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Simple Math- Need help =/

  1. May 6, 2008 #1
    Simple Math-- Need help =/

    1. The problem statement, all variables and given/known data[/b
    I am confused on how this equation:
    2h + (16/h-8 +2) (-8/(h-8)^2
    is simplified to this equation:
    2h - 128 (h-8)^3 - 16/(h-8)^2
    2. Relevant equations

    Simplifying skills.

    3. The attempt at a solution

    Well, I know that they got the number 128 from multiplying 16 and -8 together and how they got -16 by multiplying 2 and -8 together. Now, what I don't get is getting (h-8)^3 from (h-8)... SOMEONE PLEASE HELP?
  2. jcsd
  3. May 6, 2008 #2
    If you have (A + B) * -C

    and you want to expand it out you do -A*C - B*C ....do you see?

    in your question A = 16/(h-8)

    also if you have (1/y)(2/y^2)

    you'll get 2/y^3

    the top multiplied together and the bottom multiplied together... do you see?
  4. May 6, 2008 #3
    do you mean that 2h - 128/(h-8)^3 - 16/(h-8)^2? if you meant that, then the (h-8)^3 comes from multiplying (h-8) times ( h-8)^2, what you do is you add the exponents.

    It is like x times x, you add the exponents if the bases are the same, it becomes x^2, 5 times 5 is 5^2.
  5. May 6, 2008 #4
    I'll be right back. I'm at school right now and bell is going to ring soon. Thanks for the replies, I'll look into it once I get home.
  6. May 11, 2008 #5
    [tex]2h+( \frac{16}{h-8} + 2 ) ( \frac{-8}{(h-8)^2} ) [/tex]

    let [tex]x = \frac{-8}{(h-8)^2} [/tex], then:

    [tex]2h+( \frac{16}{h-8} + 2 ) x [/tex]

    [tex]2h+ \frac{16}{h-8} \times x + 2 \times x [/tex]

    sub [tex]x = \frac{-8}{(h-8)^2} [/tex] back in:

    [tex]2h+ \frac{16}{h-8} \times \frac{-8}{(h-8)^2} + 2 \times \frac{-8}{(h-8)^2} [/tex]

    [tex]2h+ \frac{16 \times (-8)}{(h-8) \times (h-8)^2} + \frac{2 \times (-8)}{(h-8)^2} [/tex]

    [tex]2h+ \frac{-128}{(h-8)^3} + \frac{-16}{(h-8)^2} [/tex]

    Does that clear it up?
    Last edited: May 11, 2008
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