We have:
f(x) = 5x2 ˗ 4 ˗ 8x
= 5x2 ˗ 8x ˗ 4
= 5x2 ˗ (10x ˗ 2x) ˗ 4
= 5x 2 ˗ 10x + 2x ˗ 4
= 5x (x ˗ 2) + 2(x ˗ 2)
= (5x + 2) (x ˗ 2)
∴ f(x) = 0 ⇒ (5x + 2) (x ˗ 2) = 0
⇒ 5x + 2= 0 or x ˗ 2 = 0
⇒ x = \(\frac{-2}5\) or x = 2
So, the zeroes of f(x) are \(\frac{-2}5\) and 2.
Sum of zeroes = \((\frac{-2}5)\) + 2 = \(\frac{-2+10}5\) = \(\frac{8}5\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,ofx^2)}\)
Product of zeroes = \((\frac{-2}5)\) x 2 = \(\frac{-4}5\) = \(
\frac{constant\,term}{(coefficient\,of\,x^2)}\)