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Math Help - Having trouble factoring

  1. #1
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    Having trouble factoring

    Hey guys..I'm having trouble factoring some functions...I can pretty easily factor and find the solutions for linear and quadratic functions, but when it comes to cubic and..um..higher-powered functions, I'm pretty lost...

    The first problem is:

    x^3 + 216 = 0

    I know to start I should set 216 to its cubic form,

    x^3 + 6^3 = 0, but I have no idea what to do after that...

    And for the next problem I have

    x^4 - x^3 + x - 1 = 0

    I worked it out like this:

    x^3(x - 1) + 1 (x - 1 )= 0
    (x^3+1) (x-1) = 0

    so I solve for x

    x^3 + 1 = 0
    (no solution) and

    x-1 = 0
    x=1....but the answer says its x = + or - 1...So I definitely did something wrong...

    So if anyone could help me I'd greatly appreciate it
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  2. #2
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    Quote Originally Posted by jonjon1324 View Post
    Hey guys..I'm having trouble factoring some functions...I can pretty easily factor and find the solutions for linear and quadratic functions, but when it comes to cubic and..um..higher-powered functions, I'm pretty lost...

    The first problem is:

    x^3 + 216 = 0

    I know to start I should set 216 to its cubic form,

    x^3 + 6^3 = 0, but I have no idea what to do after that... make use of the sum of cubes formula!

    And for the next problem I have

    x^4 - x^3 + x - 1 = 0

    I worked it out like this:

    x^3(x - 1) + 1 (x - 1 )= 0
    (x^3+1) (x-1) = 0

    so I solve for x

    x^3 + 1 = 0
    (no solution) and What's (-1)^3?!

    x-1 = 0
    x=1....but the answer says its x = + or - 1...So I definitely did something wrong...

    So if anyone could help me I'd greatly appreciate it
    .
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  3. #3
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    "x^3 + 6^3 = 0, but I have no idea what to do after that... make use of the sum of cubes formula!"



    Okay so the sum of cubes formula is (x+y)(x^2 - xy + y^2) right?
    So if I use it I get (x+6)(x^2-6x+36) ....is it normal for me not to be able to factor the
    (x^2-6x+36) part? I should use the quadratic formula for that right?
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  4. #4
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    Quote Originally Posted by jonjon1324 View Post
    "x^3 + 6^3 = 0, but I have no idea what to do after that... make use of the sum of cubes formula!"



    Okay so the sum of cubes formula is (x+y)(x^2 - xy + y^2) right?
    So if I use it I get (x+6)(x^2-6x+36) ....is it normal for me not to be able to factor the
    (x^2-6x+36) part? I should use the quadratic formula for that right?
    You won't be able to factorise it any further... check the discriminant, it will be negative...
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  5. #5
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    Oh okay..so x is just -6.

    and

    x^3 + 1 = 0
    (no solution) and What's (-1)^3?!

    You're right..I forgot you COULD find the cubed root of a negative number...it's the square root you can't find..Okay I get it now...Thanks :-D

    Can anyone give me any hints on how to start this problem?

    X^4 - 4x^2 + 3 = 0
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  6. #6
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    Quote Originally Posted by jonjon1324 View Post
    Oh okay..so x is just -6.

    and

    x^3 + 1 = 0
    (no solution) and What's (-1)^3?!

    You're right..I forgot you COULD find the cubed root of a negative number...it's the square root you can't find..Okay I get it now...Thanks :-D

    Can anyone give me any hints on how to start this problem?

    X^4 - 4x^2 + 3 = 0
    I assume that it's x^4 - 4x^2 + 3 = 0.

    I would solve this using a dummy variable, X = x^2.

    So now you get a quadratic

    X^2 - 4X + 3 = 0

    (X - 3)(X - 1) = 0

    X - 3 = 0 or X - 1 = 0

    X = 3 or X = 1.


    Now solve for x, remembering that X = x^2.

    Case 1:

    x^2 = 3

    x = \pm \sqrt{3}.


    Case 2:

    x^2 = 1

    x = \pm 1.


    So your four solutions are

    x = \{-\sqrt{3}, -1, 1, \sqrt{3}\}.
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  7. #7
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    Oh yeah I had forgotten about the whole substitution thing...this makes more sense now.


