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Math Help - Factoring Binomials

  1. #1
    Junior Member Cait's Avatar
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    Unhappy Factoring Binomials

    For college beginning algebra.

    ~Please explain every step because I'm mathmatically stupid~


    Problem:

    5k^3+40


    The answer is SUPPOSIDLY 5(K+2)(K^2-2K+4)

    but i dont see how they get it.




    Here's the example they put in the book for this type of problem.


    Factor:
    64x^3 + 1

    Solution:
    (4x)^3 + 1^3
    =
    (4x+1)[(4x)^2-(4x)(1)+1^2]
    =
    (4x+1)(16x^2-4x+1)


    and so from what im takin it is..
    if you have
    5k^3+40

    Then first step is.. you gotta make the 40 into a cubed like the 5k^3..

    but.. as far as i know, NOTHING cubed is 40. I had my dad look at this too and he has no idea what to do.. and neither do i.


    Thanks

    ~Cait
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Cait View Post
    For college beginning algebra.

    ~Please explain every step because I'm mathmatically stupid~


    Problem:

    5k^3+40


    The answer is SUPPOSIDLY 5(K+2)(K^2-2K+4)

    but i dont see how they get it.




    Here's the example they put in the book for this type of problem.


    Factor:
    64x^3 + 1

    Solution:
    (4x)^3 + 1^3
    =
    (4x+1)[(4x)^2-(4x)(1)+1^2]
    =
    (4x+1)(16x^2-4x+1)


    and so from what im takin it is..
    if you have
    5k^3+40

    Then first step is.. you gotta make the 40 into a cubed like the 5k^3..

    but.. as far as i know, NOTHING cubed is 40. I had my dad look at this too and he has no idea what to do.. and neither do i.


    Thanks

    ~Cait
    ok, first you need to know the rule:

    x^3 + y^3 = (x + y)(x^2 - xy + y^2)

    now to do the problem:

    5k^3+40 .........first thing to do is factor out the 5

    => 5(k^3 + 8) ..............now 8 = 2^3
    => 5(k^3 + 2^3) ............now apply the rule above
    => 5(k+2)(k^2 - 2k + 2^2) = 5(k+2)(k^2 - 2k + 4)
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  3. #3
    Junior Member Cait's Avatar
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    Unhappy

    awesome.
    that worked.. and also for other probs.

    which leads me to another question! yay.

    How come they want you to split the number 8 into 2 and 4 in the problem

    8a^3+1

    Ans: (2a+1)(4a^2-sa+1)

    and not split the 8 in this problem?

    8r^3-64

    Ans: 8(r-2)(r^2+2r+4)


    It's prolly because it has a +1 at the end insted of.. say.. 3 and 2..?? right??


    there's so many particlar rules about the tiniest things and i will never remember them :|
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Cait View Post
    awesome.
    that worked.. and also for other probs.

    which leads me to another question! yay.

    How come they want you to split the number 8 into 2 and 4 in the problem

    8a^3+1

    Ans: (2a+1)(4a^2-sa+1)

    and not split the 8 in this problem?

    8r^3-64

    Ans: 8(r-2)(r^2+2r+4)


    It's prolly because it has a +1 at the end insted of.. say.. 3 and 2..?? right??


    there's so many particlar rules about the tiniest things and i will never remember them :|

    8a^3+1
    remember 8 = 2^3
    also remember that x^a * y^a = (xy)^a
    now to do the problem

    8a^3 + 1
    = 2^3 * a^3 + 1
    = (2a)^3 + 1^3
    = ((2a) + 1)((2a)^2 - (2a)(1) + 1^2)
    = (2a + 1)(4a^2 - 2a + 1)


    8r^3-64
    two ways to do this one. one way is the same way we did 5k^3 + 40, another way is the way i did the problem before this one.

    First Method:

    8r^3 - 64
    = 8(r^3 - 8)
    = 8(r^3 - 2^3)
    = 8(r - 2)(r^2 + 2r + 2^2)
    = 8(r - 2)(r^2 + 2r + 4)

    Second Method:

    8r^3 - 64
    = 2^3 * r^3 - 4^3
    = (2r)^3 - 4^3
    = ((2r) - 4)((2r)^2 + 8r + 4^2)
    = (2r - 4)(4r^2 + 8r + 16) ...........factor 2 out of the first group and 4 out of the second, we get

    = 2(r - 2) * 4(r^2 + 2r + 4) ........now multiply the 4 and the 2 and put them in front
    = 8(r - 2)(r^2 + 2r + 4)
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Cait View Post
    there's so many particlar rules about the tiniest things and i will never remember them :|
    if you keep thinking like that, you won't

    how come i remember them?
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  6. #6
    Junior Member Cait's Avatar
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    Wink

    Alright, TY again. :3

    how come i remember them?
    Cuz you're smart!



    I got 2 sections left.. Hopefully I wont have any trouble with them..
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Cait View Post
    Alright, TY again. :3



    Cuz you're smart!



    I got 2 sections left.. Hopefully I wont have any trouble with them..
    not only am i not smart, but my memory is horrible, which brings me back to the advice i gave you in another thread

    it gets easier with practice and a good attitude, do a lot of problems and smile and be happy about it
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