Let f(x) = 4x2 ˗ 4x ˗ 3
= 4x2 ˗ (6x ˗ 2x) ˗ 3
= 4x^2 ˗ 6x + 2x ˗ 3
= 2x (2x ˗ 3) + 1(2x ˗ 3)
= (2x + 1) (2x ˗ 3)
To find the zeroes, set f(x) = 0
(2x + 1) (2x ˗ 3)= 0
2x + 1 = 0 or 2x ˗ 3 = 0
x = -1/2 or x = 3/2
Again,
Sum of zeroes = (-1/2) + (3/2) = (-1+3)/2 = 2/2
= -b/a
= (-Coefficient of x)/(Cofficient of x2)
Product of zeroes = (-1/2)(3/2) = (-3)/4
= c/a
= Constant term / Coefficient of x2