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Math Help - help in factoring 6 roots

  1. #1
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    help in factoring 6 roots

    i need help factoring this eqn.
    x^6 + 9x^4 + 24x^2 + 16 = 0

    how did it get
    (x^2+1)(x^2+4)^2
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    i need help factoring this eqn.
    x^6 + 9x^4 + 24x^2 + 16 = 0

    how did it get
    (x^2+1)(x^2+4)^2
    So that the large powers don't get to you (and to make long division easier), let y = x^2, then we need to solve:

    y^3 + 9y^2 + 24y + 16 = 0

    we see that y = -1 is a root, by the remainder theorem, thus y + 1 is a factor by the factor theorem

    doing y^3 + 9y^2 + 24y + 16 \div y + 1 by long or synthetic division, we find that

    y^3 + 9y^2 + 24y + 16 = (y + 1)(y^2 + 8y + 16) = (y + 1)(y + 4)^2

    now replace y with x^2 and you're done
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  3. #3
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    y:=x^2, then the polinom: f(y)=y^3+9y^2+24y+16
    Now you should notice that if we substitute -1: f(y)=-1+9-24+16=0; and for -4: f(y)=-64+144-96+16=0. So -1 and -4 are roots. Therefore: f(y)=(y+1)(y+4)^2
    And if you write back x^2 you get what you want.

    Sorry Jhevon I didn't recognize your post..
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