Question

show that x^5-5x^3+5x2+-1=0 has 3 equal roots and find that root

Answer 1

x^5 -5x^3+5x^2-1 =0

=> x^5 -x^4 + x^4 -x^3 - 4x^3 +4x^2+x^2-x+x-1 =0

=>x^4(x - 1) +x^3(x - 1) -4x^2(x - 1) + x(x - 1)+ (x - 1) = 0

=>{x^4+x^3-4x^2+x+1)(x-1)=0

=>(x-1){x^4-x^3+2x^3-2x^2-2x^2+2x-x+1)=0

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

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

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

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

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

now you see (x -1) (x -1) (x -1)
hence, it is true that above expression have three equal roots .
and two distinct roots .
equal roots , x =1

and distinct roots
x={-3+_√5}/2