Factoring Cubic Polynomials: A Confused Student's Guide

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Homework Help Overview

The discussion revolves around factoring a cubic polynomial, specifically the equation x³ - (3/4)x - (1/4) = 0. Participants express confusion regarding the factoring process and seek clarification on the steps involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore various methods for factoring the polynomial, including completing the square and applying the rational root theorem. Questions arise about the reasoning behind specific steps, such as adding and subtracting terms to facilitate factoring.

Discussion Status

Several participants have offered hints and methods for approaching the problem, including transforming the polynomial into one with integer coefficients and identifying potential rational roots. There is an ongoing exploration of different strategies without a clear consensus on the best approach.

Contextual Notes

Some participants mention the lack of detailed steps in the solution manual, which contributes to the confusion. There is also a reference to the need for creative thinking in problem-solving, indicating varying levels of comfort with the material.

torquerotates
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Well, I admit, I looked at the solution manual. They apparently solved this equation, x(cubed)-(3/4)x-(1/4)=0 by factoring it and making it look like this, (x-1)((
x+(1/2))squared. But they didn't show any steps. I'm completely confused. I have no idea how to factor cubic polynomials.
 
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x^3 - \frac 3 4 x - \frac 1 4=0

Correct?

Here is a hint:

x^3-x^2+x^2-\frac 3 4 x - \frac 1 4=0

Complete the square, then you will notice that you have difference of squares. Continue simplifying, and once again, complete the square and it's solved.
 
Last edited:
Change it into an integer coefficient polynomial by multiplying by 4. So 4x^3-3x-1=0. Now look up the rational root theorem. If p/q is a rational root of the polynomial then p divides 1 and q divides 4. So the only choices for rational roots are 1,1/2 or 1/4 or their negatives. This is important since if p/q is a root of the polynomial then (x-p/q) is a factor.
 
Wow, I don't think I could get used to that type of thinking. How did you even know that you should start by adding and subtracting 2x? Maybe I'm just not creative enough.
 
torquerotates said:
Wow, I don't think I could get used to that type of thinking. How did you even know that you should start by adding and subtracting 2x? Maybe I'm just not creative enough.
I think you meant x^2 but just making sure.

Algebra - https://www.amazon.com/dp/0817636773/?tag=pfamazon01-20

Gelfand is the man!
 
the polynomial p(x) = x^3 - \frac{3}{4}x - \frac{1}{4} is 0 at x = 1, so (x - 1) is a factor of p(x).

Now try solving this question...
 
Dick said:
Change it into an integer coefficient polynomial by multiplying by 4. So 4x^3-3x-1=0. Now look up the rational root theorem. If p/q is a rational root of the polynomial then p divides 1 and q divides 4. So the only choices for rational roots are 1,1/2 or 1/4 or their negatives. This is important since if p/q is a root of the polynomial then (x-p/q) is a factor.

this seems like a pretty cool method.. could you give the general statement please.. or maybe a link somewhere??
 
Just google 'rational root theorem'. Wikipedia has an entry.
 

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