Use Descartes rule of signs to determine the different possi

Use Descartes rule of signs to determine the different possibilities for the numbers of positive, negative, and noareal complex zeros for the function f(x) = x^5 + 3x^4 - x^3 + 2x + 3

Solution

f(x) = x^5 +3x^4 -x^3 +2x +3

First, I look at the polynomial as it stands, not changing the sign on x, so this is the \"positive\" case:

Look ate sign changes from each term to next term:   

3x^4 , -x^3 : +ve to -ve

-x^3 + 2x : -ve to +ve

There are two +ve roots which can be real or solution could be two complex roots

So, we can have real roots as 2 , 0

Now f(-x) = -x^5 +3x^4 +x^3 -2x +3

Sign changes : -x^5 , 3x^4 --- -ve to +ve

x^3 -2x ---> +ve to -ve

-2x , 3 ----> -ve to +ve

So, possible three (03) -ve roots or , one (01) -ve and two(02) comlex roots

So, there are +ve roots: 2 or 0

-ve roots : 3 or 1