# Write a polynomial function of lowest order possible in standard form that has the following zeros `x=2, x=i`

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### 1 Answer

So, we'll start by saying that we want the polynomial of smallest degree with real coefficients. If we could have complex coefficients, the problem would be trivial!

So, to start, we are given one real root and one complex root. Based on the complex conjugate theorem for polynomials, if we have one complex root and all of the coefficients of the polynomial are real, then we know that the complex conjugate of the complex root is another root of the polynomial. In other words if the following is a root:

`a + bi`

where `a,b in RR`, then the following must also be a root:

`a - bi`

In our problem, this theorem applies to the following root:

`x = i`

The existence of this root implies the existence of the following root, as well:

`x = -i`

Therefore, we have a set of 3 roots for our polynomial:

`x in {2, +-i}`

To get the polynomial, we just multiply the factors that we can derive from the roots!

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

Now, let's simplify the imaginary factors first. Remember, given an expression of the form `(a+b)(a-b)` we can simplify that by the FOIL method to make `a^2 - b^2`. Let's do the same thing to our polynomial:

`f(x) = (x-2)(x^2 - i^2)`

Recall, `i^2 = -1`, so we can simplify the equation:

`f(x) = (x-2)(x^2 + 1)`

Now, we can simply use the FOIL method to expand the rest of the polynomial:

`f(x) = x^3 -2x^2 + x -2`

There is our complete polynomial. Now, you can actually multiply it by any constant, too, if you're feeling creative! This is because constant coefficients have no bearing on the degree of the polynomial. So, our f(x) becomes:

`f(x) = c(x^3-2x^2+x-2) = cx^3 - 2cx^2 +cx -2c`

where `c` is some random constant.

I hope that helps!

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