Answer:
D. 3 + 2i and 3 - 2i
Explanation:
The polynomial function is:
f(x) = x³ - 3x² - 5x + 39
If -3 is a zero of the function, we can say that (x - (-3)) = (x + 3) is a factor of the polynomial, so, we can divide the polynomial by (x + 3) to find the other factor. Using long division, we get:
Therefore, the other zeros of the polynomial come from the factor:
(x² - 6x + 13)
So, we need to solve the following equation:
x² - 6x + 13 = 0
Using the quadratic equation, where a is 1, b is -6 and c is 13, we get:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(13)}}{2(1)} \\ x=\frac{6\pm\sqrt[]{-16}}{2} \end{gathered}[/tex]Then, simplifying, we get:
[tex]x=\frac{6\pm\sqrt[]{16}\sqrt[]{-1}}{2}=\frac{6\pm_{}4i}{2}=3\pm2i[/tex]Therefore, the solutions are:
x = 3 + 2i and x = 3 - 2i
So, the correct answer is D. 3 + 2i and 3 - 2i
