Question

How do you use Descartes rule of signs to analyze all possibilities for the complex zeroes of: `y= x^6 - 3x^5 - 5x^4 + 11x^3 - 48x^2 + 92x - 48`?

Answer 1

There are 5 or 3 or 1 positive real roots
There is 1 negative real root
There are 2 imaginary roots

Explanation:

From the given polynomial equation,
`y=x^6-3x^5-5x^4+11x^3-48x^2+92x-48`

We can see that the number of sign difference `=5`
+ to - count 1
- to + count 1
+ to - count 1
- to + count 1
+ to - count 1

So there are 5 sign differences. These means there are 5 or 3 or 1 positive real roots. The number of sign differences is equal to the number of positive real roots or less than that by an even number. So if its not 5 then it may 3 or 1.

For the negative roots

We have to substititute -x in place of x then inspect the changes in sign.
`y=x^6-3x^5-5x^4+11x^3-48x^2+92x-48`

use -x

`y=(-x)^6-3(-x)^5-5(-x)^4+11(-x)^3-48(-x)^2+92(-x)-48`

The result is

`y=x^6+3x^5-5x^4-11x^3-48x^2-92x-48`

Now let us check the sign difference
+ to - count 1

There is 1 negative real root or less than that by an even number.

The rest are imaginary roots.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Actually, the roots of the polynomial
`y=x^6-3x^5-5x^4+11x^3-48x^2+92x-48`
are

`x=1` and `x=1` and `x=4` and `x=-3` and `x=+2i` and `x=-2i`

God bless....I hope the explanation is useful.