How can I factor a polynomial without using the rational root theorem?

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Discussion Overview

The discussion revolves around the methods for factoring the polynomial $6x^4+17x^3-24x^2-53x+30$ without employing the rational root theorem. Participants explore various techniques and approaches to achieve this goal.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant presents a detailed factorization of the polynomial, breaking it down into manageable parts and ultimately arriving at the factorization $(x+2)(x+3)(2x-1)(3x-5)$.
  • Another participant reiterates the same factorization steps, indicating a shared approach to the problem.
  • A question is raised regarding the technique used to determine the pairs of numbers for rewriting the expression, suggesting a method of trial and error.
  • A later reply acknowledges that guesswork is often involved in factoring polynomials and mentions using Wolfram Alpha (W|A) as a tool to assist in the factorization process.

Areas of Agreement / Disagreement

Participants generally agree on the factorization method presented, but there is no consensus on the specific techniques used to identify the necessary pairs of numbers for rewriting the polynomial.

Contextual Notes

The discussion does not clarify the underlying assumptions or limitations of the methods used, nor does it address the potential for alternative approaches to factoring the polynomial.

paulmdrdo1
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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?

Thanks
 
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You could write:

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

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

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

And so we have:

$$6x^4+17x^3-24x^2-53x+30=(x+2)(x+3)(2x-1)(3x-5)$$
 
MarkFL said:
You could write:

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

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

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

And so we have:

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

What technique did you use to determine the pair of numbers to be used to rewrite the expression? Was it by trial and error?
 
paulmdrdo said:
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. :D
 

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