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Simple Math- Need help =/

  • Thread starter 1calculus1
  • Start date
  • #1
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Simple Math-- Need help =/

1. Homework Statement [/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

Homework Equations



Simplifying skills.

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?
 

Answers and Replies

  • #2
763
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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?
 
  • #3
39
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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.
 
  • #4
39
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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.
 
  • #5
19
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[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?
 
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