Find a polynomial f(x) of degree 3 with real coefficients and the following zeros.
4, 2+i

Question

Find a polynomial  f(x) of degree 3 with real coefficients and the following zeros. 
4, 2+i

anonymous · Answer

Well we know that the function has 2 zeroes and has is a third degree polynomial. So let's try making one by starting with the zeroes:$$y=(x-4)(x-(2+i))$$But in order for this to have real coefficients, we must in some way have to add another part that eliminates the imaginary parts from the expansion. We know for a fact that when complex numbers are multiplied by their conjugates, they result in real numbers. Can we put that knowledge to use here?

@jmprz_793

mertsj · Answer

We know that complex answers come in pairs.  If 2+i is an answer ten 2-i is also an answer.

anonymous · Answer

and so your job is to multiply 
$$(x-4)(x-(2+i))(x-(2-i))$$ which is not as hard as it seems

anonymous · Answer

one way is just to multiply it out
the other way is to work backwards to find 
$$(x-(2+i))(x-(2-i))$$ 
you know the zero is $2+i$ so you can put 
$$x=2+i$$ subtract 2 
$$x-2=i$$ square 
$$(x-2)^2=-1$$ expand 
$$x^2-4x+4=-1$$ so 
$$x^2-4x+5=0$$ is the original polynomial

anonymous · Answer

the last way is to memorize that if $a+bi$ is a zero of a quadratic, the quadratic is 
$$x^2-2ax+(a^2+b^2)$$

anonymous · Answer

i am trying to enter the answer y=x^(2)-(6+i)x+(8+4i), but it does not let me enter the answer in parentheses @satellite73

anonymous · Answer

would it be y=x^(2)-6x-ix++4i

mertsj · Answer

The question is to find a polynomial f(x)
Satellite told you that you job is to multiply : (x−4)(x−(2+i))(x−(2−i))
And then he proceeded to multiply the last two factors for you and told you that the product of (x-(2+i))(x-(2-i)) is x^2-4x+5
So finish the problem by multiplying (x-4)(x^2-4x+5)