Question

Find a polynomial f(x) of degree five with real number coefficients that has zeros of 1,5i, and 1-i.

Answer 1

complex roots come in pairs so what I did was multiply ouy (x-1)(x-5i)(x+5i)(x-(1-i))(x-(1+i)) .. but it takes a long time, so is there a faster way?

Answer 2

Exploit the fact that \[\Large (a-bi)(a+bi) = a^2+b^2\]

Answer 3

That's one useful shortcut

Answer 4

You can fabricate it, too:\[(x-1)(x^2+25)(x^2+2i)=0\]

Answer 5

wouldn't it be x^2 -2x +2? not x^2 +2i

Answer 6

yes, (x-(1-i))(x-(1+i)) = x^2-2x+2