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Factoring math trouble

  1. Nov 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Simplify the expression by factoring.

    3(x+1)1/2(2x-3)5/2+10(x+1)3/2(2x-3)3/2

    2. Relevant equations



    3. The attempt at a solution

    3(x+1)1/2(2x-3)5/2+10(x+1)3/2(2x-3)3/2
    = (x + 1)1/2(2x-3)3/2[3(2x-3) + 10(x+1)]
    = (x + 1)1/2(2x - 3)3/2(6x - 9 + 10x + 10)
    = (x + 1)1/2(2x - 3)3/2(16x + 1)

    This is an example problem in a textbook. What I don't understand however is what allows the first step to be a true statement, and the reasoning behind the subsequent steps. I would really appreciate some assistance with this.

    Thank you.
     
    Last edited by a moderator: Nov 15, 2013
  2. jcsd
  3. Nov 15, 2013 #2

    lurflurf

    User Avatar
    Homework Helper

    The first two steps use the distributive property
    a(b+c)=ab+bc

    edited to add: the rule of exponents is also used
    babc=ba+c

    in particular for the first step take
    a=(x + 1)1/2(2x-3)3/2
    b=3(2x-3)
    c=10(x+1)
    for the second step use the distributive property twice
    first
    a=3
    b=2x
    c=-3
    second
    a=10
    b=x
    c=1
    The forth step uses the associative and commutative properties of addition
    commutative
    a+b=b+a
    associative
    a+(b+c)=(a+b)+c

    now try to work through the example again:smile:
     
  4. Nov 15, 2013 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    Do you agree, or not agree, that
    [tex](2x-3)^{5/2} = (2x-3)^{3/2} \cdot (2x-3)^{2/2}
    = (2x-3)^{3/2} \cdot (2x-3)\:? [/tex]
    Do you or do you not agree that
    [tex] (x+1)^{3/2} = (x+1)^{1/2} \cdot (x+1)^{2/2} = (x+1)^{1/2} \cdot (x+1) \:? [/tex]
    If you agree with both of these, you must then agree that there is a common factor ##(2x-3)^{3/2} \, (x+1)^{1/2}## in both terms of your original expression. So, just factor out this common thing, using the distributive law ##ab + ac = a(b+c)##.
     
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