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Math Help - Factoring question

  1. #1
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    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
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    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)
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    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!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by starswept View Post
    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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    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)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    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
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  7. #7
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    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!
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    also, the book left out the 2 in front of the prw
    Quote Originally Posted by topsquark View Post
    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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    (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.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    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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