Help with factoring Edit: Figured it out:)

[gentobu]

I was going through some of my calc notes and I found this:

f = {(x+h)^3 + x^3} + 3{(x+h) - x}

= h{(x +h)^2 + (x+h)x + x^2} + 3h

= h{(x+h)^2 + (x+h)x + x^2 + 3}

Can someone explain to me how the h and 3h were factored out?

Edit: Damn I typed it out wrong it was actually {(x+h)^3 - x^3} + 3{(x+h) - x} anyway I figured it out.

He was using this rule: x^n - y^n = (x-y){x^(n-1) + x^(n-2)y + x^(n-3)y^2 + ... + y^(n-1)} for n = 1, 2, 3,...

so: f = {(x+h)^3 - x^3}
= {(x+h) - x}{(x+h)^2 + (x+h)x + x^2}
= h{(x +h)^2 + (x+h)x + x^2}

 

[her209]

what don't you get?

 

[gentobu]

Originally posted by: her209
what don't you get?

Sorry, I posted the message before I was finished with it. What I want to know is how the h and 3h were factored out.

 

[Nebor]

While I like to consider myself a math person, having elected to take a lot of Calculus in college (not needed for my IS degree) I can't really grasp it well in it's current format, it hurts my brain. YOu need pretty font. BTW, go buy an hp49... I only wish they had them when I was in college.

 

[spp]

are you sure you have this right?

i mean..... 3h is easy since x+h-x = h............... but how can you factor out h from 2x^3?

 

[gentobu]

Originally posted by: spp
are you sure you have this right?

i mean..... 3h is easy since x+h-x = h............... but how can you factor out h from 2x^3?

Oh I think I understand how 3h was factored out now. Thanks I didnt actually do this problem, it was an example the professor was doing, all I did was copy it down. I clearly remember someone asking him HTF he did all that, and all he said was <broken english>Fac-cher!</broken english> and continued on...