Question

According to Descartes' rule of signs, how many possible negative real roots could the following polynomial function have?

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

Answer 1

Substitute (-x) for x and find the new function.See how many times the sign of the coefficients changes between +ve and -ve.

If I put x = -x in you function it can be rewritten as:

f(x) = 3.(-x)^4 - 5.(-x)^3 + 5 (-x)^2 + 5(-x) -2
= 3 x^4 + 5x^3 + 5x^2 - 5x -2
Now, leading coefficient is +3, next +5, next + 5, next -5 and next -2.
Now, the sign changes from +ve to -ve between +5 and -5 which is once. So there is one negative real root. If there had been 2 times sign change then there would have been 2 negative real roots (or) zero negative real roots and 2 complex roots. Descarte's rule just gives you the possibilities. not the exact no.