List the possible zeros of the polynomial:
2x^3+5x^2-31x-15

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.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

Lol english