Master Simplifying Skills: How to Simplify Complex Equations

[1calculus1]

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?

 

[sara_87]

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?

 

[racer]

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.

 

[1calculus1]

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.

 

[misho]

2h+( \frac{16}{h-8} + 2 ) ( \frac{-8}{(h-8)^2} )

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

2h+( \frac{16}{h-8} + 2 ) x

2h+ \frac{16}{h-8} \times x + 2 \times x

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

2h+ \frac{16}{h-8} \times \frac{-8}{(h-8)^2} + 2 \times \frac{-8}{(h-8)^2}

2h+ \frac{16 \times (-8)}{(h-8) \times (h-8)^2} + \frac{2 \times (-8)}{(h-8)^2}

2h+ \frac{-128}{(h-8)^3} + \frac{-16}{(h-8)^2}

Does that clear it up?