Question

How many solutions over the complex number system does this polynomial have? 7x^5-33x^4-4x^2+3x+52=0

Answer 1

Hello : 
in : C
5 solutions

Answer 2

Answer:

Over the complex number there exist two roots.

Step-by-step explanation:

Given : Polynomial $7x^5-33x^4-4x^2+3x+52=0$

To find : How many solutions over the complex number system does this polynomial have?                

Solution :

Descartes' rule of sign is used to determine the number of real zeros of a polynomial function.

Let $f(x)=7x^5-33x^4-4x^2+3x+52$

Now, for positive roots write the sign of f(x) and note the changes,

f(x)=+ - - + + i.e, from + to - and - to + two times sign changes.

So, Two positive roots.

Now, for negative roots write the sign of f(-x) and note the changes,

f(-x)=- - - + + i.e, from - to + one time sign changes.

So, One negative roots.

Since, The polynomial is of 5 degree so five roots exist.

i.e, rest two roots are complex.

Therefore, Over the complex number there exist two roots.