Polynomial: Number of solutions

Question

Functions of polynomials often have more than one solution. For example, $x^2 = b$ with positive $b$ has two solutions for $x$.

How does that work for higher polynomials? Say, I have for positive $a,b,c$ and natural $y$

$$ (ax + b)^y = c \\ x_0 = \frac{c^{1/y}-b}{a}$$

Clearly, $x_0$ is one solution. How can I find potential real additional solutions for $y > 1$?

Are you talking about real solutions, or are you also interested in complex solutions? — Pedro M., May 30 '15 at 00:18
Hint. Look up "fundamental theorem of algebra". The wikipedia page is a good place to start. — Ethan Bolker, May 30 '15 at 00:18
Which are you interested in, finding the number of solutions or finding solutions? These are very different questions. — Robert Israel, May 30 '15 at 00:25
@RobertIsrael both, to be honest. But I guess the number of distinct solutions is a start. — FooBar, May 30 '15 at 01:09
@Meelo I am aware that it has at most $y$ roots. I'd like to check how many root it exactly has. (I thought that was the same as finding the distinct solutions, but apparently it is not) — FooBar, May 30 '15 at 01:12
For real and the given example: if y is odd than one, if even than two. If you allow complex solution, than you have y. — Moti, May 30 '15 at 06:06
@Moti I think that settles it for me here. If you'd post that as an answer, I'll accept it. — FooBar, May 30 '15 at 12:26

score 0 · Accepted Answer · answered May 31 '15 at 17:20

0

For real and the given example: if y is odd than one, if even than two. If you allow complex solution, than you have y

answered May 31 '15 at 17:20

Moti

1 Answers1