Question

Determine where the real zeros of f(x)= 2x^4 + x^2 - 3x + 3 are located

a.
between -1 & 0
c.
between 1 & 2
b.
between 0 & 1
d.
no real zeros

Answer 1

Explanation:

I used calculus to find the answer. Newton's Method is used to determine where real zeros exist in a polynomial. Couldn't factor it to find them, so I went with Newton's Method. My iterations began at a zero of .475 then jumped to 2.070168876, then to 1.545902463, then to 1.135355003, then to .7521218306, then to -.2695537736. then to .783503098. It was the last iteration there that told me there were no real zeros. When the answers jump from positive to negative and back to positive, the Method fails and there are no real zeros. So d.

Answer 2

The given polynomial is
f(x) = 2x⁴ + x² - 3x + 3
There are 2 changes of sign

From Descartes' Rule of signs, there are
(a) 2 real positive zeros, or
(b) a conjugate pair of 2 complex zeros.

f(-x) = 2x⁴ + x² + 3x + 3
There is no change in sign
There are no negative real zeros.

Let us test for real zeros in the given ranges:
x f(x)
----- --------
-1 9
-0.5 4.875
0 3
0.5 1.875
1 3
1.5 10.875
2 33
There are no changes in the sign of f(x).
Therefore no real zeros exist in (-1,0), (1,2), (0,1).
The graph of the function confirms the conclusion.

Answer: d. No real zeros.