algebra 2 question

Question

Could someone assist me to find the negative and positive zeros for this equation f(x) = 2x5-3x2+x-1

Answer 1

According to Descarte's Rule of Signs, the number of positive zeros of f(x) is equal to the number of sign changes in the coefficients of f(x) or is less than that number by an even integer.  Since there are 3 sign changes (2 to -3, -3 to 1, and 1 to -1), f(x) has either 3 positive zeros or only 1.

 

f(-x) = -2x⁵ -3x² - x - 1

 

Since f(-x) has no sign changes in its coefficients, Descarte's

Rule of Signs tells us that f(x) has no negative zeros.

 

There are 2 possibilities:

 

Case 1:  3 positive real zeros, 0 negative real zeros, and 2 imaginary zeros ( 3 + 0 + 2 = 5 = degree of f(x))

 

Case 2:  1 positive real zero, 0 negative real zeros, 4 imaginary zeros

(1 + 0 + 4 = 5 = degree of f(x))