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Math Help - factorising

  1. #1
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    factorising

    hi, could someone help me with this please:

    a polynomial is given by p(x) = x^3 - 6x^2 + 9x - 4

    a) factorise p(x) completely- i did this question and came up with (x - 1)^2 (x - 4) but then the next question i didn't get

    b) hence solve the equation y^6 - 6y^4 + 9y^2 - 4 = 0

    i'm guessing its something to do with the second equation being the same as the first one but with x squared, so y = x^2. but the answer in the book is y = \pm 1 and y = \pm2 could someone please explain this? thankyou, mark
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  2. #2
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    (y^2)^3-6(y^2)^2+9(y^2)-4 ... Do you see the similarity between the original polynomial? By taking out a common exponent, I've reproduced an equation extremely similar to p(x), just with a different variable. I've created p(y^2).

    Let's just pretend y^2 is equivalent to x, for the sake of simplicity. You found the original polynomial, factored, to be

    By re-substituting (y^2) into the factored form of the polynomial we get:
    (y^2-1)^2(y^2-4)

    Do you understand my thought process? Are you able to finish it off?
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  3. #3
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    if y = x^2 why are you making x = y^2?
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  4. #4
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    x doesn't actually equal y^2... For the sake of simplicity, it's easier to imagine y^2 as x in order to understand that y^2 can be substituted into the factored form of the p(x).
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  5. #5
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    i can see now that by the answer given in the book that that your way of putting it actually does work, i'm just struggling to see why you can put y^2 in place of x in the brackets
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  6. #6
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    I'll try to make this supremely clear!

    =

    Then,
    p(1)=(1-1)^2(1-4) correct?

    So,
    p(y^2)=(y^2-1)^2(y^2-4)






    I derived y^2 from this equation by realising that I need to manipulate it into a form which somehow relates to ... By using index laws I realised that for to become the substituted value must be y^2.


    Hence,
    p(y^2)=(y^2)^3-6(y^2)^2+9(y^2)-4
    p(y^2)=y^6-6y^4+9y^2-4

    I hope that helped
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  7. #7
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    ok thanks, that has shed a bit of light on it
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  8. #8
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    Hopefully somebody else will be able to articulate a response that best suits your frame of mind

    These questions are somewhat intuitive and are definitely tricky if you've never dealt with them before
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  9. #9
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    true, sometimes i'll read something in the book and not have a clue what it means, then someone will explain in a different way and i'll understand instantly. i do understand your explanation now though so thanks.
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  10. #10
    Senior Member pacman's Avatar
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    ,

    let y^2 = x

    x^3 - 6x^2 + 9x - 4 = 0,

    Next step is to break down the middle terms, a la "trial and error";

    x^3 - 4x^2 - 2x^2 + 8x + x - 4 = 0, then group them,

    (x^3 - 4x^2) + (-2x^2 + 8x) + (x - 4) = 0, factorize

    x^2(x - 4) - 2x(x - 4) + (x - 4) = 0,

    you will notice that we extracted the (x - 4), factor it.

    (x - 4)(x^2 - 2x + 1) = 0, but notice that (x - 1)^2 = x^2 - 2x + 1,

    then we have,

    (x - 4)(x - 1)^2 = 0. Now, it is easy to solve the roots

    but x = y^2,

    then (y^2 - 4)(y^2 - 1)^2 = 0.

    by zero-factor theorem, roots can be obtained.

    y^2 - 4 = (y + 2)(y - 2) = 0,

    y^2 - 1 = (y + 1)(y - 1) = 0,

    y = {-2, -1, 1, 2}, solution set

    Factoring a cubic polynomial is very hard, even a trained eye blinks. More practice helps . . . .
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