Answer:
Option C: there are zeros between x = 1 and x = 2, x = 0 and x = -1
Explanation:
The function is given as;
f(x) = -8x⁴ - 3x³ - 11x² + 8x + 16
Since it's the real zero of the polynomial we are looking for, it means there is a possibility of getting an imaginary zero of the polynomial.
A zero is the value of x that will make the polynomial equal to zero.
Let's try;
x = -2;
f(-2) = -8(-2)⁴ - 3(-2)³ - 11(-2)² + 8(-2) + 16
f(-2) = -128 + 24 - 44 - 16 + 16
f(-2) = -148
Let's try x = -1;
f(-1) = -8(-1)⁴ - 3(-1)³ - 11(-1)² + 8(-1) + 16
f(-1) = -8 + 3 - 11 - 8 + 16
f(-1) = -8
Let's try x = 0;
f(0) = -8(0)⁴ - 3(0)³ - 11(0)² + 8(0) + 16
f(0) = 16
Lets try x = 1
f(1) = -8(1)⁴ - 3(1)³ - 11(1)² + 8(1) + 16
f(1) = -8 - 3 - 11 + 8 + 16
f(1) = 2
Let's try x = 2;
f(2) = -8(2)⁴ - 3(2)³ - 11(2)² + 8(2) + 16
f(2) = -128 - 24 - 44 + 16 + 16
f(2) = -164
From the values of f(x) gotten from x= -2 to x =2, we can see that the zeros will be closest to x = 0 and x = - 1 and also between x = 1 and x = 2 since that is the range that produced f(x) to be close to zero.