We want to list the possible zeros of
f(x) = 2x³ + 5x² - 31x - 15
There are 3 possible zeros because the polynomial is of degree 3.
There is one change in sign, so there is one real positive zero according to Descartes' Rule of Signs
f(-x) = -2x³ + 5x² + 31x - 15
There are two changes in sign, so there are 2 real negative zeros.
According to the Rational Zeros Theorem, if f(a) = 0, then x=a is a zero of f(x).
A graph of f(x) reveals that there are indeed three real zeros (one positive, two negative) as expected, but no integral zeros of f(x) exist.
The zeros are approximately x = -5.25, -0.5, and 3.1.
Note (If you know calculus):
The zeros are in the vicinity of x = -5.25, -0.5, 3.1.
To estimate them accurately, we can use the Newton-Raphson formula. That is,
For each zero, set x₀ = -5.25, -0.5, 3.1 for the teratons.
Results:
1st zero:
n=0, x= -5.2277
n=1, x= -5.2025
n=2, x= -5.2023
n=3, x= -5.2023 (converged)
2nd zero:
n=0, x= -0.4565
n=1, x= -0.4564
n=2, x= -0.4564 (converged)
3rd zero:
n=0, x= 3.1601
n=1, x= 3.1587
n=2, x= 3.1587 (converged)
The zeros are -5.2023, -0.4564, and 3.1587.
