How do you solve the equation #6x^6+x^5-34x^4+34x^2-x-6=0# ?

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Answer 1 · 2017-06-30T05:15:22

#6x^6+x^5-34x^4+34x^2-x-6=0#

#=>(6x^6-6)+x^5-x-34x^4+34x^2=0#

#=>6(x^6-1)+(x^4-1)x-34(x^2-1)x^2=0#

#=>6(x^2-1)(x^4+x^2+1)+(x^2-1)(x^2+1)x-34(x^2-1)x^2=0#

#=>(x^2-1)[6(x^4+x^2+1)+(x^2+1)x-34x^2]=0#

#=>(x^2-1)[6((x^2+1)^2-x^2)+(x^2+1)x-34x^2]=0#

#=>(x^2-1)[6(x^2+1)^2-6x^2+(x^2+1)x-34x^2]=0#

#=>(x^2-1)[6(x^2+1)^2+(x^2+1)x-40x^2]=0#

#=>(x^2-1)[6(x^2+1)^2+16(x^2+1)x-15(x^2+1)x-40x^2]=0#

#=>(x^2-1)[2(x^2+1){3(x^2+1)+8x}-5x{3(x^2+1)x+8x}]=0#

#=>(x^2-1){3(x^2+1)+8x}{2(x^2+1)-5x}=0#

#=>(x^2-1)(3x^2+8x+3)(2x^2-5x+2)=0#

when

#(x^2-1)=0#

#=>x=pm1#

when

#3x^2+8x+3=0#

#=>x=(-8pmsqrt(8^2-4*3*3))/(2*3)#

#=>x=(-8pmsqrt28)/(2*3)#

#=>x=(-8pm2sqrt7)/(2*3)#

#=>x=(-4pmsqrt7)/3#

when

#=>2x^2-5x+2=0#

#=>x=(5pmsqrt((-5)^2-4*2*2))/(2*2)#

#=>x=(5pmsqrt9)/(2*2)#

#=>x=2,1/2#