Answer:
Step-by-step explanation:
REcall that f(x) is a polynomial whose one of its roots is -3+i. The fundamental algebra theorem states that any polynomial of degree n has n complex roots. In the real case, it can be also interpreted as any polynomial can be factored in factors of degree at most 2.
Consider that given a polynomial of degree 2 of the form
the solutions are given by
![x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
In this case, the fact that x is real or complex depends on the number
which is called the discriminant. When this number is negative, we have that x is a complex root. Let say that
and that
, so the roots are given by
this means that, whenever we have a complex root, the other root is the complex conjugate. Recall that the complex conjugate of a complex number of the form a+bi is obtained by changing the sign of the imaginary part, that is a-bi.
So, in our case since -3+i is a root, then -3-i necessarily is another root.