Find the zeros of the following quadratic polynomials

(i) x2 – 2x – 8 = (x - 4) (x + 2) The value of x2 – 2x – 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2 Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x2 Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x2                                                                                                                              (ii) 4s2 – 4s + 1 = (2s-1)2 The value of 4s2 - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2 Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2 Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s2                                                                                                                                         
  (iii) 6x2 – 3 – 7x = 6x2 – 7x – 3 = (3x + 1) (2x - 3) The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2 Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2 Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2                                                                                                                                                                                                                                               (iv) 4u2 + 8u = 4u2 + 8u + 0 = 4u(u + 2) The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2 Therefore, the zeroes of 4u2 + 8u are 0 and - 2. Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2 Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of                                                                                                                                                                                                                      (v) t2 – 15 = t2 - 0.t - 15                                                                                   = (t - √15) (t + √15) The value of t2 - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15 Therefore, the zeroes of t2 - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2 Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.                                                                                                                              (vi) 3x2 – x – 4 = (3x - 4) (x + 1) The value of 3x2 – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1 Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -1.Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2 Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.