Using the given zero, find one other zero of f(x). Explain the process you used to find your solution. (3 points)
1  6i is a zero of f(x) = x4  2x3 + 38x2  2x + 37.
3 Answers

f(x) = x⁴  2x³ + 38x²  2x + 37
If (1  6i) is a zero, it means that (1 + 6i) is a zero too.
So you can factorize: [x  (1  6i)].[x  (1 + 6i)]
= [x  (1  6i)].[x  (1 + 6i)]
= (x  1 + 6i).(x  1  6i)
= x²  x  6xi  x + 1 + 6i + 6ix  6i  36i²
= x²  2x + 1  36i² → where: i² =  1
= x²  2x + 37 → abd to obtain x⁴, it's necessary the first term of the other factor to be x²
= (x²  2x + 37).(x² + ax + b) → you expand
= x⁴ + ax³ + bx²  2x³  2ax²  2bx + 37x² + 37ax + 37b → you group
= x⁴ + x³.(a  2) + x².(b  2a + 37)  x.(2b  37a) + 37b → you compare with: x⁴  2x³ + 38x²  2x + 37
37b = 37 → b = 1
(2b  37a) = 2 → 37a = 2b  2 → 37a = 0 → a = 0
(b  2a + 37) = 38 → (1  2a + 37) = 38 → 38  2a = 38 →  2a = 0 → a = 0 ← of course
(a  2) =  2 → a = 0 ← of course
Resart:
= (x²  2x + 37).(x² + ax + b) → we've just seen that: a = 0
= (x²  2x + 37).(x² + b) → recall:: b = 1
= (x²  2x + 37).(x² + 1)
We've seen that (x²  2x + 37) = 0 corespnds to: [x  (1  6i)].[x  (1 + 6i)] = 0
The other case is:
(x² + 1) = 0
x² + 1 = 0
x² =  1
x² = i²
x = ± i
The other zero is (+ i) and ( i)

You have a polynomial with real coefficients. Therefore any complex roots occur in conjugate pairs, (a+ib) and (aib). You are given one zero that is one member of a conjugate pair of roots.
It is trivial that another zero, such as you are asked to find, is the other member of that conjugate pair.
Ans: Given that x =16i is a zero of f(x), another zero is x = 1+6i

If a+bi then abi is a root.
Lets call these r1 & r2.
Thus (xr1) & (xr2) are factors of f(x). Therefore, just factor these out of your f(x).
Or equivalently, expand (xr1)*(xr2) and factor out that resulting expression from f(x) by synthetic division, say.
You will be left with a quadratic, which you know how to find its roots (your last two roots!).
Show your steps here if need be and we can verify ur work.
Done!