no. of real roots of equation is equal to

x⁸-x⁵+x²-x+1=0

a) 0

b) 2

c) 4

d) 6

Let f(x) = x⁸-x⁵+x²-x+1

f(0) = 1 = f(1)

Its obvious that for x<0, f(x)>0

For x>1 we have x⁸>x⁵ and x²>x so that f(x)>1

Again for 0<x<1, we write f(x) = x⁸ +(x²-x⁵) + (1-x) and we know that in this domain we have (x²>x⁵) and (1>x).

So again f(x)>0

Thus f(x)>0 for all real x and hence this polynomial has no real roots