Quadratic equation choose roots are 1+2i and 1−2i will be ‌x2−x(2)+(1+4)=0 ‌x2−2x+5=0 So, (x4−3x3+8x2−7x+5) ‌=(x2−2x+5)×(x2+ax+b) ⇒x2+ax+b=x2−x+1 ⇒f(x)=(x2−2x+5)(x2−x+1) Roots of f(x)=0 will be α,β,γ and δ Roots of x2−2x+5=0 be γ and δ. γ=1+2i,‌‌δ=1−2i Roots of x2−x+1=0 be α,β⇒α=−ω, β=−ω2 Sum of squares of other roots ‌=(1−2i)+(−ω)2+(−ω2)2 ‌=1−4−4i+ω2+ω ‌=−4−4i
