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Math Help - solve x^3 - 12x + 16 = 0

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    solve x^3 - 12x + 16 = 0

    Hi

    I can't see how to factorise this. And can't think of another method. Can someone please help?

    x^3 - 12x + 16 = 0

    Angus
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    This polynomial is \displaystyle P(x) = x^3 - 12x + 16.

    Notice that \displaystyle P(2) = 2^3 - 12\cdot 2 + 16 = 8 - 24 + 16 = 0. So \displaystyle (x - 2) is a factor.

    Long divide to find the quadratic factor and factorise the quadratic if possible.
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    Quote Originally Posted by angypangy View Post
    Hi

    I can't see how to factorise this. And can't think of another method. Can someone please help?

    x^3 - 12x + 16 = 0

    Angus
    Another way....

    12x-x^3=16\Rightarrow\ x\left(12-x^2\right)=16

    16 factorises to 16(1)=8(2)=4(4) so we can quickly discover
    whether or not the function has an integer root.
    If it does, the factoring is made simple.

    x=2\Rightarrow\ 2(12-4)=2(8) works


    x=4 gives -16, but x=-4 gives (-4)(-4)=16


    x^3-12x+16=(x-2)(x+4)(x+c)\Rightarrow\ -8c=16\Rightarrow\ c=-2

    Check that this holds.


    Or

    (x-2)\left(x^2+bx+c\right)=x^3-12x+16\Rightarrow\ c=-8,\;\;bx^2-2x^2=0\Rightarrow\ b=2

    as there is no x^2 term.
    Last edited by Archie Meade; January 22nd 2011 at 05:13 AM.
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    Archie Meade's method assumes that the roots will be integer since those are the only factors he is trying.

    Another method is to use the "rational root theorem"- that if m/n is a rational number root of a polynomial with integer coefficients then the denominator, n, must divide the leading coefficient and the numerator, m, must divide the constant term. Here, the leading coefficent is 1 so the denominator must be 1 or -1 (any rational root must be an integer just as Archie Meade thought) and the constant term is 16 so any rational root must evenly divide 16: 1, -1, 2, -2, 4, -4, 8, -8, 16,or -16 are possible roots. Trying each of those numbers in the polynomial we see that (2)^3- 12(2)+ 16= 8- 24+ 16= 0 and that (-4)^3- 12(-4)+ 16= -64+ 48+ 16= 0
    Using x-2 and x+ 4 as factors, we see that the third factor is x- 2 so that x= 2 is a double root.

    Of course, the rational root theorem can only point out possible rational roots. It may be that a polynomial has no rational number roots but in that case, no method is going to be easy!
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