Question

PLEASE Help, Completely lost!
Find a polynomial f(x) of degree 3 with real coefficients and the following zeros.

−3 , +2i

Answer 1

Let's start by looking at a simpler example...
What if it said the polynomial has zeros of -3, 5 and -9?
What would we do to get the polynomial?

Answer 2

Since x is equal to -3 and +2i, Then itll start off as (x+3)(x-2i). Since there is an imaginary number you must need the opposite of it. So itll be (x+3)(x-2i)(x+2i)=f(x) Solve from there.

Answer 3

Must start off with the imaginary numbers. (x-2i)(x+2i)(x+3)=(x^2-2i+2i-4i^2)(x+3)=(x^2+4)(x+3)=x^3+3x^2+3x^2+12=x^3+6x^2+12

Answer 4

@Miracrown You just plug them in and factor them right? Resulting in (x+3)(x-5)(x+9)? Which = \[x ^{3}+7x ^{2}-33x-135\] Can't remember which step to take next...please keep in mind it's been a long day and I'm not at my sharpest right now haha

@marie.serina09 I just tried to work it out on scratch paper and with some help from some videos, does that end up = \[x ^{3}+3x ^{2}+4x+12\]

Is that my answer?

Answer 5

Im pretty sure. I mightve missed a step cx

Answer 6

Hold on..

Answer 7

No worries!

Answer 8

Yup youre answer is correct

Answer 9

Woohoo! Awesome, thank you so much! Have a good night! :)

Answer 10

Youre welcome and you too! :)