# simplifying help

• May 23rd 2012, 11:26 AM
Tweety
simplifying help
Can someone please explain, how $\frac{k^{2}(k+1)^{2} +4(k+1)^{3}}{4}$ simplifies to

$\frac{({k+1}^{2})k^{2}+4(k+1)}}{{4}}$
• May 23rd 2012, 11:35 AM
Plato
Re: simplifying help
Quote:

Originally Posted by Tweety
Can someone please explain, how $\frac{k^{2}(k+1)^{2} +4(k+1)^{3}}{4}$ simplifies to

$\frac{({k+1}^{2})k^{2}+4(k+1)}}{{4}}$

Well it does not!

$\frac{k^{2}(k+1)^{2} +4(k+1)^{3}}{4}=\frac{(k+1)^{2}[k^{2} +4(k+1)]}{4}=\frac{(k+1)^{2}[(k+2)^2]}{4}$
• May 23rd 2012, 11:51 AM
Tweety
Re: simplifying help
Quote:

Originally Posted by Plato
Well it does not!

$\frac{k^{2}(k+1)^{2} +4(k+1)^{3}}{4}=\frac{(k+1)^{2}[k^{2} +4(k+1)]}{4}=\frac{(k+1)^{2}[(k+2)^2]}{4}$

but your second expression is exactly the same as my second expression. I know it simplifies to your last expression, but I dont understand the second part.
• May 23rd 2012, 11:58 AM
Plato
Re: simplifying help
Quote:

Originally Posted by Tweety
but your second expression is exactly the same as my second expression. I know it simplifies to your last expression, but I dont understand the second part.

No it is not.
Look at the grouping symbol $[~~]$
• May 23rd 2012, 11:59 AM
Tweety
Re: simplifying help
I think I got it, we factor out a (k+1) term, from $(k+1)^{2}$ and from $(k+1)^{3}$
• May 24th 2012, 06:06 AM
Wilmer
Re: simplifying help
Make your life easier: remove the division by 4; let x = k+1