# If you have a function with imaginary roots, how can you create a function that will help you find the roots based on the roots of the second functionExample would be this function...

If you have a function with imaginary roots, how can you create a function that will help you find the roots based on the roots of the second function

Example would be this function

(x^4)-14(x^3)+93(x^2)-320(x)+500

The imaginary roots are 3 +/- 4i and 4 +/- 2i

### 1 Answer | Add Yours

You need to remember that you may write the factored form of a polynomial when you know its roots such that:

`f(x) = (x - x_1)(x - x_2)(x - x_3)(x - x_4)`

Since the problem provides four roots, hence the polynomial will be of fourth order.

Substituting 3 + 4i for`x_1, 3 - 4i ` for `x_2, 4 + 2i` for `x_3` and `4 - 2i ` for `x_4` yields:

`f(x) = (x - 3 - 4i)(x - 3 + 4i)(x - 4 - 2i)(x - 4 + 2i)`

You may use the differences of squares instead of products `(x - 3 - 4i)(x - 3 + 4i) ` and `(x - 4 - 2i)(x - 4 + 2i) ` such that:

`f(x) = ((x-3)^2 - (4i)^2)((x-4)^2 - (2i)^2)`

You need to use complex number theory such that `i^2 = -1` :

`f(x) = ((x-3)^2 + 16)((x-4)^2 + 4)`

Opening the brackets yields:

`f(x) = (x-3)^2*(x-4)^2 + 4(x-3)^2 + 16(x-4)^2 + 64`

Expanding the binomials yields:

`f(x) = (x^2 - 6x + 9)*(x^2 - 8x + 16) + 4(x^2 - 6x + 9) + 16(x^2 - 8x + 16) + 64`

`f(x) = x^4 - 8x^3 + 16x^2 - 6x^3 + 48x^2 - 96x + 9x^2 - 72x^2 + 256 + 4x^2 - 24x + 36 + 16x^2 - 128x + 256 + 64`

`f(x) = x^4 - 14x^3 + 21x^2 - 248x + 612`

**Hence, evaluating the fourth order polynomial, under the given conditions, yields that `f(x) = x^4 - 14x^3 + 21x^2 - 248x + 612.` **