f(x) = 2x2 ˗ 11x + 15
= 2x2 ˗ (6x + 5x) + 15
= 2x2 ˗ 6x ˗ 5x + 15
= 2x (x ˗ 3) ˗ 5 (x ˗ 3)
= (2x ˗ 5) (x ˗ 3)
∴ f(x) = 0 ⇒ (2x ˗ 5) (x ˗ 3) = 0
⇒ 2x ˗ 5= 0 or x ˗ 3 = 0
⇒ x = \(\frac{5}2\) or x = 3
So, the zeroes of f(x) are \(\frac{5}2\) and 3.
Sum of zeroes = \(\frac{5}2\) + 3 = \(\frac{5+6}2\) = \(\frac{11}2\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)
Product of zeroes = \(\frac{5}2\) x 3 = \(\frac{-15}2\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)