Question

1

Answer 1

f(x) = 7x^5 + 15x^4 – x^3 + 4x^2 – 6x – 11

There are 3 sign changes so the possible number of positive real roots are 3 and 1.

f(-x) = 7(-x)^5 + 15(-x)^4 - (-x)^3 + 4(-x)^2 - 6(-x) - 11
= 7(-x^5) + 15x^4 - (-x^3) + 4x^2 + 6x - 11
= -7x^5 + 15x^4 + x^3 + 4x^2 + 6x - 11

There are 2 sign changes so the possible number of negative real roots are 2 and 0.

Since there are total of 5 roots, both real and imaginary, and imaginaries come in pair,

Possible roots are

Positive..Negative..Imaginary
3............2............0...........…
3............0............2...........…
1............2............2...........…
1............0............4...........…