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**Short Answer Type Questions****Ques 1: Find the sum and product of zeroes of 3x ^{2} - 5x + 6.**

Comparing it with ax

a = 3, b = - 5, c = 6

âˆ´ Sum of the zeroes =

and, Product of the zeroes = **Ques 2: Find the sum and product of the zeroes of polynomial p (x) = 2x ^{3} - 5x^{2} - 14x + 8.**Comparing p (x) = 2x

Sol:

a = 2, b = â€“5,

c = - 14 and d = 8

âˆ´ Sum of the zeroes =

Product of zeroes

Sol:

Product of roots (P)

Since, the required Quadratic polynomial

= k(x

= k

Thus, the required polynomial is

= k (x^{2} - 2x - 1/4)**Ques 4: If Î± and Î² are the zeroes of a Quadratic polynomial x ^{2} + x - 2 then find the value of **

Sol:

a = 1, b = 1, c = - 2

Thus, **Ques 5: If a and b are the zeroes of x ^{2} + px + q then find the value of **

Sol:

a =1, b = p and c = q

âˆ´ Sum of zeroes, a + b = - b/a

â‡’

and Î±Î² = c/a

â‡’ Î±Î² = q/1 = q

Now,

Thus, the value of is **Ques 6: Find the zeroes of the quadratic polynomial 6x ^{2} - 3 - 7x.Sol: **We have:

= 6x

= 6x

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

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

For 6x

either (3x + 1) = 0 or (2x - 3) = 0

â‡’ 3x = - 1 or 2x = 3

â‡’

Thus, the zeroes of and 3/2.

Sol:

2x

= 2x (x - 3) - 2 (x - 3)

= (2x - 2) (x - 3)

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

For 2x

Either, x - 1 = 0 â‡’ x = 1

or x - 3 = 0 â‡’ x = 3

âˆ´ The zeroes of 2x

Sol:

p (x) = 3x

= 3x

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

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

For p (x) = 0, we get

Either x + 2 = 0 â‡’ x = - 2

or 3x - 1 = 0 â‡’ x = 1/3

Thus, the zeroes of 3x

Sol:

âˆ´ p (2) = 3 (2)

[2 is a zero of p (x)]

or 12 - 2p + 2 or 14 - 2p = 0

or p = 7

Next g (x) = 4x

âˆ´ g (2) = 4(2)

[2 is a zero of g (x)]

or 4 Ã— 4 - 2q - 10 = 0

or 16 - 2q - 10 = 0

or 6 - 2q = 0

â‡’ q = 6/2 â‡’ q = 3

Thus, the required values are p = 7 and q = 3.

Sol:

= 3x

Comparing p (x) with ax

a = 3, b = (2k + 1),

c = - (k + 5)

âˆ´ Sum of the zeroes

Product of the zeroes

According to the condition,

Sum of zeroes = 1/2 (product of roots)

â‡’ - 2 (2k + 1) = - (k + 5)

â‡’ 2 (2k + 1) = k + 5

â‡’ 4k + 2 = k + 5

â‡’ 4k - k = 5 - 2

â‡’ 3k = 3

â‡’ k = 3/3 = 1**Ques 11: On dividing p (x) by a polynomial x - 1 - x ^{2}, the Quotient and remainder were (x - 2) and 3 respectively. Find p (x).Sol:** Here,dividend = p (x)

Divisor, g (x) = (x - 1 - x

Quotient, q(x) = (x - 2)

Remainder, r (x) = 3

âˆµ Dividend = [Divisor Ã— Quotient] +Remainder

âˆ´ p (x)= [g (x) Ã— q(x)] + r (x)

= [(x - 1 - x

= [x

= 3x

= - x

Sol:

= (âˆš2 )

Sol:

âˆ´ (x - 5), (x - 3) and (x + 2) are the factors of p (x)

â‡’ p (x) = k (x - 5) (x - 3) (x + 2)

= k (x

= k (x

= k (x

= k (x

Thus, the required polynomial is k (x

Sol:

Î±b + Î²Î³ + Î³Î± = - 10

Î±Î²Î³ = - 24

âˆµ A cubic polynomial having zeroes as Î±,Î²,Î³ is

p (x) = x

âˆ´The required cubic polynomial is

= k {x

= k(x

Note:If Î±, Î² and Î³ be the zeroes of a cubic polynomial p (x) then

p (x) = x^{3}- [Sum of the zeroes] x^{2}+ [Product of the zeroes taken two at a time] x - [Product of zeroes]

i.e., p (x) = k {x^{3}- (Î± + Î² + Î³) x^{2}+ [Î±Î² + Î²Î³ + Î³Î±] x - (Î±Î²Î³).

