Polynomial Functions

Question

Write a polynomial functionſ of least degree that has rational coefficients, a leading coefficient of 1 and the given zeros

3, -2, 2i

Answer 1

Hi Layla B.,

First thing to realize: anytime you have an imaginary or complex root, you have AT LEAST one other root, either a multiple root of the one you are given already, or the complex conjugate of the root you are given already.

So, you are dealing with ± 2i as roots! Even though only +2i was given! (Explanation: the polynomial can't have imaginary terms, so you MUST have the "other" root to remove the i term in the expanded form). Of course, you could have (x^2+4)^n where n>1, but that would not give the polynomial with the least possible degree.

So, that gives you a factor of (x^2+4) since x^2 must be -4 for this to be a root (i.e. have value overall of 0).

Your other 2 factors in the polynomial are (x-3) and (x+2) since these will respectively yield zeros on the total polynomial (remember, if any of the factors of the polynomial are value 0, the product using that factor must be 0).

So mutiply (x^2+4), (x-3), and (x+2). What do you get?

Was any of this new material to you? If so, you need to keep up in your textbook reading -- and if not, what crucial connection didn't you make to what you already knew? You might wish to make a few problems specifically on that, just on your own, and practice solving them. Doing math is the best way to really learn it cold!

-- Cheers, -- Mr. d.

Answer 2

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

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

x^4-x^3 -2x^2 -4x -24

if 2i is a zero, so is -2i

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