Factoring Equations with Real Roots

[Benzoate]

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

 

[th3plan]

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

 

[RyanSchw]

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 =(

 

[physixguru]

observe the degradation in powers carefully...

 

[Kurret]

It doesn't have any real factors, since:
$$x^4-x^3-x^2+4=(x^2-\frac{x}{2}-\frac{6}{8})^2+~(\frac{x}{2}-\frac{6}{8})^2+\frac{184}{64} > 0$$
:)

 

[tiny-tim]

But every real quartic equation has real quadratic factors (because the complex roots come in conjugate pairs):

(x² + ax + b)(x² + cx + d) ​

 

[Kurret]

uhm didnt know that! but i mistyped a little, i meant real roots, not real factors.