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    #1
    I'm just curious as to how to go about factoring a polynomial like this one $6x^4+17x^3-24x^2-53x+30$ without using rational root theorem?

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  2. Pessimist Singularitarian
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    #2
    You could write:

    $ \displaystyle 6x^4+17x^3-24x^2-53x+30=\left(6x^4+12x^3\right)+\left(5x^3+10x^2\right)-\left(34x^2+68x\right)+\left(15x+30\right)=6x^3(x+2)+5x^2(x+2)-34x(x+2)+15(x+2)=(x+2)\left(6x^3+5x^2-34x+15\right)$

    $ \displaystyle 6x^3+5x^2-34x+15=\left(6x^3+18x^2\right)-\left(13x^2+39x\right)+\left(5x+15\right)=6x^2(x+3)-13x(x+3)+5(x+3)=(x+3)\left(6x^2-13x+5\right)$

    $ \displaystyle 6x^2-13x+5=(2x-1)(3x-5)$

    And so we have:

    $ \displaystyle 6x^4+17x^3-24x^2-53x+30=(x+2)(x+3)(2x-1)(3x-5)$

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    #3 Thread Author
    Quote Originally Posted by MarkFL View Post
    You could write:

    $ \displaystyle 6x^4+17x^3-24x^2-53x+30=\left(6x^4+12x^3\right)+\left(5x^3+10x^2\right)-\left(34x^2+68x\right)+\left(15x+30\right)=6x^3(x+2)+5x^2(x+2)-34x(x+2)+15(x+2)=(x+2)\left(6x^3+5x^2-34x+15\right)$

    $ \displaystyle 6x^3+5x^2-34x+15=\left(6x^3+18x^2\right)-\left(13x^2+39x\right)+\left(5x+15\right)=6x^2(x+3)-13x(x+3)+5(x+3)=(x+3)\left(6x^2-13x+5\right)$

    $ \displaystyle 6x^2-13x+5=(2x-1)(3x-5)$

    And so we have:

    $ \displaystyle 6x^4+17x^3-24x^2-53x+30=(x+2)(x+3)(2x-1)(3x-5)$
    What technique did you use to determine the pair of numbers to be used to rewrite the expression? Was it by trial and error?

  4. Pessimist Singularitarian
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    #4
    Quote Originally Posted by paulmdrdo View Post
    What technique did you use to determine the pair of numbers to be used to rewrite the expression? Was it by trial and error?
    Yes, there is generally guesswork involved in factoring polynomials. What I actually did was used W|A to factor it, and then constructed my post based on that.

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