a root of function f(x) is -1-2i. Which could be an equation for this function?

Question

A. f(x)= x^2-2x-3

B. f(x)=x^2-2x+5

C. f(x)=x^2+2x-3

D. f(x)=x^2+2x+5

Answer 1

I will tell you HOW to do this.  f(x) = (x-root)(x-other root); one root is known (given)--a complex number.  So you should know that its conjugate is also a root. 

 

So f(x) = [x-(-1-2i)]•[x-(-1+2i)] 

So just simplify the above & do the multiplication, recalling that i² = -1.

 

Note: the conjugate of a + ib is a - ib & vice versa.

Answer 2

Assuming that f(x) has real coefficients, complex roots occur in conjugate pairs.  Also, if c is a root of f(x), then x-c is a factor.  

 

So, a function with -1-2i as a root, is

 

      f(x) = [x-(-1-2i)][x-(-1+2i)] = [(x+1) + 2i][(x+1) - 2i]

 

                                             = (x+1)² - (2i)²

 

                                             = x²+2x+1+4 = x²+2x+5