# Factoring question

• Jul 8th 2007, 05:46 PM
starswept
Factoring question
In my endless assignment of factoring questions, I've come across another that I can't quite get:

3(x+2w)^3 - 3p^3r^3

Thanks for any help :)
• Jul 8th 2007, 05:59 PM
Jhevon
Quote:

Originally Posted by starswept
In my endless assignment of factoring questions, I've come across another that I can't quite get:

3(x+2w)^3 - 3p^3r^3

Thanks for any help :)

$\displaystyle 3(x + 2w)^3 - 3p^3 r^3$ .........factor out the 3

$\displaystyle \Rightarrow 3 \left( (x + 2w)^3 - (pr)^3 \right)$

Now the inside of the brackets is the difference of two cubes, which you learned a formula for in class recently (i think), so you can take it from here

(if not say so)
• Jul 8th 2007, 06:53 PM
starswept
Quote:

Originally Posted by Jhevon
$\displaystyle 3(x + 2w)^3 - 3p^3 r^3$ .........factor out the 3

$\displaystyle \Rightarrow 3 \left( (x + 2w)^3 - (pr)^3 \right)$

Now the inside of the brackets is the difference of two cubes, which you learned a formula for in class recently (i think), so you can take it from here

(if not say so)

Thanks again Jhevon :) I factored out the three, and used the formula I know, which is (a^3-b^3) = (a-b)(a^2+ab+b^2).

I ended up with 3(x+2w-pr)(x^2-prx-2prw+p^2r^2). However, the answer should, according to the text, be 3(x+2x-pr)(x^2 + 4xw + 4w^2 + prx + prw + p^2r^2).

This textbook is notorious for having mistakes in the answers, but it can't be that off, so I must have made a mistake with the formula somewhere, but I can't spot it. Can anyone point out my error? Thanks in advance!
• Jul 8th 2007, 07:02 PM
topsquark
Quote:

Originally Posted by starswept
Thanks again Jhevon :) I factored out the three, and used the formula I know, which is (a^3-b^3) = (a-b)(a^2+ab+b^2).

I ended up with 3(x+2w-pr)(x^2-prx-2prw+p^2r^2). However, the answer should, according to the text, be 3(x+2x-pr)(x^2 + 4xw + 4w^2 + prx + prw + p^2r^2).

This textbook is notorious for having mistakes in the answers, but it can't be that off, so I must have made a mistake with the formula somewhere, but I can't spot it. Can anyone point out my error? Thanks in advance!

There is a slight problem in the book's answer, but I suspect it's typos on your part. (See the red x above and the coefficient on the wpr term.) I'd say you expanded wrong somewhere, but I can't figure out where.
$\displaystyle 3(x+2w)^3 - 3p^3r^3$

$\displaystyle = 3[(x + 2w)^3 - p^3r^3]$

$\displaystyle = 3[(x + 2w) - (pr)][(x + 2w)^2 + (x + 2w)(pr) + (pr)^2]$

$\displaystyle = 3(x + 2w - pr)(x^2 + 4xw + 4w^2 + xpr + 2wpr + p^2r^2)$

-Dan
• Jul 8th 2007, 07:03 PM
Jhevon
Quote:

Originally Posted by starswept
Thanks again Jhevon :) I factored out the three, and used the formula I know, which is (a^3-b^3) = (a-b)(a^2+ab+b^2).

I ended up with 3(x+2w-pr)(x^2-prx-2prw+p^2r^2). However, the answer should, according to the text, be 3(x+2x-pr)(x^2 + 4xw + 4w^2 + prx + prw + p^2r^2).

This textbook is notorious for having mistakes in the answers, but it can't be that off, so I must have made a mistake with the formula somewhere, but I can't spot it. Can anyone point out my error? Thanks in advance!

both of those are incorrect i'm afraid.

the answer should be $\displaystyle 3(x + 2w - pr) \left( x^2 + 4xw + 4w^2 + prx + 2wpr + p^2 r^2 \right)$
• Jul 8th 2007, 07:04 PM
Jhevon
Quote:

Originally Posted by topsquark
There is a slight problem in the book's answer, but I suspect it's typos on your part. (See the red x above and the coefficient on the wpr term.) I'd say you expanded wrong somewhere, but I can't figure out where.
$\displaystyle 3(x+2w)^3 - 3p^3r^3$

$\displaystyle = 3[(x + 2w)^3 - p^3r^3]$

$\displaystyle = 3[(x + 2w) - (pr)][(x + 2w)^2 + (x + 2w)(pr) + (pr)^2]$

$\displaystyle = 3(x + 2w - pr)(x^2 + 4xw + 4w^2 + xpr + 2wpr + p^2r^2)$

-Dan

also, the book left out the 2 in front of the prw
• Jul 8th 2007, 07:21 PM
starswept
I see my mistake. I conveniently squared only the x of (x+2w), and forgot the 2w existed. Stupid errors like that are why taking accelerated calculus after no math for six months is dangerous :rolleyes: And as sad as it is, Dan, those errors in the book's equation were not typos, but from the book straight-on. This book has so many errors that having the answers can be more of a hindrance than a help. It makes getting the right answer a fun game of telephone tag for people in my math class ;)

Anyways, thanks again for the help, both of you!
• Jul 8th 2007, 07:22 PM
topsquark
Quote:

Originally Posted by Jhevon
also, the book left out the 2 in front of the prw

Quote:

Originally Posted by topsquark
There is a slight problem in the book's answer, but I suspect it's typos on your part. (See the red x above and the coefficient on the wpr term.)

(Chuckles) As wpr = prw, I believe I mentioned that...

In need of some caffeine Jhevon? :D

-Dan
• Jul 8th 2007, 07:25 PM
Jhevon
Quote:

Originally Posted by topsquark
(Chuckles) As wpr = prw, I believe I mentioned that...

In need of some caffeine Jhevon? :D

-Dan

nah, i don't drink anything with caffeine, well, not that i know of, they put caffeine into a lot of things. i need some sleep though. :D
• Jul 8th 2007, 07:32 PM
topsquark
Quote:

Originally Posted by Jhevon
nah, i don't drink anything with caffeine, well, not that i know of, they put caffeine into a lot of things. i need some sleep though. :D

Hmmm.... Sounds almost Mormon. (I can get away with saying that because I am one. ;) )

-Dan