​ Last one. help asap!!!

5.

For Descartes rule you must look at the sign changes of rthew coefficient

\(f(x)=-2x^3+3x^2-5x-2\\ \mbox{-2 then +3 first sign change}\\ \mbox{+3 then -5 second sign change.}\\ \mbox{-5 then -2 Sign did not change}\\ \)

So there is a maximum of 2 roots on the positive x axis. but they could be complex roots so there are either 2 roots or 0 roots on positive x axis.

Now look at f(-x)

\(f(-x)=2x^3+3x^2+5x-2\\ \mbox{+2 then +3 Sign did not change}\\ \mbox{+3 then +5 Sign did not change.}\\ \mbox{+5 then -2 first and only sign change}\\ \)

So there is 1 root on the negative x axis.

So this graph will have  0 or 2 positive roots and 1 negative root.

I've never seen Descarte's rule before.  I like it to find the number of positive roots BUT once you have the number of positive roots it is easy to know how many negative roots there could be.

For the example above the degree is 3.  This means that the graph will change directions up to three times and it MUST finish in the 4th and 2nd quadrants because the leading coefficient is negative.  -2 to be precise.

Since it changes direction up to 3 times, if it has up to 2 positive roots it MUST have 1 negative root.  I don't need descarte's rule to tell me that.  :)

Here is the graph

https://www.desmos.com/calculator/zjb0n3rvrt

And here is where I learned about Descartes rule

http://www.purplemath.com/modules/drofsign.htm

3.     6 - √6

4.    [x +4 ] [x - (6 + i)] [x - ( 6 - i) ]

       [x + 4 ] [ x^2 -x(6 + i) - x(6 -i) + (6 + i)(6 - i) ]

       [x + 4]  [ x^2  - 6x - ix - 6x + ix + 36 - i^2 ]

       [ x + 4] [ x^2 - 12x + 37]

      x^3  - 12x^2  + 37x

                 4x^2   -  48x  + 148

P(x)  = x^3 - 8x^2 - 11x + 148  = 0

5.   -2x^3 + 3x^2 - 5x - 2  = 0

Possible positive roots =  2 or 0

Possible negative roots test

f(-x)   =  2x^2 + 3x^2 + 5x - 2

So...the number of possible negative roots = 1

The last answer is correct

Thank you so much CPill & Melody!