**(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** 6x**^{2} +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.ofx^{2} =** −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 x^{2} = **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 s^{2} = **−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 s^{2} = 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 x^{2} = **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 x^{2} = **−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 u^{2}** = −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 u^{2} =** −0/4 = 0**

**Hence, the relationship is verified.**

**(v) t**^{2} – 15

**t**^{2} – 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 x^{2}** = −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 t^{2} **= −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 x^{2}= −(−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 x^{2 }**= −4/3**

**Hence, the relationship is verified.**