Question

please help
I need to write a polynomial function with rational coefficients so that P(x)=0 has the given roots
3i+9

Answer 1

let's think about this
Pretend we have P(x)=Ax^2+Bx+C
where A,B,and C to be real
---
So we know to solve a quadratic equation that we can use
\[x=\frac{-B \pm \sqrt{B^2-4AC}}{2A} \\ x=\frac{-B}{2A} \pm \frac{\sqrt{B^2-4AC}}{2A}\\ \text{ we are given } x=9+3i \\ \]
so what do you think the other zero is

Answer 2

looking at the quadratic formula

Answer 3

another hint:
we have two equations the quadratic formula gives...
one with a plus in between and one with a minus in between

Answer 4

If your given root is 3i+9

Since it is imaginary, the conjugate 3i-9 is also a root.

Answer 5

So now, you just need to multiply out (3i+9)(3i-9), to get back your equation.

Answer 6

well if you take the conjugate of a+bi that would be right

Answer 7

like what I'm saying is the other zero would be the conjugate of 9+3i

Answer 8

\[x=\frac{-B \pm \sqrt{B^2-4AC}}{2A} \\ x=\frac{-B}{2A} \pm \frac{\sqrt{B^2-4AC}}{2A}\\ \text{ we are given } x=9 \pm 3i \\\]

Answer 9

I knew to order 3i+9 as 9+3i because in the quadratic formula the only thing that can give us an imaginary number is the sqrt part