I need to find a polynomial P(x) that satisfies these conditions:

Question

a) minimal degree

b) coefficients are real

c) P(2)=0

d) P(i)=0

e) P(0)=1

you may leave the polynomial in factored form but combine the complex conjugate pairs of roots into quadratic factors with real coefficients.

Answer 1

We can start this problem by first looking at conditions c, d, and e.

To satisfy P(2) = 0, you need a root at P(2). This gives us P(x) = (x-2).

Next, we want a root at P(i). This is complex, so there will be a second root at P(-i). So now we have P(x) = (x+i)(x-i) = (x^2 + i).

Now we want P(0) = 1. For this, we can look at what we already have: P(x) = (x^2 + 1)(x-2). If we plug 0 in, we get P(0) = -2. To resolve this issue, we can simply multiply a constant in front. This gives P(x) = -(1/2)(x^2+1)(x-2).

This satisfies all of the conditions. We know we have minimal degree because the prompt asked for three roots and our polynomial is to the third degree. We can also see that all of our coefficients are real.