Question

form a polynomial whose zeros and degree are given. write in standard form. zeros: -3, multiplicity 2; 2i; degree 4

Answer 1

zeros: -3, -3, 2i, -2i

If -3 is a root of multiplicity 2, then -3 appears twice as a root.
If 2i is a root, then its conjugate -2i is also a root.

The polynomial is the product of (x - (-3))* (x - (-3))* (x - 2i) * (x - (-2i)).

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

Crank out the product of these four factors to find P(x).

@svance2

Answer 2

That's where I get stuck with the i and ix @directrix

Answer 3

@svance2

Do the product in parts. Do this first (x+3) (x + 3) = and post your results.

Do what you can on this product and post your results.

(x - 2i) (x+2i)

I'll check back later to help you put this together.

Answer 4

(x^2-4xi+4)(x^2+6x+9)
That's after doing it in parts.
X^4-6x^3+9x^2-4ix^3-24ix^2-36xi+4x^2+24x+36 (not sure if that is right but here is where I am stuck)
@directrix

Answer 5

(x - 2i) (x+2i) =(x^2-4xi+4) --> Wrong [-2ix + 2ix = 0]
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(x - 2i) (x+2i) = (x^2 + 4)

(x+3) (x + 3) = (x^2+6x+9) --> Correct

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Polynomial:

P(x) = (x^2 + 4)* (x^2+6x+9)

P(x) = x^4 + 6x^3 + 13x^2 + 24x + 36 --> Check my work.

@svance2