Find a polynomial $()$ of degree 5 with zeros $3i, 1+ i$ and $2$ that satisfies $(0) = −18$ . Do not need to multiply it out
A problem like this is simple, start with $p(x)=(x-3i)(x-(1+i))(x-2)$ . Now I’m assuming these are the only zeros we’re allowed to have, and we must have a degree of 5. So two of these must be squared. But there is no way to combine these to get $p(0)=-18$. I feel like there’s a mistake in the problem.
Here’s another approach $|-3i|=3, |-1-i|=\sqrt 2 $, and $|2|=2$. It’s impossible to combine $\sqrt 2, 2,$ and $3 $to get $-18$
You can have $$p(x)=\frac{i}{2}(x-3i)^2(x-(1+i))^2(x-2)$$
But this is a complex polynomial. It doesn't mention that these are the only zeros you are allowed to have, so we can use the complex conjugates to obtain $$p(x)=\frac{1}{2}(x-3i)(x+3i)(x-(1+i))(x-(1-i))(x-2).$$
This gives a degree $5$ polynomial with real coefficients that satisfies $p(0)=-18.$