Factoring Equations with Real Roots

AI Thread Summary
The discussion centers on the difficulty of factoring the polynomial equation x^4 - x^3 - x^2 + 4 = 0, with participants noting that it appears to have no real roots since the graph does not intersect the x-axis. One contributor explains that while the polynomial does not have real roots, every quartic equation can be expressed as a product of real quadratic factors. The conversation highlights the confusion between real roots and real factors, clarifying that the original poster was specifically inquiring about real roots. Ultimately, the consensus is that the polynomial does not yield real roots, supported by an analysis of its structure. Understanding these concepts is crucial for tackling higher-degree polynomials effectively.
Benzoate
Messages
418
Reaction score
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
 
Physics news on Phys.org
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
 
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 =(
 
observe the degradation in powers carefully...
 
It doesn't have any real factors, since:
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
:)
 
But every real quartic equation has real quadratic factors (because the complex roots come in conjugate pairs):

(x² + ax + b)(x² + cx + d) :smile:
 
uhm didnt know that! but i mistyped a little, i meant real roots, not real factors.
 
Back
Top