**Ques 15: Find all the zeroes of the polynomial 4x ^{4} - 20x^{3} + 23x^{2} + 5x - 6 if two of its zeroes are 2 and 3.** Here, p (x) =4x

Sol:

Since, 2 and 3 are the zeroes of p (x),

âˆ´ (x - 2) and (x - 3) are the factors of p(x)

â‡’ (x - 2) (x - 3) is a factor of p (x)

â‡’ x

Now, using the division algorithm for x

get:

âˆ´ We get (x^{2} - 5x + 6) (4x^{2} - 1) = p (x)

â‡’ (x - 3) (x - 2) [(2x)^{2} - (1)2] = p (x)

â‡’ (x - 3) (x - 2) (2x - 1) (2x + 1) = p (x)

â‡’

Thus, all the zeroes of p (x) are:

**Ques 16: If 1 is a zero of x ^{3} - 3x^{2} - x + 3 then find all other zeroes.** Here,p (x) = x

Sol:

âˆµ 1 is a zero of p (x)

âˆ´ (x - 1) is a factor of p (x).

Now, dividing p (x) by (x - 1), we have:

âˆ´ p (x) = (x - 1) (x^{2} - 2x - 3)

â‡’ p (x) = (x - 1) [(x^{2} - 3x + x - 3)]

= (x - 1) [x (x - 3) + 1 (x - 3)]

= (x - 1) [(x - 3) (x + 1)]

i.e., (x - 1), (x - 3) and (x + 1) are the factors of p (x).

â‡’ 1, 3, and - 1 are the zeroes of p (x).**Ques 17: Find all the zeroes of 2x ^{4} - 3x^{3} - 3x^{2} + 6x - 2, if two of its zeroes are 1 and 1/2.** Here, p(x) = 2x

Sol:

âˆµ 1 and are the zeroes of p (x)

âˆ´ (x - 1) and are the factors of p (x)

â‡’ (x - 1) (2x - 1) is a factor of p (x)

â‡’ (2x

Now, dividing p (x) by 2x

âˆ´ p (x) = (2x^{2} âˆ’ 3x + 1) (x^{2} âˆ’ 2)

are the zeroes of p (x).**Ques 18: On dividing 4x ^{3} - 8x^{2} + 8x + 1 by a polynomial g (x), the Quotient and remainder were (2x^{2} - 3x + 2) and (x + 3) respectively. Find g (x).**âˆµ Dividend = Divisor Ã— Quotient + Remainder

Sol:

i.e., p (x) = g (x) Ã— Q (x) + r (x)

âˆ´ g (x) =

Thus, the required polynomial g (x) = 2x - 1.**Ques 19: If Î± and Î² are the zeroes of the quadratic polynomial p (x) = kx ^{2} + 4x + 4 such that Î±^{2} + Î²^{2} = 24, find the value of k.**Here, p (x) = kx

Sol:

Comparing it with ax

a = k; b = 4; c = 4

âˆ´ Sum of the zeroes = -b/a

â‡’ Î± + Î² = -4/k

and Product of the zeroes = c/a

â‡’ Î±Î² = 4/k

âˆµ Î±

âˆ´ (Î± + Î²)

[âˆµ (x + y)

â‡’

â‡’

â‡’ 16 âˆ’ 8k âˆ’ 24k

â‡’ 24k

â‡’ (3k âˆ’ 2) (k + 1) = 0

â‡’ 3k âˆ’ 2= 0 or k + 1 = 0

â‡’ k = 2/3 or k = -1

Sol:

= 6x

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

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

=

âˆ´ Zeroes of p (x) are 3/2 and

To verify the relationship:

Sum of the zeroes =

â‡’

â‡’

â‡’ 7/6 = 7/6

L.H.S = R.H.S â‡’ Relationship is verified.

Product of the zeroes =

â‡’

â‡’

i.e., L.H.S = R.H.S â‡’ Relationship is verified.

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