    So would substitution be appropriate to use in a problem such as
    x^3 - 6x^2 - 9x = 0? Because I'm not seeing a way lol

    When exactly is it useful to substitute for polynomials and when is it not?
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  8. #8
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    Quote Originally Posted by jonjon1324 View Post
    Oh yeah I had forgotten about the whole substitution thing...this makes more sense now.


    So would substitution be appropriate to use in a problem such as
    x^3 - 6x^2 - 9x = 0? Because I'm not seeing a way lol

    When exactly is it useful to substitute for polynomials and when is it not?
    Try taking out a common factor of x first...
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    Okay now I feel really dumb about not seeing that last one lol

    I'm trying to also do substitution here but it doesn't seem possible:

    3x(x-1)^1/2 + 2 (x-1)^3/2 = 0

    It's not coming out right so I'm thinking I'm not supposed to use substitution?
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  10. #10
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    Quote Originally Posted by jonjon1324 View Post
    Okay now I feel really dumb about not seeing that last one lol

    I'm trying to also do substitution here but it doesn't seem possible:

    3x(x-1)^1/2 + 2 (x-1)^3/2 = 0

    It's not coming out right so I'm thinking I'm not supposed to use substitution?
    Note that (x - 1)^{3/2} = (x - 1)^{1/2} (x - 1). So take out the common factor of (x - 1)^{1/2}.

    And please start a new thread for a new question next time.
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  11. #11
    Senior Member pacman's Avatar
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    x^3 - 6x^2 - 9x = 0,

    x(x^2 - 6x - 9) = 0,

    a quadratic formula awaits you . . . .
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    Note that (x - 1)^{3/2} = (x - 1)^{1/2} (x - 1). So take out the common factor of (x - 1)^{1/2}.

    And please start a new thread for a new question next time.
    Oh sorry about that.

    But I don't get what you did...What happened to the 3x? and the 2? I'm confused
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  13. #13
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    Quote Originally Posted by pacman View Post
    x^3 - 6x^2 - 9x = 0,

    x(x^2 - 6x - 9) = 0,

    a quadratic formula awaits you . . . .
    What did you do? I don't know how you got that
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  14. #14
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    Quote Originally Posted by jonjon1324 View Post
    [snip]
    I'm trying to also do substitution here but it doesn't seem possible:

    3x(x-1)^1/2 + 2 (x-1)^3/2 = 0

    It's not coming out right so I'm thinking I'm not supposed to use substitution?
    Quote Originally Posted by mr fantastic View Post
    Note that (x - 1)^{3/2} = (x - 1)^{1/2} (x - 1). So take out the common factor of (x - 1)^{1/2}.
    And please start a new thread for a new question next time. Mr F adds: If you look over this thread I'm sure one of the reasons why this is required is becoming more and more obvious as the thread becomes more and more convoluted and difficult to follow. (And it gets more and more difficult to post replies in such a way as to try and avoid that problem).
    Quote Originally Posted by jonjon1324 View Post
    Oh sorry about that.

    But I don't get what you did...What happened to the 3x? and the 2? I'm confused
    I have told you how to re-write one of the terms in your expression so that it makes it easier for you to see what the common factor is.
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    Quote Originally Posted by mr fantastic View Post
    I have told you how to re-write one of the terms in your expression so that it makes it easier for you to see what the common factor is.
    Oh okay so that's not the actual problem, that's you showing what to factor out..I get it...So I tried it, and I think I did it correctly..can you let me know if I did?

    3x(x-1)^1/2 + 2(x-1)^3/2 = 0
    I factor out (x-1)^1/2

    (x-1)^1/2 (3+2(x-1)) = 0
    (x-1)^1/2 (3+2x-1) = 0
    (x-1)^1/2 (1+2x) = 0

    I then squared everything to get rid of the ^1/2
    and got
    (x-1) (1+4x+4x^2) = 0

    x-1 = 0, x=1

    and factored 4x^2+4x+1 = 0 t get x = -1/2, but I plugged it in the original equation and found that it didn't work out...

    Did I do this correctly?

    Edit: and would it be better to start a new thread for every question I have, or a thread for similar questions? Because I have a lot of questions but I don't wanna fill the whole first page with my questions
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