Here's the question you clicked on:
caroo.salinaas19
Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree: 4; zeros: 2i and -3i
In general, if you have a polynomial with real coefficients, and some complex root \(\alpha i\) where \(\alpha\) is a real number, the you also have the complex root \(-\alpha i\). Using this, can you tell me what all the roots of your polynomial will be?
2i, -2i, -3i, and 3i?
Bingo. So that means your factored polynomial will be \[(x-2i)(x+2i)(x-3i)(x+3i).\]Now you just have to expand it out.
i got this answer: x^3 -3ix^2-2ix^2+6ix^2-4xi^2+12i^2.
Hmmm. I've definitely got something different. I'll walk you through the first few steps I did. \[(x-2i)(x+2i)=x^2-2ix+2ix-(2i)^2=x^2-4(-1)=x^2+4\]Note that \(i^2=-1\) by definition. Similarly, \[(x-3i)(x+3i)=x^2-3ix+3ix-(3i)^2=x^2-9(-1)=x^2+9\]Using this, can you find \[(x^2+4)(x^2+9)\]on your own?