Respuesta :
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 [tex]ax^2+bx+c[/tex] the solutions are given by
[tex] x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}[/tex]
In this case, the fact that x is real or complex depends on the number [tex]b^2-4ac[/tex] which is called the discriminant. When this number is negative, we have that x is a complex root. Let say that [tex]b^4-4ac<0[/tex] and that [tex]\sqrt[]{b^4-4ac}=di[/tex], so the roots are given by
[tex] x_1 = \frac{-b + di}{2a}, x_2 = x_1 = \frac{-b - di}{2a}[/tex]
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.
