Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x² - 2x - 8 (ii) 4s² - 4s + 1 (iii) 6x² - 3 - 7x (iv) 4u² + 8u (v) t² - 15 (vi) 3x² - x - 4

(i) x² - 2x - 8
x² - 2x - 8 = x² - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x + 2)(x - 4)
Factorize the equation, we got (x + 2)(x - 4)
So, the value of x² - 2x - 8 is zero when x + 2 = 0, x - 4 = 0, i.e., when x = -2 or x = 4.
Therefore, the zeros of x² - 2x - 8 are -2 and 4.
Now, ⇒Sum of zeroes = -2 + 4 = 2 = -(-2)/1 = -Coefficient of x / Coefficient of x²
⇒Product of zeros = (-2) × (4) = -8 = -8/1 = Constant term / Coefficient of x²

(ii) 4s² - 4s + 1
4s² - 4s + 1 = 4s² - 2s - 2s + 1 = 2s(2s - 1) - 1(2s - 1) = (2s - 1)(2s - 1)
Factorize the equation, we got (2s - 1)(2s - 1)
So, the value of 4s² - 4s + 1 is zero when 2s - 1 = 0, 2s - 1 = 0, i.e., when s = 1/2 or s = 1/2.
Therefore, the zeros of 4s² - 4s + 1 are 1/2 and 1/2.
Now, ⇒Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -Coefficient of s / Coefficient of s²
⇒Product of zeros = (1/2) × (1/2) = 1/4 = 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)
Factorize the equation, we got (3x + 1)(2x - 3)
So, the value of 6x² - 3 - 7x is zero when 3x + 1 = 0, 2x - 3 = 0, i.e., when x = -1/3 or x = 3/2.
Therefore, the zeros of 6x² - 3 - 7x are -1/3 and 3/2.
Now, ⇒Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -Coefficient of x / Coefficient of x²
⇒Product of zeros = (-1/3) × (3/2) = -1/2 = -3/6 = Constant term / Coefficient of x²

(iv) 4u² + 8u = 4u(u + 2)
Factorize the equation, we got 4u(u + 2)
So, the value of 4u² + 8u is zero when 4u = 0, u + 2 = 0, i.e., when u = 0 or u = -2.
Therefore, the zeros of 4u² + 8u are 0 and -2.
Now, ⇒Sum of zeroes = 0 - 2 = -2 = -8/4 = -Coefficient of u / Coefficient of u²
⇒Product of zeros = 0 × -2 = 0 = 0/4 = Constant term / Coefficient of u²

(v) t² - 15 = t² - (√15)² = (t + √15)(t - √15)
So, the value of t² - 15 is zero when t + √15 = 0, t - √15 = 0, i.e., when t = √15 or t = -√15.
Therefore, the zeros of t² - 15 are ±√15.
Now, ⇒Sum of zeroes = √15 - √15 = 0 = -0/1 = 0 = -Coefficient of t / Coefficient of t²
⇒Product of zeros = √15 × -√15 = -15 = -15/1 = Constant term / Coefficient of t²

(vi) 3x² - x - 4 = 3x² - 3x + 4x - 4 = 3x(x -1) + 4(x - 1) = (x -1)(3x + 4)
Factorize the equation, we get (x - 1)(3x + 4)
So, the value of 3x² - x - 4 is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x = 4/3.
Therefore, the zeros of 3x² - x - 4 are -1 and 4/3.
Now, ⇒Sum of zeroes = -1 + 4/3 = 1/3 = -(-1)/3 = -Coefficient of x / Coefficient of x²
⇒Product of zeros = -1 × 4/3 = -4/3 = -4/3 = Constant term / Coefficient of x²