the polynomial $6x^3-18x^2-6x-6$ can be factored as $6(x-r)(x^2+ax+b)$ for some $a,b \in \Bbb{R}$ and where $r$ is a real root fo the polynomial. How would you prove that the polynomial $x^2+ax+b$ has no real roots. I know that you can do polynomial long division to get concrete values for $a$ and $b$ but $r$ is a decimal so the long division would be messy and inaccurate, but other then that method I have no idea how i would prove that it has no real roots.
Subscribe to:
Post Comments (Atom)
real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
$$ 3x+6y+5z=7 $$ The general solution to this linear Diophantine equation is as described here (Page 7-8) is: $$ x = 5k+2l+14 $$ $$ y = -l $...
-
I'm having trouble understanding a certain property of CDFs for negative random variables. Let $Y$ be an exponential random variable and...
-
I am trying to solve the following two problems: 1) Prove that the functions $\cos(z)$, $\sin(z)$ are surjective over the complex numbers. 2...
No comments:
Post a Comment