(i) 6x² + 11x + 5 = 0

6x2 + 11x + 5 = 6x2 + 6x + 5x + 5

= 6x(x +1) + 5(x +1)

= (x +1) (6x +5)

∴ zeroes of polynomial equation 6x² +11x +5 are { −1, −5/6 }

Now, Sum of zeroes of this given polynomial equation = −1+( −56 ) = −11/6

But, the Sum of zeroes of any quadratic polynomial equation is given by = −coeff.of x / coeff.ofx² = −11/6

And Product of these zeroes will be = −1×−5/6 = 5/6

But, the Product of zeroes of any quadratic polynomial equation is given by = constant term / coeff.of x² = 5/6

Hence the relationship is verified.

(ii) 4s2 – 4s + 1

4s2 – 4s + 1 = 4s2 – 2s – 2s + 1

= 2s (2s – 1) –1(2s – 1)

= (2s – 1) (2s – 1)

∴ zeroes of the given polynomial are: {1/2,1/2}

∴ Sum of these zeroes will be =  = 1.

But, The Sum of zeroes of any quadratic polynomial equation is given by = −coeff. of s / coeff.of s² = −4/4 = 1

And the Product of these zeroes will be = 1/2 × 1/2 =1/4

But, Product of zeroes in any quadratic polynomial equation is given by = constant term/ coeff.of s² = 1/4.

Hence, the relationship is verified.

(iii) 6x2 – 3 – 7x

6x2 – 7x – 3 = 6x2 – 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3)

= (3x + 1) (2x – 3)

∴ zeroes of the given polynomial are: – (−1/3,3/2)

∴ sum of these zeroes will be =  −1/3 + 3/2 =7/6

But, The Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x² = 7/6

And Product of these zeroes will be = −1/3 × 3/2=−1/2
Also, the Product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of x² = −3/6 = −1/2

Hence, the relationship is verified.

(iv) 4u2 + 8u

4u2 + 8u = 4u (u+2)

Clearly, for finding the zeroes of the above quadratic polynomial equation either: – 4u=0 or u+2=0

Hence, the zeroes of the above polynomial equation will be (0, −2)

∴ Sum of these zeroes will be = −2

But, the Sum of the zeroes in any quadratic polynomial equation is given by = −coeff.of u / coeff.of u² = −8/4 = −2

And product of these zeroes will be = 0 × −2 = 0

But, the product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of u² = −0/4 = 0

Hence, the relationship is verified.

(v) t² – 15

t² – 15 = (t+ √15) (t − √15)

Therefore, zeroes of the given polynomial are: – {√15, −√15}

∴ sum of these zeroes will be = √15 −  √15 = 0

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x² = −0/1 = 0   

And the product of these zeroes will be = (√15) × (−√15)  = −15

But, the product of zeroes in any quadratic polynomial equation is given by

= constant term / coeff.of t² = −15/1  = −15

Hence, the relationship is verified.

(vi) 3x2 – x – 4

3x2 − x − 4   = 3x2 – 4x + 3x − 4

= x (3x – 4) +1(3x – 4)

= ( x + 1) (3x – 4)

∴ zeroes of the given polynomial are: – {−1, 4/3 }

∴ sum of these zeroes will be = −1 +4/3 = 13

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x²= −(−1)/3 = 1/3

And the Product of these zeroes will be = {−1 × 4/3 }= −4/3

But, the Product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of x² = −4/3

Hence, the relationship is verified.