Given -
f(x) = x⁵ - 4x³ + 7x² + 3x - 5
To Find -
How many positive real zeroes does f(x) have =?
Step-by-Step Explanation -
We have
f(x) = x⁵ - 4x³ + 7x² + 3x - 5
where the signs of coefficients are:
+ - + + -
We can see there are three changes in the sign in f(x).
So, from
Descarte's rule,
there are either 3 0r 1 positive real roots
Now,
f(-x) = -x⁵ + 4x³ + 7x² - 3x - 5
Signs: - + + - -
So, here f(-x) has two sign changes.
From this we can conclude there is at least 1 real root and either 4, 2 or 0 imaginary roots.
Final Answer -
Positive real zeroes f(x) have =
Minimum = 1
Maximum = 3