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Math Help - Factorise a polynomial without given the equation

  1. #1
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    Question Factorise a polynomial without given the equation

    P(x) is an even monic polynomial of degree 4 with integer coefficients. One zero is 2i and the product of the zeros is -8. Factorise P(x) over R.

    the answer is -2i, 2i, root 2, - root2

    i know that if 2i is a root, then -2i is also another root, but I dont get how they got root 2 and - root 2

    I can let the other roots be "a" and "b"
    where ab = -2 (since product of roots is -8), however I don't understand how you could get root 2, - root 2 from that

    Thanks
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  2. #2
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    Quote Originally Posted by differentiate View Post
    P(x) is an even monic polynomial of degree 4 with integer coefficients. One zero is 2i and the product of the zeros is -8. Factorise P(x) over R.

    the answer is -2i, 2i, root 2, - root2

    i know that if 2i is a root, then -2i is also another root, but I dont get how they got root 2 and - root 2

    I can let the other roots be "a" and "b"
    where ab = -2 (since product of roots is -8), however I don't understand how you could get root 2, - root 2 from that

    Thanks
    I can let the other roots be "a" and "b"
    Then the product of roots is 2i*-2i*ab = -8 . So ab = -2.
    In an even monic polynomial odd powered x terms are missing.
    So the sum of the roots must be zero.
    So a+b = 0 and ab = -2. Now find a and b.
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by differentiate View Post
    P(x) is an even monic polynomial of degree 4 with integer coefficients. One zero is 2i and the product of the zeros is -8. Factorise P(x) over R.

    the answer is -2i, 2i, root 2, - root2

    i know that if 2i is a root, then -2i is also another root, but I dont get how they got root 2 and - root 2

    I can let the other roots be "a" and "b"
    where ab = -2 (since product of roots is -8), however I don't understand how you could get root 2, - root 2 from that

    Thanks
    If the polynomial is even, then for every root a\neq 0 you have another root b=-a (of the same order). Since you have only two roots left, and their product must be -2, they must have absolute value \sqrt{2}; and they must be real, because if they were a pair of conjugate complex numbers x\pm i\cdot y their product would be x^2+y^2>0.
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