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**Exercise 2.2****Ques 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.****(i) x ^{2} - 2x - 8 **

We have p (x) = x

= x

= (x - 4) (x + 2)

For p (x) = 0, we have

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

Either x - 4 = 0 â‡’ x = 4

or x + 2 = 0 â‡’ x = - 2

âˆ´ The zeroes of x

Now, sum of the zeroes

2 = 2 [L.H.S. = R.H.S]

âˆµ

4 Ã— (- 2) = -8/1

â‡’ - 8 = - 8 [L.H.S = R.H.S]

Thus, relationship between zeroes and the coefficients in x

(ii) 4s^{2 }- 4s + 1

We have p (s) = 4s^{2} - 4s + 1

= 4s^{2} - 2s - 2s + 1 = 2s (2s - 1) - 1 (2s - 1)

= (2s - 1) (2s - 1)

For p (s) = 0, we have,

(2s - 1) = 0 â‡’ s = 1/2

âˆ´ The zeroes of 4s^{2} - 4s + 1 are 1/2 and 1/2

Now,

Sum of the zeroes =

â‡’

â‡’ 1 = 1 [L.H.S = R.H.S]

and Product of zeroes =

â‡’

â‡’ 1/4 = 1/4 [L.H.S. = R.H.S]

Thus, the relationship between the zeroes and coefficients in the polynomial

4s^{2} - 4s + 1 is verified.

(iii) 6x^{2} - 3 - 7x

We have

p (x) = 6x^{2} - 3 - 7x

= 6x^{2} - 7x - 3

= 6x^{2} - 9x + 2x - 3

= 3x (2x - 3) + 1 (2x - 3)

= (3x + 1) (2x - 3)

For p (x) = 0, we have,

Either

or (2x - 3) = 0 â‡’ x =3/2

Thus, the zeroes of 6x^{2 }- 3 - 7x are

Now,

â‡’

â‡’ and 7/6 = 7/6 [L.H.S = R.H.S]

and

â‡’

â‡’ [L.H.S = R.H.S]

Thus, the relationship between the zeroes and the coefficients in the polynomial

6x^{2} - 3 - 7x is verified.

(iv) 4u^{2} + 8u

We have, f(u) = 4u^{2} + 8u = 4u (u + 2)

For f (u) = 0,

Either 4u = 0 â‡’ u = 0

or u + 2 = 0 â‡’ u = - 2

âˆ´ The zeroes of 4u^{2} + 8u are 0 and - 2.

Now, 4u^{2} + 8u can be written as 4u^{2} + 8u + 0.

â‡’ 0 + (- 2) = -(8)/4

â‡’ - 2 = - 2 [L.H.S. = R.H.S]

and the product of the zeroes

â‡’ 0 Ã— (- 2) = 0/4

â‡’ 0 = 0 [L.H.S. = R.H.S]

Thus, the relationship between the zeroes and the coefficients in the polynomial

4u^{2} + 8u is verified.

(v) t^{2} - 15

We have,

f (t) = t^{2} - 15 = (t)^{2} - (âˆš15)^{2}

(t+âˆš15) = (t- âˆš15)

For f (t) = 0, we have

Either ( t+âˆš15) = 0 â‡’ t = -âˆš15

or (t- âˆš15) = 0 â‡’ t = âˆš15

âˆ´The zeroes of t^{2} - 15 are -âˆš15 and âˆš15.

Now, we can write t^{2} - 15 as t^{2 }+ t - 15.

âˆ´

â‡’

â‡’ 0 = 0 [L.H.S = R.H.S]

and

â‡’

â‡’ âˆ’ (15) = âˆ’ (15) [L.H.S = R.H.S]

Thus, the relationship between the zeroes and the coefficients in the polynomial

t^{2} - 15 is verified.

(vi) 3x^{2} â€“ x â€“ 4

We have, f (x) = 3x^{2} - x - 4 = 3x^{2 }+ 3x - 4x - 4

= 3x (x + 1) - 4 (x + 1)

= (x + 1) (3x - 4)

For f (x) = 0 â‡’ (x + 1) (3x - 4) = 0

Either (x + 1) = 0 â‡’ x = - 1

or 3x - 4 = 0 â‡’ x = 4/3

âˆ´The zeroes of 3x^{2 }- x - 4 are 1 and

Now

â‡’

â‡’ 1/3 = 1/3 [L.H.S = R.H.S]

and Product of zeroes

â‡’

â‡’ [L.H.S. = R.H.S]

Thus, the relationship between the zeroes and the coefficients in 3x^{2} - x - 4 is verified.**Ques 2: Find a quadratic polynomial each with the given numbers as sum and product of its zeroes respectively:**

Note:A quadratic polynomial whose zeroes are Î± and Î² is given by

p (x) = {x^{2}-(Î± + Î²) x + Î±Î²}

i.e., p (x) = {x^{2}- (sum of the zeroes) x + (product of the zeroes)}

(i) Since, sum of the zeroes, (Î± + Î²) =

Product of the zeroes, ab = - 1

âˆ´ The required quadratic polynomial is

x^{2} - (Î± + Î²) x + ab

Since, and (4x^{2} - x - 4) have same zeroes, (4x^{2} - x - 4) is the required quadratic polynomial.

(ii) Since, sum of the zeroes, (Î± + Î²) = âˆš2

Product of zeroes, Î±Î² = 1/3

âˆ´The required quadratic polynomial is

Since, have same zeroes, is the required quadratic polynomial.

(iii) Since, sum of zeroes, (Î± + Î²) = 0

Product of zeroes, Î±Î² = âˆš5

âˆ´ The required quadratic polynomial is

x^{2} - (Î± + Î²) x + Î±Î²

= x^{2} - (0) x + âˆš5

= x^{2} + âˆš5

(iv) Since, sum of the zeroes, (Î± + Î²) = 1

Product of the zeroes = 1

âˆ´ The required quadratic polynomial is

x^{2} - (Î± + Î²) x + Î±Î²

= x^{2} - (1) x + 1

= x^{2 }- x + 1

(v) Since, sum of zeroes,

Product of zeroes, Î±Î² = 1/4

âˆ´ The required quadratic polynomial

Since, 1/4 (4x^{2} + x +1) and (4x^{2} + x +1) have same zeroes, (4x^{2} + x +1)the required quadratic polynomial is .

(vi) Since, sum of the zeroes, (Î± + Î²) = 4

Product of the zeroes, Î±Î² = 1

âˆ´ The required quadratic polynomial is

= x^{2} - (Î± + Î²) x + Î±Î²

= x^{2} - (4) x + 1

= x^{2} - 4x + 1

**Division Algorithm for Polynomials**

Let f (x) and g (x) be any two polynomials with g (x) = 0, then we can find polynomials q (x) and r (x) such that

f (x) = [g (x) Ã— q (x)] + r (x)

where [degree of r (x)] < [degree of g (x)], where r (x) may also be zero.

This may be written as

Dividend = [Divisor Ã— Quotient] + Remainder

Note:If Î± be a zero of the polynomial p (x), then (x âˆ’ Î±) is a factor of p (x).

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