To find the zeros of the function f(x) = x^3 - 4x^2 + x + 6, we need to find the values of x for which f(x) = 0.
One way to do this is to use synthetic division or long division to divide the polynomial by its linear factors. However, this function does not appear to have any linear factors that can be easily factored out.
Another approach is to use the Rational Root Theorem to test possible rational zeros of the function. According to the Rational Root Theorem, any rational zero of the function must be of the form p/q, where p is a factor of the constant term (6) and q is a factor of the leading coefficient (1).
The possible factors of 6 are ±1, ±2, ±3, ±6, and the possible factors of 1 are ±1. Therefore, the possible rational zeros are ±1, ±2, ±3, and ±6.
To check if any of these possible zeros are actually zeros of the function, we can substitute each value into the function and see if the result is equal to zero.
When x = 1, f(1) = (1)^3 - 4(1)^2 + (1) + 6 = 1 - 4 + 1 + 6 = 4 ≠ 0
When x = -1, f(-1) = (-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0 ✓
When x = 2, f(2) = (2)^3 - 4(2)^2 + (2) + 6 = 8 - 16 + 2 + 6 = 0 ✓
Therefore, the zeros of the function f(x) = x^3 - 4x^2 + x + 6 are x = -1 and x = 2.
Find the zeros of the function
f(x) = x^3 - 4x^2 + x + 6
1 answer