(i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x (iv) 4u2 + 8u (v) t2 – 15 (vi) 3x2 – x – 4
Solution:
(i) x2 – 2x – 8
= x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4)(x + 2)
Therefore, zeroes of polynomial equation x2 – 2x – 8 are (4, -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
= 4s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
Therefore, zeroes of polynomial equation 4s2 – 4s + 1 are (1/2, 1/2)
Sum of zeroes = (½) + (1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s2)
Product of zeros = (1/2) × (1/2) = 1/4 = (Constant term)/(Coefficient of s2 )
(iii) 6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
Therefore, zeroes of polynomial equation 6x2 – 3 – 7x are (-1/3, 3/2)
Sum of zeroes = -(1/3) + (3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = -(1/3) × (3/2) = -(3/6) = (Constant term) /(Coefficient of x2 )
(iv) 4u2 + 8u
= 4u(u + 2)
Therefore, zeroes of the polynomial equation 4u2 + 8u are (0, -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 u2 )
(v) t2 – 15
= t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√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 t2 )
(vi) 3x2 – x – 4
= 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation 3x2 – x – 4 are (4/3, -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 )