Given that (z + 2 - 3i) is a factor of f(z) = z^3 + 9z^2 + 33z + 65,

Express f(z) in the form f(z) = (z + a)(z^2 + bz + c),

Where a, b and c are integers.

QUESTION:
Given that (z + 2 - 3i) is a factor of f(z) = z^3 + 9z^2 + 33z + 65,
Express f(z) in the form f(z) = (z + a)(z^2 + bz + c),
Where a, b and c are integers.

Since a and c are integers, then an expansion of f(z) = (z + a)(z^2 + bz + c) would show that a*c = 65.

The only factors of 65 are 5 and 13. So a = 5 and c = 13, or vice versa.

Expanding f(z) = (z + a)(z^2 + bz + c) we get,

f(z) = z^3 + (b+a)z^2 + (c+ab)z + ac

comparing coefficients with f(z) = z^3 + 9z^2 + 33z + 65, this gives us,

b + a = 9

c + ab = 33

ac = 65

Test: if a = 5

b + a = 9 ---> b = 4

ac = 65 ---> c = 13

c + ab = 33 ---> a = 5 (this verifies that a equal 5)

a = 5, b = 4, c = 13

Then f(z) = z^3 + 9z^2 + 33z + 65 = (z + 5)(z^2 + 4z + 13)

BTW: if you use (z + 2 - 3i) as a factor. you end up with f(z) = (z + 2 - 3i)(z^2 + (7+3i)z + (10+15i))

by Level 11 User (80.5k points)
thank you very much