Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point

Question

Degree: 4. Zeros: -3, 1, i. Solution point: f(0) = -12

Answer 1

I like to remember these two closely related forms for a polynomial (in this case, degree 4):

f(x) = a(x-r1)(x-r2)(x-r3)(x-r4) = a(x³+bx²+cx+d)

The left side is:

f(x) = a(x+3)(x-1)(x-i)(x-r4)

From these we can come up with some information about the polynomial.

We know that the coefficients must be real, so the last zero has to be the complex conjugate of i. Therefore, r4 = -i, so

f(x) = a(x+3)(x-1)(x-i)(x+i)

We know f(0) = 12, so plugging that in we get

f(0) = a*3*-1*-i*i = ad

-3a = ad

d = -3

We also have

ad = -12

so

a = 4

Now that we know a and all the roots, we can write

f(x) = 4(x+3)(x-1)(x-i)(x+i)

Which we can expand (if necessary):

f(x) = 4(x²+2x-3)(x²+1)

f(x) = 4(x⁴+2x³-2x²+2x-3)

f(x) = 4x⁴+8x³-8x²-12