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Math Help - Further factorisation

  1. #1
    M.R
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    Further factorisation

    I am stuck in this two question. I have gone as far as i can, and i provided the answer. So if someone can fill in the blank it would be great:

    (a) x^7 + x
    = x (x^6 + 1)
    = x ((x^2)^3 + 1^3)
    = x (x^2 + 1) (x^4 - x^2 + 1)
    .
    .
    .
    = x (x^2 + 1) (x^2 - \sqrt{3}x + 1) (x^2 + \sqrt{3}x + 1)


    (b) x^{12} - y^{12}
    = (x^4)^3 - (y^4)^3
    = (x^4 - y^4) (x^8 + x^4 y^4 + y^8)
    = (x^2 - y^2) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)
    = (x - y) (x + y) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)
    .
    .
    .
    = (x - y) (x + y) (x^2 + y^2) (x^2 + xy + y^2) (x^2 - xy + y^2) (x^2 - \sqrt{3}xy + y^2) (x^2 + \sqrt{3}xy + y^2)

    Note: The questions are from Cambridge Mathematics 3 Unit (year 11). Exercise 1E, question 11
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  2. #2
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    Hello, M.R!

    (a)\;\;x^7 + x
    . . = x (x^6 + 1)
    . . = x ((x^2)^3 + 1^3)
    . . = x (x^2 + 1) (x^4 - x^2 + 1)

    Answer: . x (x^2 + 1) (x^2 - \sqrt{3}x + 1) (x^2 + \sqrt{3}x + 1)
    Normally, we could stop where you did.
    . . Usually, "factoring" means integer coefficients.
    But they pulled a cute trick and took it a step further.

    That quartic polynomial, x^4 - x^2 + 1, can be factored
    . . if we're desperate enough.


    We have: . x^4 - x^2 + 1

    Add and subtract 3x^2\!:\quad x^4 - x^2 \,{\color{red}+\, 3x^2} + 1 \,{\color{red}-\, 3x^2} \;=\;x^4 + 2x^2 + 1 - 3x^2

    So we have: . (x^2+1)^2 - (\sqrt{3}x)^2 . . . a difference of squares!

    Factor: . \left(x^2 + 1 - \sqrt{3}x\right)\left(x^2+1 + \sqrt{3}x\right) \;= \;\left(x^2 - \sqrt{3}x + 1\right)\left(x^2 + \sqrt{3}x + 1\right)

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  3. #3
    M.R
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    Thanks Soroban. And i have figured out part (b) also. Here it is:

    (b) x^{12} - y^{12}
    = (x^4)^3 - (y^4)^3
    = (x^4 - y^4) (x^8 + x^4 y^4 + y^8)
    = (x^2 - y^2) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)
    = (x - y) (x + y) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)

    Now as for (x^8 + x^4 y^4 + y^8):
    =x^8+2x^4y^4+y^8 - x^4y^4
    =(x^4+y^4)^2 - (x^2y^2)^2
    =(x^4+y^4 - x^2y^2) (x^4+y^4 + x^2y^2)

    As for =(x^4+y^4 + x^2y^2:
    =x^4+2x^2y^2+y^4 - x^2y^2
    =(x^2+y^2)^2 - (xy)^2
    =(x^2-xy+y^2) (x^2+sy+y^2)

    As for (x^4+y^4 - x^2y^2):
    =x^4+2x^2y^2+y^4 - 3x^2y^2
    =(x^2+y^2)^2 - (\sqrt{3}xy)^2
    =(x^2-\sqrt{3}xy+y^2) (x^2+\sqrt{3}xy+y^2)

    Putting it all together it =
    (x - y) (x + y) (x^2 + y^2) (x^2 + xy + y^2) (x^2 - xy + y^2) (x^2 - \sqrt{3}xy + y^2) (x^2 + \sqrt{3}xy + y^2)
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