Let f(x) = x2 ˗ 2x ˗ 8
= x2 ˗ 4x + 2x ˗ 8
= x(x ˗ 4) + 2(x ˗ 4)
= (x ˗ 4) (x + 2)
To find the zeroes, set f(x) = 0, then
either (x ˗ 4) = 0 or (x+2) = 0
x = 4 or x = -2
Again,
Sum of zeroes = (4 – 2) = 2 = 2/1
= -b/a
= (-Coefficient of x)/(Cofficient of x2)
Product of zeroes = (4) (-2) = -8/1
= c/a
= Constant term / Coefficient of x2