Ask a New Question
Menu

Could someone please check the first one and help me with the second one. I'm stuck-Thank you

1.What are theroots of the following polynomial equation?
(x+2i)(x-5)(x-2i)(x+8) = 0
I think the roots are -2i,5,2i,-8-you take out the roots as opposites of what are in the equaion correct?
2. Write a polynomial of the smallest degree with roots 3,3i and -3i
Final answer can't have parentheses
(x-3)(x-3i)(x+3i)=0 and then I'm stuck I don't know what to do when you multiply the x times the -3i

4 answers

First find (x-3i)(x+3i)=x^2+9 and then
(x-3)(x^2+9)=x^3-3x^2+9x-27
Do the last two first: (x-3i)(x+3i) You can do FOIL (the anwer to x*(-3i) is -3xi), however, notice x+9 is the answer. (x^2+9) is the difference of two squares, x and 3i, so (x^2+9)=(x+3i)*(x-3i)

so you now have (x-3)(x^2+9)
= x^3+9x -3x^3-27 Using the FOIL method.
Whenn I get to (x-3)(x^2+9) and FOIL wouldn't it be x^3 +9x-3x^2 -27? I get (-3x^2) not (3x^3) when I multiply -3 times x^2 or am I missing something
Thank you
correc, typo. -3x^2