# Simple Factoring question that is bugging me

• Jun 12th 2008, 03:54 PM
Charles Dexter Ward
Simple Factoring question that is bugging me
I'm trying to factor

s^3 + 3s^2 + 3s + 1

From pascals triangle I know the answer by looking at it. But when I go to try and factor it showing all work I get stumped.. I can't seem to get (s + 1)^3 when factoring the long way and not just by looking at it.

I've tried grouping, long division, I can't get it. Can anyone show me what I'm missing?
• Jun 12th 2008, 04:33 PM
topsquark
Quote:

Originally Posted by Charles Dexter Ward
I'm trying to factor

s^3 + 3s^2 + 3s + 1

From pascals triangle I know the answer by looking at it. But when I go to try and factor it showing all work I get stumped.. I can't seem to get (s + 1)^3 when factoring the long way and not just by looking at it.

I've tried grouping, long division, I can't get it. Can anyone show me what I'm missing?

\$\displaystyle s^3 + 3s^2 + 3s + 1\$

\$\displaystyle = s^3 + (2s^2 + s^2) + (2s + s) + 1\$

\$\displaystyle = (s^3 + 2s^2 + s) + (s^2 + 2s + 1)\$

\$\displaystyle = s(s^2 + 2s + 1) + (s^2 + 2s + 1)\$

\$\displaystyle = (s + 1)(s^2 + 2s + 1) = (s + 1)(s + 1)^2 = (s + 1)^3\$
but you'd probably have to be psychic to see something like that in a more complicated example.

Of course, you can always use the rational root theorem.

-Dan
• Jun 12th 2008, 04:58 PM
Charles Dexter Ward
ahhh, I didn't think about splitting up the s^2 and s like that. Tricky!

ahh I see I was trying to use the rational root theorem( I wasn't aware of it's name before now though ) I just messed up with dividing by s + 1. I also see that I wasn't properly finding the roots, though it just happened to work out this time because the leading coefficient is 1. I wasn't aware of p / n. I was only using the numbers the constant was divisible by.

Thanks for the help, I appreciate it!