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Math Help - Factorising cubic polynomials through short division

  1. #1
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    Thumbs up Factorising cubic polynomials through short division

    Use short division to factorise x^3 - 5x^2 - 2x + 24.



    I don't know how to use short division to factorise this cubic polynomial. I only know how to factorise it through long division. If anyone could show a very simple worked out solution to this it will be appreciated!
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  2. #2
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    Hello, Joker37!

    Use short division to factorise: . x^3 - 5x^2 - 2x + 24
    I'm not familiar with a "short division method."

    I would use the Factor Theorem and then use long division.

    Given a polynomial p(x), if p(a) = 0, then (x-a) is a factor.


    In this problem, the possible roots are factors of 24: . \pm1, \pm2, \pm3, \pm4,\:\hdots

    We find that p(3) = 0\!:\;\;3^3 - 5(3^2) - 2(3) + 24 \:=\:0
    . . Hence, (x-3) is a factor.


    Using long division: . x^3 - 5x^2 - 2x + 24 \;=\;(x-3)(x^2-2x-8)

    . . which factors further: . \boxed{(x-3)(x+2)(x-4)}

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  3. #3
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    Quote Originally Posted by Joker37 View Post
    Use short division to factorise x^3 - 5x^2 - 2x + 24.



    I don't know how to use short division to factorise this cubic polynomial. I only know how to factorise it through long division. If anyone could show a very simple worked out solution to this it will be appreciated!
    Hi Joker37,

    x^3 - 5x^2 - 2x + 24.

    The first thing I would do is try to find a rational root if f(x)=0. According to the Rational Root Theorem, the possibilites are +/-1, +/-2, +/-3, +/-4, +/-6, +/-8, +/-12, +/-24.

    I found that f(3)=0, so I will use synthetic division to find the depressed quadratic which should be easier to factor.

    Code:
    3 | 1  -5  -2  24
            3  -6 -24
       --------------
        1  -2  -8  0
    The resulting quadratic is x^2-2x-8 which can easily be factored into

    (x-4)(x+2)

    So
    x^3 - 5x^2 - 2x + 24=(x-3)(x-4)(x+2).
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Joker37!

    I'm not familiar with a "short division method."
    Yes...it doesn't seem to be quite a popular method.
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  5. #5
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    Quote Originally Posted by Joker37 View Post
    Yes...[short division] doesn't seem to be quite a popular method.
    Could you tell us what it is, so we'll know in the future?

    Thank you!
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    Quote Originally Posted by stapel View Post
    Could you tell us what it is, so we'll know in the future?

    Thank you!
    I,too, am curious. I found Short division - Wikipedia, the free encyclopedia regarding whole number "short" division, but nothing on polynomial "short" division.
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    Quote Originally Posted by Soroban View Post
    Hello, Joker37!

    I'm not familiar with a "short division method."

    I would use the Factor Theorem and then use long division.

    Given a polynomial p(x), if p(a) = 0, then (x-a) is a factor.


    In this problem, the possible roots are factors of 24: . \pm1, \pm2, \pm3, \pm4,\:\hdots
    The possible rational roots, not the possible roots.

    CB
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  8. #8
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    Quote Originally Posted by masters View Post
    I,too, am curious. I found Short division - Wikipedia, the free encyclopedia regarding whole number "short" division, but nothing on polynomial "short" division.
    Maybe it can refer to Shorthand division of polynomials called Synthetic Division or Ruffin's rule
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