Question

resolver x³-9x²+52x-102=0,sabiendo que 3+5i es un de sus raices

Answer 1

$x^3-9x^2+52x-102=0 \\ x^3-3x^2-6x^2+18x+34x-102=0 \\ (x^3-3x^2)-(6x^2-18x)+(34x-102)=0 \\ (x-3)x^2-(x-3)(6x)+34(x-3)=0 \\ (x-3)(x^2-6x+34)=0 \\ \\ x-3=0 \\ x=3 \\ \\ x^2-6x+34=0 \\ b^2-4ac=(-6)^2-4\times1\times34=-100 \\ \\ \boxed{x=3}$