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

1 Answer
Jun 30, 2017

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