Given that #1-i# is a solution of the given equation. We can say that the conjugate pair #1+i# must be another solution .
Hence the quadratic factor of the given Polynomial #(P(x))# will be
#=x^2-("sumof the roots")x+("product of the roots")#
#=x^2-(1-i+1+i)x+(1-i)(1+i)#
#=x^2-2x+(1-i^2)#
#=x^2-2x+2#
Hence we can write
#P(x)=0#
#=>9x^6+x^4+26x^3= 24x^5+22x^2+32x+8=0#
#=>9x^6+x^4+26x^3-24x^5-22x^2-32x-8=0#
#=>9x^6-24x^5+x^4+26x^3-22x^2-32x-8=0#
#=>9x^6-18x^5+18x^4-6x^5+12x^4-12x^3-29x^4+58x^3-58x^2-20x^3 +40x^2-40x-4x^2+8x-8=0#
#=>9x^4(x^2-2x+2)-6x^3(x^2-2x+2)-29x^2(x^2-2x+2)-20x(x^2-2x+2)-4(x^2-2x+2)=0#
#=>(x^2-2x+2)(9x^4-6x^3-29x^2-20x-4)=0#
Now we are solve the Quartic equation
#(9x^4-6x^3-29x^2-20x-4)=0# to get other solutions
#(9x^4-6x^3-29x^2-20x-4)=0#
#=>(3x^2)^2-2xx3x^2xx x+x^2-30x^2-20x-4=0#
#=>(3x^2-x)^2-10x(3x+2)-4=0#
#=(3x^2-x)^2-2^2-10x(3x+2)=0#
#=>(3x^2-x+2)(3x^2-x-2)-10x(3x+2)=0#
#=>(3x^2-x+2)(3x^2-3x+2x-2)-10x(3x+2)=0#
#=>(3x^2-x+2)(3x(x-1)+2(x-1))-10x(3x+2)=0#
#=(3x^2-x+2)(3x+2)(x-1))-10x(3x+2)=0#
#=>(3x+2){(3x^2-x+2)(x-1)-10x}=0#
#=(3x+2)(3x^3-x^2+2x-3x^2+x-2-10x)=0#
#=>(3x+2)(3x^3-4x^2-7x-2)=0 # #color(red)" please see below"#
#=>(3x+2)(3x^3+2x^2-6x^2-4x-3x-2)=0#
#=>(3x+2){(x^2(3x+2)-2x(3x+2)-1(3x+2)}=0#
#=>(3x+2)^2(x^2-2x-1)=0#
So solutions are
for #(3x+2)^2=0#
#=>x=-2/3,-2/3#
for #(x^2-2x-1)=0#
#=>x=(2pmsqrt((-2)^2-4xx1xx(-1)))/(2xx1)#
#=>x=1pmsqrt2#
So solution set
#{(1pmi),-2/3,-2/3,1pmsqrt2}#
# color(red)"Please Note"#
The linear factor #(3x+2)# of the polynomial #P'(x)=3x^3-4x^2-7x-2# has been found out by trial putting #x=pm2/3,pm1/3 ,pm2,pm1# in the polynomial and got #P'(-2/3)=0# .
