Question

find the zeros of quadratic polynomial and verify the relationship between the zeroes and cofficant
(i)x2-2n-8​

Answer 1

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 x2Read more on Sarthaks.com - https://www.sarthaks.com/694809/zeros-quadratic-polynomialsx-8and-verify-relationship-between-zeros-coefficients

Answer 2

2x-2n-8 ,-------------(dividing by 2)

x-n-4