Question

How many complex roots does the equation

Answer 1

we are given

$3x^5-x^4-5x^2+1=0$

We can use Descartes rule of sign

This is rule is used to find number positive real roots and negative real roots

Positive real roots:

$3x^5-x^4-5x^2+1=0$

We will find number of sign changes in this equation

We can it changes from +3 to -1  and -5 to +1

so, number of sign changes =2

so, total number of positive real roots =2

Negative real roots:

We can replace x as -x

$3(-x)^5-(-x)^4-5(-x)^2+1=0$

now, we can simplify it

$-3x^5-x^4-5x^2+1=0$

now, we will find number of sign changes

we can see that it changes from -5 to +1 only

so, number of sign changes =1

so, number of negative real roots =1

Complex roots:

We are given

degree of polynomial =5

so, total number of roots=5

we know that total root is sum of all roots

Total number of roots = number of positive real roots + number of negative real roots + number of complex roots

now, we can plug values

5=2+1+number of complex roots

so, we get

Number of complex roots =2.........Answer

Answer 2

Here to confirm that the answer is "5", just took the test and got it right so I hope I could help :)