# Find the polinomial with lowest degree having the following zeros : 2 and 2i

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To find the polynomial of the lowest degree with zeros 2 and 2i.

A polynomial which has a complex root x+yi, also the conjugate root x-yi.

Therefore , the polynomial which has the root aroot 2i or 0+2i also has the conjugate root 0-2i = -2i.

Therefore , the lowest degree of the polynomial which has the roots 2 , 2i and -2i is a cubic and is given by:

f(x) = (x-2)(x-2i)(x+2i)

f(x) = (x-2) {x^2 -(2i)^2}

f(x) = (x-2)(x^2+4)

f(x) = x(x^2+4) -2(x^2+4)

f(x) = x^3 +4x -2x^2-6)

f(x) = x^3 -2x^3 +4x -6 is the lowest degree polynomial with the roots 2 and 2i.

The zeroes of a polynomial are the roots of that polynomial. We also know that the complex solutions come in complex conjugate pairs.

So, if 2i is a solution of the equation, that means that -2i is also the solution of the equation.

We also know that a polynomial could be written as a product of linear factors, if we know it's solutions.

We'll note the solutions of the polynomial as:

x1 = 2

x2 = 2i

x3 = -2i

The polynomial is:

P(x) = a(x - x1)(x - x2)(x - x3)

We'll substitute x1,x2 and x3:

P(x) = (x - 2)(x - 2i)(x + 2i)

We'll write the product (x - 2i)(x + 2i) as a difference of squares:

(x - 2i)(x + 2i) = x^2 + 4

P(x) = (x - 2)(x^2 + 4)

We'll remove the brackets:

P(x) = x^3 + 4x - 2x^2 - 8

**The lowest degree polynomial, having as zeros the values 2, 2i, -2i, is: P(x) = x^3 - 2x^2 + 4x - 8.**