Given that x=-1+4i is a zero of f (x) = x^3+x^2+15x-17 find all the zeroes of f

All the zeroes of f (x) are x = 1, x = - 1 + 4i and x = - 1 - 4i.

Step-by-step explanation:

Given that f (x) = x³ + x² + 15x - 17

Now, we have to find all the zeroes of the function.

Given that x = - 1 + 4i is a zero of the function.

So, x = - 1 - 4i must be another zero of the function.

Therefore, (x + 1 - 4i) (x + 1 + 4i) will be factor of the function.

Hence, (x + 1 - 4i) (x + 1 + 4i)

= x² + 2x + (1 - 4i) (1 + 4i)

= x² + 2x + [1² - (4i) ²]

= x² + 2x + 17

Assume that (x + a) is another factor of f (x).

Therefore, we can write f (x) = x³ + x² + 15x - 17 = (x + a) (x² + 2x + 17)

⇒ x³ + x² + 15x - 17 = x³ + (a + 2) x² + (2a + 17) x + 17a

Hence, comparing the coefficients we can write

a + 2 = 1

⇒ a = - 1

Therefore, f (x) = x³ + x² + 15x - 17 = (x - 1) (x² + 2x + 17)

So, all the zeroes of f (x) are x = 1, x = - 1 + 4i and x = - 1 - 4i (Answer)