Given,
x5(x3−x2−x+1)+x(3x3−4x2−2x+4)−1=0
Now on rearranging we get
⇒x8−x7−x6+x5+3x4−4x3−2x2+4x−1=0
⇒x7(x−1)−x5(x−1)+3x3(x−1)−x(x2−1)−2x(x−1)+x−1=0
⇒(x−1)(x7−x5+3x3−x(x+1)−2x+1)=0
⇒(x−1)(x5(x2−1)+3x(x2−1)−1(x2−1))=0
⇒(x2−1)(x−1)(x5+3x−1)=0
From above equation it is clearly visible that the root are x=±1 and for x5+3x−1
Let f(x)=x5+3x−1, f′(x)=5x4+3
f′(x)>0∀x∈R, hence it is monotonic in nature so it will have only one root other than 1&-1.
Hence 3 real distinct roots.