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Math Help - Factor the expression

  1. #1
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    Factor the expression

    First question:


    From the book... whomever wrote this doesn't know HOW to explain things.

    Solve:
    p(m-n)-q(n-m)


    p(m-n)-q(-1)(-n+m)
    From the second two terms, pull out a ‘-1’. You can double check that this is correct by distributing the “-1” back in.

    p(m-n)-q(-1)(m-n)
    Inside the parenthesis on the far right, rearrange terms.

    p(m-n)+q(m-n)
    Now, in the second group of terms, we have a “-1” times a “-1” which gives a “+1”.
    Answer:
    (m-n)(p+q)



    Can someone explain how the answer is obtained better? I'm fumbling on factoring with -1 on these types of equations.



    -----------------------------------------
    Second question:

    Solve:
    3x(c-3d)+6y(c-3d)




    Answer:
    3(c-3d)(x+2y)

    If you apply the order of distribution it comes out to 3c-9d...


    Why can't the answer be
    (c-3d)(x+2y)3
    or
    3(x+2y)(c-3d)



    -----------------------------------------



    Anyone have a better explanation? I'm starting to cram for a COMPASS test this coming Friday and still need to refresh my math skills from algebra, adv. algebra, geo, and trig.
    Last edited by pychon; August 3rd 2010 at 11:33 AM.
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  2. #2
    A Plied Mathematician
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    It can be either of those. I would hesitate to write the first one, though, because a standard convention is to write constants first. Also, you might get someone thinking that the 3 is an exponent, which is not the case.
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  3. #3
    Senior Member eumyang's Avatar
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    Quote Originally Posted by pychon View Post
    Solve:
    3x(c-3d)+6y(c-3d)
    Answer:
    3(c-3d)(x+2y)
    If you apply the order of distribution it comes out to 3c-9d...
    Why can't the answer be
    (c-3d)(x+2y)3
    or
    3(x+2y)(c-3d)
    Usually when you multiply a number by a variable or expression you write the number first, so that there is no ambiguity. I could look at
    (c-3d)(x+2y)3
    and thought that you forgot to make the 3 an exponent, like this:
    (c-3d)(x+2y)^3
    ... which is certainly wrong. So keep the 3 in front.

    This:
    3(x+2y)(c-3d)
    is just fine. Commutative property of multiplication says that the order you multiply doesn't matter. So it doesn't matter if you wrote 3(c-3d)(x+2y) or 3(x+2y)(c-3d) as your answer.

    EDIT: Ack! Too slow!
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  4. #4
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    Ok, so is this legal to check the answer?


    3(c-3d)(x+2y)

    distribute

    3x-9d+3x+6y<br />

    grrr... that doesn't look right.
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  5. #5
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    Quote Originally Posted by pychon View Post
    From the book... whomever wrote this doesn't know HOW to explain things.

    Solve:
    p(m-n)-q(n-m)




    p(m-n)-q(-1)(-n+m)



    p(m-n)-q(-1)(m-n)



    p(m-n)+q(m-n)


    Answer:
    (m-n)(p+q)


    -----------------------------------------


    Solve:
    3x(c-3d)+6y(c-3d)




    Answer:
    3(c-3d)(x+2y)

    If you apply the order of distribution it comes out to 3c-9d...


    Why can't the answer be
    (c-3d)(x+2y)3
    or
    3(x+2y)(c-3d)



    -----------------------------------------



    Anyone have a better explanation?
    They are all the same really!

    3x(c-3d)+6y(c-3d) is (c-3d) times both 3x and 6y

    so (c-3d)(3x+6y)=(c-3d)3(x+2y)=(3c-9d)(x+2y)

    All give the exact same result if you pick values for x, y, c, d.


    Also.

    (m-n)=-(n-m)

    because 1=(-1)(-1)

    m-n=(-1)(-1)(m-n)=(-1)((-1)m-(-1)n)=(-1)(-m--n)=(-1)(n-m)

    It's simplest to realise you only need to switch the terms and change the sign.
    You are really multiplying by -1 twice.
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  6. #6
    A Plied Mathematician
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    I don't think you distributed correctly. You lost a c in there, among other things. Double-check your distribution, and go one step at a time.
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  7. #7
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    Quote Originally Posted by pychon View Post
    Ok, so is this legal to check the answer?


    3(c-3d)(x+2y)

    distribute

    3x-9d+3x+6y<br />

    grrr... that doesn't look right.
    Assuming you meant

    3c-9d+3x+6y

    you multiplied out (distributed) 3(c-3d+x+2y)
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  8. #8
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    Ok, so is this legal to check the answer?


    3(c-3d)(x+2y)

    distribute

    3c-9d+3x+6y



    3c-9d+3x+6y

    3 is common...

    c-3d+x+2y

    so, now.. I'm failing to see where this can validate (c-3d)(x+2y) since (*)(*) implies multiplication.
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  9. #9
    Senior Member eumyang's Avatar
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    No.

    Quote Originally Posted by pychon View Post
    3(c-3d)(x+2y)
    distribute
    3c-9d+3x+6y
    First, you between the 3d and x you changed from a multiplication to an addition. Can't do that.

    You can only distribute the 3 over EITHER the (c - 3d) factor OR the (x + 2y) factor, NOT both. If you try to distribute over both, you would be multiplying by 3 TWICE. Since in the original problem I see a 3x and a 6y, that tells me that I should distribute the 3 over the (x + 2y) factor:
    3(c-3d)(x+2y)
    (c-3d)(3)(x+2y)
    (c-3d)(3x+6y)

    Now distribute the (c - 3d) over the (3x + 6y):
    (c-3d)(3x) + (c-3d)(6y)
    3x(c-3d) + 6y(c-3d)
    ...which what we started with.
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  10. #10
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    Quote Originally Posted by pychon View Post
    Ok, so is this legal to check the answer?


    3(c-3d)(x+2y)

    distribute

    3c-9d+3x+6y



    3c-9d+3x+6y

    3 is common...

    c-3d+x+2y

    so, now.. I'm failing to see where this can validate (c-3d)(x+2y) since (*)(*) implies multiplication.
    Hi pychon,

    think in terms of numbers

    3(4)5=60 as it's either 3(20) or (12)5, however it's also 4(3)5=4(15)

    You can multiply them in any order you like.
    3(4)5=3(4)+3(4)+3(4)+3(4)+3(4) or 4(5)+4(5)+4(5) or 3(5)+3(5)+3(5)+3(5).

    Alternatively 3(6-2)(8-3) must also equal 60.
    We can "distribute" that..

    3[(6-2)(8-3)]=3[6(8)+6(-3)-2(8)-2(-3)]=3[48-18-16+6]=3[30-16+6]=3[36-16]=3[20]=60

    That's the idea, then the algebra is easy from there.
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  11. #11
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    Quote Originally Posted by pychon View Post
    Ok, so is this legal to check the answer?


    3(c-3d)(x+2y)

    distribute

    3x-9d+3x+6y<br />

    grrr... that doesn't look right.
    No, that isn't right because you have changed the product to a sum.

    You can distribute the 3 into the first term only: 3(c-3d)(x+2y)= (3c- 3d)(x+ 2y) and now separate that 3c- 3d into parts:
    3c(x+ 2y)- 3d(x+ 2y)= 3cx+ 6cy- 3dx+ 6dy.
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