3x2 + 4x – 4
Splitting the middle term, we get,
3x2 + 6x – 2x – 4
Taking the common factors out, we get,
3x(x+2) -2(x+2)
On grouping, we get,
(x+2)(3x-2)
So, the zeroes are,
x+2=0 ⇒ x= -2
3x-2=0⇒ 3x=2⇒x=2/3
Therefore, zeroes are (2/3) and -2
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = – b/a
– 2 + (2/3) = – (4)/3
= – 4/3 = – 4/3
Product of the zeroes = constant term ÷ coefficient of x2
α β = c/a
Product of the zeroes = (- 2) (2/3) = – 4/3