Find a polynomial of degree 2 with integer coefficients and zeros 5-i and 5+i.

Question

anonymous · Answer

you have lots of choices of methods here

anonymous · Answer

one way is to multiply out
$$(x-(5+i))(x-(5-i))$$ which is not as hard as it seems. but is kind of a pain

anonymous · Answer

next choice is to work backwards 
you know 
$$x=5+i$$
$$x-5=i$$
$$(x-5)^2=i^2=-1$$
$$x^2-10x+25=-1$$
$$x^2-10x+26=0$$ is the equation you started with

anonymous · Answer

the other is to memorize that if $a+bi$ is the zero of a polynomial of degree two with leading coefficient 1, that quadratic equation is 
$$x^2-2ax+(a^2+b^2)=0$$

anonymous · Answer

pick whichever method you like

anonymous · Answer

Thank you for the very helpful and speedy response!

anonymous · Answer

yw