# Factoring and the Distributive Property Question

• Aug 6th 2012, 02:56 PM
Srengam
Factoring and the Distributive Property Question
Hi all,

I've been struggling a little with my algebra so during the summer break I've been working through the book Algebra Demystified and it's really helping.

But I've come across a question half way through that I can't solve, I'm not sure how to tackle it, the question and answer are

\$\displaystyle x^{3}+5x^{2}-x-5=(x-1)(x+1)(x+5)\$

Any tips or a starting point would be greatly appreciated.

Thanks.
• Aug 6th 2012, 03:06 PM
emakarov
Re: Factoring and the Distributive Property Question
Quote:

Originally Posted by Srengam

\$\displaystyle x^{3}+5x^{2}-x-5=(x-1)(x+1)(x+5)\$

This is not a question; this is an equality.
• Aug 6th 2012, 03:08 PM
GJA
Re: Factoring and the Distributive Property Question
Hi, Srengam. First, kudos for working through this stuff!

Typically cubics (when the highest power of x is 3) are not easy to factor. However, the one you have can be done if you group the first two terms together \$\displaystyle x^{3}\$ and \$\displaystyle 5x^{2}\$, and group the second two terms \$\displaystyle -x\$ and -5 together, then factoring these groupings individually.

If that's not clear, let me know and I will write up what I mean more explicitly. Good luck!
• Aug 6th 2012, 03:25 PM
Plato
Re: Factoring and the Distributive Property Question
Quote:

Originally Posted by Srengam
But I've come across a question half way through that I can't solve, I'm not sure how to tackle it, the question and answer are
\$\displaystyle x^{3}+5x^{2}-x-5=(x-1)(x+1)(x+5)\$

Note that \$\displaystyle x^{3}+5x^{2}-x-5=x^2(x+5)-(x+5)=(x^2-1)(x+5)\$
• Aug 7th 2012, 01:14 AM
Srengam
Re: Factoring and the Distributive Property Question
Thanks all,

I can see now what I need to "pull out" to simplify it all.

Thanks again.