Question

polynomial equation of least root degree having 1,square root 3i

Answer 1

Answer:

To find a polynomial with those roots first you must know that it a polynomial has a root of 2 then (x-2) is a factor of the polynomial. For example say your factors of your polynomial were 3, 2i and -2i. Then we know that (x-3)(x-2i)(x+2i) are factors of f(x). All you have to do now is foil out the factors so you get f(x) in standard form. I like to factor two at a time and then foil in the thrid. I'll start with (x-2i)(x+2i). Foil this (First, outside, inside last) and you get x^2 + 2ix - 2ix -4i^2. The great thing is that 2ix and -2xi cancel out and you are left with x^2 -4i^2. Now i^2 =-1, so if you substitute that in you get x^2 - 4(-1) = x^2 + 4. Now foil in the last factor. (X^2 + 4)(x-3) = x^3 -3x^2 +4x -12. So f(x) = x^3 - 3x^2 +4x -12. Apply thsee same steps to your problem and you should be able to get f(x). Hope this helps!

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Answer 2

Answer:

$1 ,\ \sqrt[]{3i} \\\\ax^{2} +bx+c = x^{2} +(\alpha +\beta )x+\alpha \beta \\= x^{2} + (1+\sqrt{3i})x +1(\sqrt{3i})\\= x^{2} + (1+\sqrt{3i})x +\sqrt{3i}$