How do I find out how many roots a polynomial has.?

Question

austinl · Answer

Factor it or use the quadratic equation. Assuming that you are meaning something like.

$x^2+10x+25=0$
Factor it,
(x+5)(x+5)

One root, -5

anonymous · Answer

quadratic equation as in formula.?

anonymous · Answer

this is the problem I'm looking at on my practice sheet, 
x(x^2 + 4)(x^2 - x - 6) = 0 
so 2*2* = 4
2*3= 6

so there's only 2 solutions .?

anonymous · Answer

2 roots I mean

austinl · Answer

Made that way harder than it needed to be.
$x(x^2 + 4)(x^2 - x - 6) = 0$

$\color{blue}{x=0}$

$x^2+4=0$ 
$x^2=-4$
$\color{blue}{x=2i}$

$x^2-x-6=0$
$(x-3)(x+2)=0$
$\Rightarrow~x-3=0,~\color{blue}{x=3}$
$\Rightarrow~x+2=0,~\color{blue}{x=-2}$

Therefore, 3 real roots, 1 complex root!

anonymous · Answer

Okay I'm lost

ganeshie8 · Answer

use this :-
A polynomial of degree $n$ will have $n$ roots

ganeshie8 · Answer

x(x^2 + 4)(x^2 - x - 6) = 0 

clearly its degree is $5$
so it will have exactly $5$ roots

anonymous · Answer

3 roots was the correct one. 

but I'm confused by the x, it also contains a root.?

austinl · Answer

@gjhfdfg I made a slight error above in notation, @ganeshie8 is correct, there are 5 roots to this problem.

It is,
$x=\pm2i ~\Leftrightarrow~2i~\&~-2i$

That is how there are 5 roots,

anonymous · Answer

In my practice test answer key it said there was 3, hmm