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Thread: need to factorize x^3 - 6x^2 + 11x - 6

  1. #1

    need to factorize x^3 - 6x^2 + 11x - 6

    x^3 - 6x^2+11x - 6

    I want to factorize it.

    what are easiest and quicktest way to find the factors ?

  2. #2
    Elite Member
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    Try the factors of the constant, 6: [tex]\pm{1}, \;\ \pm{2}, \;\ \pm{3}, \;\ \pm{6}[/tex]

    Try these. Upon dividing, when your cubic reduces to a quadratic you found one.

    For instance, divide by x-3. Does it reduce to a quadratic?.

  3. #3
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    Re: need factors

    Hello, defeated_soldier!

    Factor: [tex]f(x)\:=\:x^3\,-\,6x^2\,+\,11x\,-\,6[/tex]

    There is the Rational Roots Theorem.

    If a polynomial has a rational root, then it is of the form [tex]\frac{n}{d}[/tex]
    . . where [tex]n[/tex] is a factor of the constant term and [tex]d[/tex] is a factor of the leading coefficient.

    The constant term is [tex]6[/tex] with factors: [tex]\,\pm1,\:\pm2,\;\pm3,\:\pm6[/tex]
    The leading coefficient is [tex]1[/tex] with factors: [tex]\,\pm1[/tex]
    . . Hence, the possible roots are (as Galactus pointed out) are: [tex]\,\pm1,\:\pm2,\:\pm3, \:\pm6[/tex]


    Then there is the Factor Theorem.

    If [tex]f(a)\,=\,0[/tex], then [tex](x\,-\,a)[/tex] is a factor of [tex]f(x).[/tex]

    Get it?
    Plug in a number for [tex]x[/tex] ... If it comes out to zero, we've found a factor.


    Try [tex]x\,=\,1:\;\;f(1)\:=\:1^3\,-\,6\cdot1^2\,+\,11\cdot1\,-\,6\:=\:0[/tex] . . . Bingo!
    . . So, we know that [tex](x\,-\,1)[/tex] is a factor.

    Use long (or synthetic) division to get: [tex]\,x^3\,-\,6x^2\,+\,11x\,-\,6\;=\;(x\,-\,1)(x^2\,-\,5x\,+\,6)[/tex]

    Then we can factor the quadratic factor: [tex]\:(x\,-\,1)(x\,-\,2)(x\,-\,3)[/tex]

    I'm the other of the two guys who "do" homework.

  4. #4
    thanks i knew that . In fact i do that way only .....i was thinking there might be some other method .

    Anyway , Thanks for the nice explanation . very helpful

    BTW, just q small question here ,


    you mentioned , "possible roots are +- blha blah " from rational root theorem .

    can I conclude the following thing ? ( i made this conclusion , will you please validate)

    If the polynomoal has real root then All real roots must be from that set ONLY. There cant be any real root other than that set .

  5. #5
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    Quote Originally Posted by defeated_soldier
    thanks i knew that . In fact i do that way only .....i was thinking there might be some other method .

    Anyway , Thanks for the nice explanation . very helpful

    BTW, just q small question here ,


    you mentioned , "possible roots are +- blha blah " from rational root theorem .

    can I conclude the following thing ? ( i made this conclusion , will you please validate)

    If the polynomoal has real root then All real roots must be from that set ONLY. There cant be any real root other than that set .
    NO, NO and NO.

    Rational roots of such an equation must be of the form p/q where "p" is a factor of the constant term of the polynomial and "q" is a factor of the leading coefficient of the polynomial. A polynomial function may have REAL roots which are NOT rational.....

    Just because you can't find any rational roots, that certainly does NOT imply that there are no real roots.

  6. #6
    umm...ok ,

    Actually , my statement was a bit misleading .

    what i actually wanted to say is ,

    If the polynomiaL HAS rational root , then all of them will belong to "rational roots theorem".

    there are no OTHER rational root which does not belong to the "ration root test " .

    is this correct ?

    i made a mistake by stating "real".


    However, do you believe that all the rational roots will be from this test and there are no other rational root exists outside ?

  7. #7
    Senior Member
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    Quote Originally Posted by defeated_soldier
    umm...ok ,


    However, do you believe that all the rational roots will be from this test and there are no other rational root exists outside ?
    I most certainly DO believe that the Rational Roots Theorem identifies all possible rational roots, and that there are NO rational roots that exist outside of the set of possibles identified by the theorem.

  8. #8
    thank you....i understand now.

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