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)

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64