- #1
Benzoate
- 420
- 0
Homework Statement
Not sure how to factor this equation:
x^4-x^3-x^2+4=0
btw , I do not have a TI 83 calculator
Factoring Equations with Real Roots
- Thread starter Benzoate
- Start date
-
- Tags
- Algebra Basic algebra
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Homework Help Overview
The discussion revolves around the factoring of a quartic polynomial equation, specifically x^4 - x^3 - x^2 + 4 = 0. Participants are exploring the nature of its roots and the possibility of factoring it.
Discussion Character
- Exploratory, Assumption checking, Conceptual clarification
Approaches and Questions Raised
- Participants express uncertainty about the factoring process for a quartic polynomial and question the existence of real roots. Some suggest that the polynomial may not have real factors or roots based on graphical observations.
Discussion Status
The discussion is ongoing, with various interpretations being explored. Some participants have offered insights into the nature of quartic equations and their roots, while others are questioning assumptions about the polynomial's behavior.
Contextual Notes
There is mention of the original poster's lack of a calculator, which may influence their approach. Additionally, a participant corrects their earlier statement regarding factors to clarify they meant real roots instead.
Not sure how to factor this equation:
x^4-x^3-x^2+4=0
btw , I do not have a TI 83 calculator
- #2
th3plan
- 93
- 0
lol this isn't a simple equation to factor its a 4 exponent polynomial, good luck with that, but there is a method i learn in college, don't remember though
- #3
RyanSchw
- 36
- 0
I'd be curious to know this as well; if I had to guess I’d say that is has no factors given that the graph never crosses the x-axis. I don't know =(
- #4
physixguru
- 335
- 0
observe the degradation in powers carefully...
- #5
Kurret
- 141
- 0
It doesn't have any real factors, since:
[tex]x^4-x^3-x^2+4=(x^2-\frac{x}{2}-\frac{6}{8})^2+~(\frac{x}{2}-\frac{6}{8})^2+\frac{184}{64} > 0[/tex]
:)
[tex]x^4-x^3-x^2+4=(x^2-\frac{x}{2}-\frac{6}{8})^2+~(\frac{x}{2}-\frac{6}{8})^2+\frac{184}{64} > 0[/tex]
:)
- #6
tiny-tim
Science Advisor
Homework Helper
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But every real quartic equation has real quadratic factors (because the complex roots come in conjugate pairs):
(x² + ax + b)(x² + cx + d) 
- #7
Kurret
- 141
- 0
uhm didnt know that! but i mistyped a little, i meant real roots, not real factors.
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