Help with Quartic Polynomials and Imaginary Numbers?

Question

Hi guys, I could use some help with math homework.

A QUARTIC polynomial P(x) with real coefficients has zeroes (2+i) and (3-2i). If P(0)=13, find a rule for P(x).

Additionally, why is only knowing the zeroes of P(x) not enough to find a rule for P(x)?

Thanks, I appreciate it.

Answer 1

first for every imagery solution it comes in pairs, so the solution (2 + i) would also have (2 - i) as another solution. So (3 - 2i), (3 + 2i), (2 + i), and (2 - i) will be your solutions.

As far as setting up the equation. it would be
[x - (2 + i)] [x - (2 - i)] [x - (3 - 2i)] [x - (3 + 2i)], since if your solution is 2 then the equation would be x-2.
great, multiply it -.-.
the first one is like
-ix + 2i + 1 - 2x + 4 - 2i + x^2 - 2x +ix = x^2 - 4x + 5
the second one = x^2 - 6x + 13

continue multiplying
(x^2 - 4x + 5)(x^2 - 6x + 13)

then you get
x^4 - 10x^3 + 42x^2 - 82x + 65
since you want P(0) = 13, so divide the equation by 5
(x^4) / 5 - 2x^3 + (42x^2) / 5 - (82x) / 5 + 13

and the only knowing zeros for a equation its like there are different equations with the same zeros such as
y = x^2
y = 2x^2
so, you need something to identify the equation from another

Answer 2

It's not enough because the leading coefficient of a polynomial doesn't have to be 1. Let c be the leading coefficient.

If a+bi is a zero, so is a-bi the complex conjugate.

P(x)=c*(x-(2+i))*(x-(2-i))
*(x-(3-2i))*(x-(3+2i))

Multiply this out and plug in the point (0, 13) for x and y to solve for c.

Answer 3

Glad you asked.

Zeros come in complex conjugate pairs. Therefore

If 2+i is a zero then 2-i is also a zero.

Same for the other zero.

So now you know 4 roots.

So, you know:
p(x) = A(x-r1)(x-r2)(x-r3)(x-r3)(x-r4)
where A is a real constante

You know all 4 of these roots, if you multiply it all out, the constant term will be
A*r1*r2*r3*r4 = p(x=0) = 13

That's enough to get A