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?

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,
$y = {x}^{6} - 3 {x}^{5} - 5 {x}^{4} + 11 {x}^{3} - 48 {x}^{2} + 92 x - 48$

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.
$y = {x}^{6} - 3 {x}^{5} - 5 {x}^{4} + 11 {x}^{3} - 48 {x}^{2} + 92 x - 48$

use -x

$y = {\left(- x\right)}^{6} - 3 {\left(- x\right)}^{5} - 5 {\left(- x\right)}^{4} + 11 {\left(- x\right)}^{3} - 48 {\left(- x\right)}^{2} + 92 \left(- x\right) - 48$

The result is

$y = {x}^{6} + 3 {x}^{5} - 5 {x}^{4} - 11 {x}^{3} - 48 {x}^{2} - 92 x - 48$

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.
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Actually, the roots of the polynomial
$y = {x}^{6} - 3 {x}^{5} - 5 {x}^{4} + 11 {x}^{3} - 48 {x}^{2} + 92 x - 48$
are

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

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