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Factorising a complicated Quadratic

  1. Apr 3, 2010 #1
    Hello, im revising eigenvalues in matrices and ive come across a part where i need to factorise the following quadratic equation; let lander = x
    36-36x+11x^2-x^3 = 0
    I know that the answer is (x-2)(-x^2+9x-18), but i dont know how he got to it.
    Ive look through google but i cant find any information on quadratics of this type.
    Thanks in advance
  2. jcsd
  3. Apr 3, 2010 #2


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    Staff: Mentor

    Quadratic suggests polynomial of 2nd degree, that is one with the highest power 2. Here you have x3, so it is a cubic polynomial.
  4. Apr 3, 2010 #3
    aaah i see, i know it = 0 however, thats probably why i havent been able to find search results, ill have a look again, thank you
  5. Apr 3, 2010 #4
    hhhm, im still not finding any decent material. I will continue this in the morning.
  6. Apr 4, 2010 #5
    can someone please go through how this is done, i cant find an easy way to do it, or any help or google
  7. Apr 4, 2010 #6
    You can try to use the Rational Roots theorem. If x = p/q with p and q integers that don't have any factors in common, then p will divide the constant term and q will divide the coefficent of the highest power of x.

    In this case we have:

    f(x) = 0


    f(x) = x^3 - 11 x^2 + 36 x - 36

    A rational zero of f thus has to be an integer that divides 36. They can thus be:

    x= ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36

    Since were too lazy to try out all these 18 candidates, we're going to use the following trick. Let's substitute for x some random integer r such that f(r) is some ugly number with only a few divisors. How will that help?

    Consider the polynomial g(y) defined as:

    g(y) = f(r+y)

    We apply the Rational Roots theorem to g(y). We note that to do that you don't need to expand out f(r+y), all you need are the coefficients of y^3 and the constant term of f(1+y). The constant term is the value for y = 0, which is f(r) = ugly number with only a few divisors. The coefficient of y^3 is the same as the coefficient of x^3, so the rational roots of g(y) are those few divisors of the ugly number f(r)

    Since g(y) = f(r+y), adding r to these candidates gives the possible rational roots for f.

    E.g., we have that f(5) = -6. Tis means that the possible rational roots of g(y) = f(5+y) are:

    y = ±1, ±2, ±3, ±6

    The possible rational roots of f are thus:

    x = 5+y = -1, 2, 3, 4, 6, 7, 8, 11

    But the rational roots also have to be divisors of 36, therefore we can strike out any items in the list that are not divisors of 36, leaving us with:

    x = -1, 2, 3, 4, 6,

    If we try f(-1), we see that f(-1) = -84. If we then take r = -1 and play the same game as above, we see that the list of candidates is reduced to:

    x = 2, 3, 6

    These are in fact the three roots of f.
  8. Apr 4, 2010 #7
    okay, thanks for the reply, but how does knowing the roots help me in factoring the expression.
    What i am trying to do is but it into two brackets.
    Sorry if i sound stupid, but im not seeing as too how this helps. X can be treated as any value when i put it into brackets, so why am i finding the roots? that was all way too complicated for me im afraid.
  9. Apr 4, 2010 #8


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