1
$\begingroup$

question

What is a polynomial that has the roots: 3 and 5-i and also crosses the origin with integer coefficients?

My thoughts

on my first instinct, i wrote this:

$(x-(5-i))(x+(5-i))(x-3)(x)$

But then i've realized that when you factor out the complex parts, you dont get integer coefficients so I was really confused if this is even possible

2 Answers 2

2

If you want real coefficients, which integers implies, you need conjugate imaginary roots. To cross the origin you need a factor $x$. The simplest choice is then $$x(x-3)(x-5+i)(x-5-i)=x^4 - 13 x^3 + 56 x^2 - 78 x$$ You can multiply this by any polynomial with integer coefficients that you like. You are close but did not get the conjugate right.

1

Note that the conjugate of $5-i$ is $5+i$.

So,$$(x-5+i)(x-5-i)(x-3)x=x(x-3)(x^2-10x+26)$$ Is the simplest polynomial that meets the conditions.