zeros of polynomials?

Question

how do i find a zero with this equation if it has real coefficients?

7 + 2i

i am really bad with imaginary numbers. -_-
7 + 2i is a zero of q(x)

Answer 1

A second-degree polynomial equation can have a zero with an imaginary number, even though all coefficients are real. Remember the quadratic:

zeros = (-b +/- sqrt(b^2 - 4ac))/2a

if b^2-4ac is negative, then the zeros of the equation will have the square root of a negative number in them, meaning they are imaginary numbers.

We can even construct a second-degree polynomial with real coefficients which has a zero at 7+2i.

Let's say a=1, for simplicity.

Since the non-imaginary portion of the answer is -b/2a, and we want that to be equal to 7, and we know a=1, we know that b...

-b/2a = 7
-b/(2*1) = 7
-b = 14
b = -14

... must equal -14.

And we want the imaginary portion to be 2i, so we know that:

sqrt(b^2 - 4ac)/2a = 2i

And since we know a and b already, we can solve that for c:

sqrt(b^2 - 4ac)/2a = 2i
(b^2 - 4ac)/4a^2 = -4
((-14)^2 - 4*1*c)/(4*1*1) = -4
196 - 4c = -16
-4c = -212
c = 53

So... the equation:

q(x) = x^2 -14x + 53

... has one zero at 7+2i. (And the other is 7-2i.)

And now let's test it:

q(x) = x^2 -14x + 53
q(7+2i) = (7+2i)^2 - 14(7+2i) + 53
q(7+2i) = 49 + 28i - 4 - 98 - 28i + 53
q(7+2i) = 49 + 53 - 4 - 98 + 28i - 28i
q(7+2i) = 102 - 102 + 28i - 28i
q(7+2i) = 0