Factoring question

**starswept** · July 8th 2007, 05:46 PM

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

**Jhevon** · July 8th 2007, 05:59 PM

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

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

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

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)

**starswept** · July 8th 2007, 06:53 PM

Originally Posted by Jhevon

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

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

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!

**topsquark** · July 8th 2007, 07:02 PM

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

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

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

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

-Dan

**Jhevon** · July 8th 2007, 07:03 PM

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 $3(x + 2w - pr) \left( x^2 + 4xw + 4w^2 + prx + 2wpr + p^2 r^2 \right)$

**Jhevon** · July 8th 2007, 07:04 PM

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

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

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

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

-Dan

also, the book left out the 2 in front of the prw

**starswept** · July 8th 2007, 07:21 PM

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

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!

**topsquark** · July 8th 2007, 07:22 PM

Originally Posted by Jhevon

also, the book left out the 2 in front of the prw

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?

-Dan

**Jhevon** · July 8th 2007, 07:25 PM

Originally Posted by topsquark

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

In need of some caffeine Jhevon?

-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.

**topsquark** · July 8th 2007, 07:32 PM

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.

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

)

-Dan

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