Let f(x) = 2x2 ˗ 11x + 15
= 2x2 ˗ (6x + 5x) + 15
= 2x2 ˗ 6x ˗ 5x + 15
= 2x (x ˗ 3) ˗ 5 (x ˗ 3)
= (2x ˗ 5) (x ˗ 3)
To find the zeroes, set f(x) = 0
(2x ˗ 5) (x ˗ 3) = 0
2x ˗ 5= 0 or x ˗ 3 = 0
x = 5/2 or x = 3
Again,
Sum of zeroes = 5/2 + 3 = (5+6)/2 = 11/2
= -b/a
= (-Coefficient of x)/(Cofficient of x2)
Product of zeroes = 5/2 x 3 = 15/2
= c/a
= Constant term / Coefficient of x2