Why is this required for finding zeros in Polynomials

Question

I believe this question is very simple, I just would like to know why in this specific example (2x³+2x²-19x+20=0) once I am done plugging in the trinomial into the quadratic formula and get 6 +- 2i / 4 as my answer, why do I then split it into

3+i/2 and 3-i/2 getting 2 separate zeros instead of just leaving it as 6+- 2i / 4. Also a separate question, when a zero/root has an i in it does that mean it's an imaginary zero?

Answer 1

As an addition to Michael's answer;

 

A)  Yes, having the i  makes it a complex root.

 

B)  When you go from (6 ± 2i)/4 to (3 ±i)/2, you are just simplifying a fraction.

 

Let's say you had 5 ± 7i.  That is really 2 answers.  They are the answers 5 + 7i and 5 - 7i.

Answer 2

Since 19 is a prime coefficient, we would have to use synthetic division to find a factor of the polynomial.

 

-4 |  2       2        -19          20

 

               -8         24         -20

___________________________

       2      -6           5            0

 

 

(x + 4)(2x² - 6x + 5)

 

If you got his factorization on your own, then you are fantastic.

 

 

In a quadratic, you have two solutions.  So you have to break up the complex solution into two parts.  The negative part provides one solution and the positive part provides the other.  Also, you should know that complex solutions come in conjugates.