How to factor 3rd degree polynomial with 3 terms

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Discussion Overview

The discussion revolves around the challenges of factoring a third-degree polynomial with three terms, specifically the polynomial -x^3 + 12x + 16. Participants explore various techniques and express uncertainty regarding the applicability of these methods.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses frustration that grouping does not seem to work for this polynomial due to its structure.
  • Another participant suggests checking online resources for factoring techniques.
  • A participant claims that +4 is a root of the polynomial based on their observation.
  • It is proposed that the rational root theorem could be useful, listing possible rational roots and suggesting synthetic or polynomial division to find them.
  • A participant shares their experience of being unprepared for the cubic polynomial in a linear algebra homework context, noting a lack of prior examples in class.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for factoring the polynomial, and multiple approaches are discussed without resolution.

Contextual Notes

Participants mention various methods for finding roots and factoring, but there is uncertainty regarding the effectiveness of these methods for the specific polynomial in question.

leroyjenkens
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-x^3+12x+16

Every single technique I read about online of how to factor 3rd degree polynomials, it says to group them. I don't think grouping works with this. I tried but it didn't work, since there's only 3 terms. Apparently I'm not supposed to have a cubic variable without a squared variable? I don't know. But how is this done?

Thanks.
 
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By <eyesight> , +4 is a root of the polynomial.
 
The rational root theorem is a good place to start. For your problem, the only possible candidates for rational roots are ±1, ±2, ±4, ±8, or ±16. You can check each one very quickly by using synthetic division, or a bit more laboriously by using ordinary polynomial division.

Once you find one root of a cubic, the other factor is a quadratic, so you can use the quadratic formula to find the other roots.
 
Thanks for the responses. This was part of my linear algebra homework, and the teacher's answers just shows it factored, as if it's a simple factoring procedure that everyone should know how to do. The only examples gone over in class were the typical quadratic factoring. Math teachers are usually pretty dirty, so it's not surprising she would throw in a cubic and expect us to remember how to do synthetic division or whatever.

Thanks.
 

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