Three zeroes of a fifth degree polynomial function are 1/3, 4 - 6i, and -2 + 11i. Determine the remaining zeroes of the function.

4 + 6i, -2 - 11i
-1/3, 4 + 6i, 2 + 11i
-4 + 6i, 2 - 11i
3, 4 + 6i, -2 - 11i

Can I have some guidance for this please? I am unsure how to solve this. Thanks

A. 4+6i,-2-11i :)

I believe I am incorrect since it's a fifth degree function and I do not think -1/3 is pliable. I think it's the first option.

complex roots (containing "i") occur in conjugate pairs

... if a+bi is a root, then a-bi is also

Of course, I'm here to help! To solve this problem, we need to use the fact that complex zeros always come in pairs: if a complex number a + bi is a zero of a polynomial with real coefficients, then its conjugate a - bi is also a zero.

Given that the zeros are 1/3, 4 - 6i, and -2 + 11i, we can determine the remaining zeros using the complex conjugate rule.

The first zero, 1/3, is a real number, so we don't have to apply the rule to it.

Now, let's apply the rule to the second zero, 4 - 6i. Its complex conjugate is 4 + 6i. Therefore, our new zeros are 4 - 6i and 4 + 6i.

Finally, let's apply the rule to the third zero, -2 + 11i. Its complex conjugate is -2 - 11i. Therefore, our new zeros are -2 + 11i and -2 - 11i.

So, the remaining zeros of the polynomial function are 4 - 6i, 4 + 6i, -2 + 11i, and -2 - 11i.

I hope this helps! Let me know if you have any more questions.

To determine the remaining zeros of the fifth-degree polynomial function, we need to use the conjugate pairs theorem for complex zeros. According to this theorem, if a polynomial has complex zeros, they must occur in conjugate pairs.

Given that the zeros of the polynomial are:
1/3, 4 - 6i, and -2 + 11i,

we can deduce the following conjugate pairs:
1) -2 + 11i and its conjugate -2 - 11i (since they have the same real part and opposite imaginary parts)
2) 4 - 6i and its conjugate 4 + 6i (same reasoning as above)

So, the remaining zeros of the function are 4 + 6i and -2 - 11i.

Therefore, the correct answer is: 4 + 6i, -2 - 11i.

Okay, so would it be the second option?