Zeros of polynomial functions

Question

Factor f(x) into linear factors given that k is a zero of f(x):

 

1.)    f(x)=x^3+(13-3i)x^2+(40-39i)x-120i; k=3i

 

 

2.)   f(x)=x^4+3x^3-30x^2-124x-120; k=-2 (multiplicity 2)

 

3. Find a polynomial of degree 3 with real coefficients and zeros of -3, -1, and 4, for which f(-2)=30

Answer 1

1. f(x)=x^3+(13-3i)x^2+(40-39i)x-120i=(x-3 i)(x+5)(x+8)

2.f(x)=x^4+3x^3-30x^2-124x-120=(x-6)(x+2)^2(x+5)

3.

f(x)=a(x+3)(x+1)(x-4).

From f(-2)=30, follows that a=5.

[30=a(1)(-1)(-6) --< a=5],

So

f(x)=5(x^3-13x-12).