Distribution in Algebra: Calculating 16(1+h)^2 - 16(1)^2/h

[robertjford80]

Homework Statement

This is a calc problem but it's the algebra part I'm having trouble with:

The Attempt at a Solution

[16(1+h)² - 16(1)²]/h

[16 + 32h + 16h² - 16²]/h

= 16/h + 32 + 16h - 16²

this is a calc problem so h approaches zero

= 32 + 16h - 16²

= 16(h + 2) - 16²

I can't figure out why - 16² disappears.

 

[robertjford80]

<deleted>

 

[Infinitum]

= 16/h + 32 + 16h - 16²

this is a calc problem so h approaches zero

= 32 + 16h - 16²

What happened to that 16/h?? Surely, as h -> 0 16/h is -not- equal to 0.



Try taking the 16 common out of the original expression, giving you

$$\frac{16((1+h)^2-1)}{h}$$

Simplify this further, that should lead you to your answer.