Let f(x) = 4 s2 – 4s + 1
By splitting the middle term, we get
f(x) = 4 s2 – (2 – 2)s + 1 [∵ – 4 = – (2 + 2) and 2×2 = 4]
= 4 s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
On putting f(x) = 0 , we get
(2s – 1) (2s – 1) = 0
⇒ 2s – 1 = 0 or 2s – 1 = 0
s = 1/2 or s = 1/2
Thus, the zeroes of the given polynomial 4s2 – 4s + 1 are 1/2 and 1/2.
Verification
So, the relationship between the zeroes and the coefficients is verified.