Need to form a polynomial function f(x) with a degree of 3; zeros -9, and 3-i. I am super confused!!! I think the problem should look something like this
(X+9)(x-3-i)(x+3+i) can someone help me please :-)

Question

Need to form a polynomial function f(x) with a degree of 3; zeros -9, and 3-i. I am super confused!!! I think the problem should look something like this
(X+9)(x-3-i)(x+3+i) can someone help me please :-)

anonymous · Answer

yes, close

anonymous · Answer

it should be 
$$(x+9)(x-(3-i))(x-(3+i))$$ and your first job is to multiply out 
$$(x-(3-i))(x-(3+i))$$ which is not nearly as hard as it looks

anonymous · Answer

we can use one of three methods to find that quadratic, the one with zeros $3+i$ and $3+i$ one method is to work backwards 
$$x=3+i$$
$$x-3=i$$
$$(x-3)^2=-1$$
$$x^2-6x+9=-1$$
$$x^2-6x+10=0$$ so it is 
$$x^2-6x+10$$

anonymous · Answer

another it to recall that if the zero is $a+bi$ the quadratic is 
$$x^2-2ax+(a^2+b^2)$$

anonymous · Answer

in your case $a=3,b=1$ so you get 
$$x^2-2\times 3x+3^2+1^2=x^2-6x+10$$