information is given about a polynomial f(x) whose coefficients are real numbers. find the remaining zeros of f degree 5; zeros 4,i,-2i must show work

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You should remember that if you know a complex solution to a polynomial equation, then you know the complex conjugate solution, hence, the complex solutions come in pairs.

Since the problem provides the complex solutions `x = i ` and `x = -2i,`  hence, the polynomial has the complex conjugates `x = -i ` and `x = 2i.`

You need to write the factored form of the polynomial such that:

`f(x) = a(x - x_1)(x - x_2)(x - x_3)(x - x_4)(x - x_5)`

You need to substitute the given values of the roots for `x_1,x_2,x_3,x_4,x_5 ` such that:

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

You need to convert the products into differences of squares such that:

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

You need to remember that `i^2 = -1`  such that:

`f(x) = a(x^2+ 1)(x^2+ 4)(x - 4)`

Hence, evaluating the  polynomial, under the given conditions yields `f(x) = a(x^2+ 1)(x^2+ 4)(x - 4).`