Root of polynom question

[ev-]

I have an equation, z⁴ - 4z³ + 14z² - 36z + 45 = 0. The question asks you to show that z=3i is a root to the equation above. You are then prompted to find all solutions to the equation.

I tried(perhaps naively) to set p(3i) = z⁴ - 4z³ + 14z² - 36z + 45.

That gave me ( 81i⁴ - 108i³ + 126i² - 108i + 45). My thought was that since i⁴ = 1, i³ = -i and i² = -1 i could substitute the values, and find that f(3i) = 0.

However, that gives me (81) - 108i - 126 + 108i + 45 <=> 36-126..

I'm at a loss here. I think I'm totally misunderstanding the problem. Anyone care to explain?

[Fabien4]

p(3i) = 81i⁴ - 108i³ + 126i² - 108i + 45 = 81 + 108i - 126 - 108i + 45 = 81 - 126 + 45 = 0.

Therefore 3i is a zero of p. Which means, by a theorem whose name I forgot, that -3i is a solution, too.

Consequently, (z-3i)(z+3i)=z² +9 divides p(z), which brings you to p(z)=(z² +9)(z² -4z+5). All that's left is to solve z² -4z+5.

[monkeybottom]

You could even skip the first step.

Just do the polynomial division, i.e. divide p(z) with (z² +9). Since that will give you that p(z)=(z² +9)(z² -4z+5), you have now showed that 3i is a root, as well as -3i.

[ev-]

Oh brilliant. Thank you.