Let f(x) = 4x2–3x–1
= 4x2–4x+x–1 [by splitting the middle term]
= 4x(x–1)+1(x–1)
= (x–1)(4x+1)
So, the value of 4x2–3x–1 is zero when x – 1 = 0 or 4x + 1 = 0 i.e., when x = 1 or x= −14
So, the zeroes of 4x2–3x–1 are 1 and −14
Now to verify the relations between the zeroes and the coefficient:
Sum of zeroes = (−ba)=1−14=34=−(−3)4
=(−1)Coefficient of xCoefficient of x2.
And product of zeroes = (ca)=(1)(−14)=(−14)
= (−1)2constant termCoefficient of x2
Hence, verified.