3x-4, and b= 5x-1. Then factor and replace like the last problem.
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  3. #3
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    I am still not sure about the answer the the second problem
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  4. #4
    Senior Member MacstersUndead's Avatar
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    5(3-4x) - 8(3-4x)(5x-1)
    (3 - 4x) [ 5(3-4x) - 8(5x-1)] ... can you finish?
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by WilliamR View Post
    I am still not sure about the answer the the second problem
    Can you factor

    5a^2-8ab ?
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    Can you factor

    5a^2-8ab ?
    a(5a-8b)


    Quote Originally Posted by MacstersUndead View Post
    5(3-4x) - 8(3-4x)(5x-1)
    (3 - 4x) [ 5(3-4x) - 8(5x-1)] ... can you finish?
    Erm....240x -272x+69....?
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  7. #7
    Senior Member MacstersUndead's Avatar
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    EDIT:// yes, your answer is right. sorry

    (3-4x) [ 5(3-4x) - 8(5x-1)]
    (3-4x) [ (15 - 20x) - 40x + 8)]
    (3-4x) [ -60x + 23 ]

    and we are done factoring.
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  8. #8
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    Quote Originally Posted by MacstersUndead View Post
    EDIT:// yes, your answer is right. sorry

    (3-4x) [ 5(3-4x) - 8(5x-1)]
    (3-4x) [ (15 - 20x) - 40x + 8)]
    (3-4x) [ -60x + 23 ]

    and we are done factoring.
    I see now, that is clear, thank you .
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  9. #9
    No one in Particular VonNemo19's Avatar
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    Now substitute back for a and b.

    The point that I'm trying to make here is that factors can be a distraction if they contain more than one term like (z+5). But if you look at it like the way that I've showed you, you will become really good really fast.

    Sometimes its better not to just give someone a solution....

    But actually teach them something.
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  10. #10
    Senior Member pacman's Avatar
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    VonNemo19, i agree with you. That is what MATH HELPER FORUM intends to, guiding them. Voila!
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