Pre Calculus Help

Question

What is the equation of a polynomial function f(x) with real coefficients that has zeros -3i and -1, that has multiplicity 2.

Answer 1

zeros: 3i, -3i, -1, -1

f(x) = (x^2 + 9)(x+1)^2

f(x) = (x^2 + 9)(x^2 +2x + 1)

f(x) = x^4 +2x^3 + 10x^2 +18x + 9

Answer 2

For the zero if -3i, the factor would be (x+3i). Since this is a complex number, they come in pairs, the other factor would be the complex conjugate, (x-3i) For the factor of -1, the factor would be (x+1) Since there is a multiplicity of 2, it would be (x+1)²

Bringing these all together, this is the result.(x+3i)(x-3i)(x+1)(x+1). Next, we have to multiply these all out. You can multiply in any order you want. I will do the first 2 and the last 2 together as a first step using FOIL to multiply.

(x+3i)(x-3i) =

x²+3ix-3ix-9i² The middle terms cancel. The i²=-1

x²-9(-1) Simplify

x²+9

(x+1)(x+1) =

x²+x+x+1 Combine like terms

x²+2x+1

Now, these need to multiply together (x²+9)(x²+2x+1) =

(x²)(x²)+(x²)(2x)+(x²)(1)+(9)(x²)+9(2x)+9(1) Multiply

x⁴+2x³+x²+9x²+18x+9 Combine like terms

x⁴+2x³+10x²+18x+9