(i) f(x) = x² - 2x - 8

factorize the given polynomial by splitting the middle term:

⇒ x² - 4x + 2x – 8

⇒ x (x - 4) + 2 (x - 4)

For zeros of f(x), f(x) = 0

⇒(x + 2) (x - 4) = 0x+2 = 0x = - 2x - 4 = 0x = 4

⇒x = - 2, 4

Therefore zeros of the polynomial are - 2 & 4

In a polynomial the relations hold are as follows:

sum of zeroes is equal to - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Constant\,term}{coefficient\,of\,x^2}\)

For the given polynomial,

Sum of zeros = - 2 + 4 = 2

And - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) is - (- 2) = 2

Hence the value of - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) and sum of zeroes are same.

Product of zeros = - 2 × 4 = - 8

\(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) is - 8.

Hence the value of \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) and product of zeroes are same.

(ii) g(s) = 4s² - 4s + 1 

factorize the given polynomial by splitting the middle term:

⇒ 4s² - 2s - 2s + 1

⇒ 2s (2s - 1) - 1 (2s - 1)

For zeros of g(s), g(s) = 0 (2s - 1) (2s - 1) = 0

2s - 1 = 0 s = \(\frac{1}{2}\)

Therefore zeros of the polynomial are \(\frac{1}{2}\), \(\frac{1}{2}\)

In a polynomial the relations hold are as follows:

sum of zeroes is equal to - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Constant\,term}{coefficient\,of\,x^2}\)

For the given polynomial,

Sum of zeros = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1  \(\frac{-Coefficient\,of\,s}{Coeficient\,of\,s^2}\) = \(\frac{4}{4}\) = 1

Hence the value of - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) and sum of zeroes are same.

Product of zeros = - \(\frac{1}{2}\) × - \(\frac{1}{2}\) = \(\frac{1}{4}\) \(\frac{Constant\,term}{coefficient\,of\,s^2}\) = \(\frac{1}{4}\)

Hence the value of \(\frac{Constant\,term}{coefficient\,of\,x^2}\) and product of zeroes are same. 

(iii) h(t) = t² - 15

use the formula a² - b² = (a+ b)(a - b) to solve the above equation,

Here a is t and b is √15.

Solve the given expression as:

t² - (√15)² = (t + √15)(t - √15)

For zeros of h(t), h(t) = 0

t + √15 = 0

t = - √15

t - √15 = 0

t = √15

Therefore zeros of the given polynomial are t = √15 & -√15

In a polynomial the relations hold are as follows:

sum of zeroes is equal to - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\)

For the given polynomial,

Sum of zeros = √15 + (- √15) = 0

The value of - \(\frac{Coefficient\,of\,t}{Coeficient\,of\,t^2}\) is 0.

Hence, the value of - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\)and sum of zeroes are same.

Product of zeros = - √15 x √15

The value of \(\frac{Constant\,term}{coefficient\,of\,t^2}\) is - √15

Hence the value of \(\frac{Constant\,term}{coefficient\,of\,x^2}\) and product of zeroes are same.

(iv) f(x) =  6x² - 3 - 7x

Write the equation in the form of ax² +bx+c as:

6x² - 7x -3

factorize the given polynomial by splitting the middle term:

⇒ 6x² - 9x + 2x - 3

⇒ 3x(2x - 3) +1(2x - 3)

⇒ (3x + 1) (2x - 3)

For zeros of f(x),f(x) = 0

⇒ (3x + 1) (2x - 3) = 0

x = \(\frac{-1}{3}\), \(\frac{3}{2}\)

Therefore zeros of the polynomial are   \(\frac{-1}{3}\), \(\frac{3}{2}\)

In a polynomial the relations hold are as follows:

sum of zeroes is equal to -  \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Constant\,term}{coefficient\,of\,x^2}\)

Sum of zeros = \(\frac{-1}{3}\) + \(\frac{3}{2}\) = \(\frac{-2+9}{2}\) = \(\frac{7}{6}\) = \(\frac{7}{6}\) = - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\)

Product of zeros = \(\frac{-1}{3}\)× \(\frac{3}{2}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\) = \(\frac{Constant\,term}{coefficient\,of\,x^2}\)