Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Short Answer Questions: Polynomials

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q1: Find the sum and product of zeroes of 3x2 - 5x + 6.

Here, p (x) = 3x2 - 5x + 6
Comparing it with ax2 + bx + c, we have
a = 3, b = - 5, c = 6
∴ Sum of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

and, Product of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q2: Find the sum and product of the zeroes of polynomial p (x) = 2x3 - 5x2 - 14x + 8.

Comparing p (x) = 2x3 - 5x2 - 14x + 8 with ax3 + bx2 + cx + d, we have
a = 2, b = –5,
c = - 14 and d = 8
∴ Sum of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Product of zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q3: Find a Quadratic polynomial whose zeroes are Class 10 Maths Chapter 2 Question Answers - Polynomials - 1.

Sum of zeroes (S)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Product of roots (P)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Since the required Quadratic polynomial
= k(x2 - Sx + P) ; where k is any real number.

= k Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Thus, the required polynomial is
= k (x2 - 2x - 1/4)

Q4: If α and β are the zeroes of a Quadratic polynomial x2 + x - 2 then find the value of Class 10 Maths Chapter 2 Question Answers - Polynomials - 1.

Comparing x2 + x - 2 with ax2 + bx + c, we have:
a = 1, b = 1, c = - 2

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Thus, Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q5: If a and b are the zeroes of x2 + px + q then find the value of Class 10 Maths Chapter 2 Question Answers - Polynomials - 1.

Comparing x2 + px + q with ax2 + bx + c
a =1, b = p and c = q
∴ Sum of zeroes, a + b = - b/a
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
and αβ = c/a
⇒ αβ = q/1 = q
Now,
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Thus, the value of Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 is Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q6: Find the zeroes of the quadratic polynomial 6x2 - 3 - 7x.

We have,
= 6x2 - 3 - 7x = 6x2 - 7x - 3
= 6x2 - 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (3x + 1) (2x - 3)
For 6x2 - 3 - 7x to be equal to zero,
either (3x + 1) = 0 or (2x - 3) = 0
⇒ 3x = - 1 or 2x = 3
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Thus, the zeroes ofClass 10 Maths Chapter 2 Question Answers - Polynomials - 1 and 3/2.

Q7: Find the zeroes of 2x2 - 8x + 6.

We have,
2x2 - 8x + 6 = 2x2 - 6x - 2x + 6
= 2x (x - 3) - 2 (x - 3)
= (2x - 2) (x - 3)
= 2 (x - 1) (x - 3)
For 2x2 - 8x + 6 to be zero,
Either, x - 1 = 0 ⇒ x = 1
or x - 3 = 0 ⇒ x = 3
∴ The zeroes of 2x2 - 8x + 6 are 1 and 3.

Q8: Find the zeroes of the quadratic polynomial 3x2 + 5x - 2.

We have,
p (x) = 3x2 + 5x - 2
= 3x2 + 6x - x - 2
= 3x (x + 2) - 1 (x + 2)
= (x + 2) (3x - 1)
For p (x) = 0, we get
Either x + 2 = 0 ⇒ x = - 2
or 3x - 1 = 0 ⇒ x = 1/3
Thus, the zeroes of 3x2 + 5x - 2 are - 2 and 1/3.

Q9: If the zero of a polynomial p (x) = 3x2 - px + 2 and g (x) = 4x2 - q x - 10 is 2, then find the value of p and q.

∵ p (x) = 3x2 - px + 2
∴ p (2) = 3 (2)2 - p (2) + 2 = 0
[2 is a zero of p (x)]
or 12 - 2p + 2 or 14 - 2p = 0
or p = 7
Next g (x) = 4x2 - q x - 10
∴ g (2) = 4(2)2 - Q (2) - 10 = 0
[2 is a zero of g (x)]
or 4 × 4 - 2q - 10 = 0
or 16 - 2q - 10 = 0
or 6 - 2q = 0
⇒ q = 6/2 ⇒ q = 3
Thus, the required values are p = 7 and q = 3.

Q10: Find the value of ‘k’ such that the quadratic polynomial 3x2 + 2kx + x - k - 5 has the sum of zeroes as half of their product.

Here, p (x) = 3x2 + 2kx + x - k - 5
= 3x2 + (2k + 1) x - (k + 5)
Comparing p (x) with ax2 + bx + c, we have:
a = 3, b = (2k + 1),
c = - (k + 5)
∴ Sum of the zeroes
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Product of the zeroes
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
According to the condition,
Sum of zeroes = 1/2 (product of roots)

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒ - 2 (2k + 1) = - (k + 5)
⇒ 2 (2k + 1) = k + 5
⇒ 4k + 2 = k + 5
⇒ 4k - k = 5 - 2
⇒ 3k = 3
⇒ k = 3/3 = 1

Q11: On dividing p (x) by a polynomial x - 1 - x2, the Quotient and remainder were (x - 2) and 3 respectively. Find p (x).

Here,dividend = p (x)
Divisor, g (x) = (x - 1 - x2)
Quotient, q(x) = (x - 2)
Remainder, r (x) = 3
∵ Dividend = [Divisor × Quotient] +Remainder
∴ p (x)= [g (x) × q(x)] + r (x)
= [(x - 1 - x2) (x - 2)] + 3
= [x2 - x - x3 - 2x + 2 + 2x2] + 3
= 3x2 - 3x - x3 + 2 + 3
= - x3 + 3x- 3x + 5

Q12: Find the zeroes of the polynomial f (x) = 2 - x2.

We have f (x)= 2 - x2
= (√2 )- x2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q13: Find the cubic polynomial whose zeroes are 5, 3 and - 2.

∵ 5, 3 and - 2 are zeroes of p (x)
∴ (x - 5), (x - 3) and (x + 2) are the factors of p (x)
⇒ p (x) = k (x - 5) (x - 3) (x + 2)
= k (x2 - 8x + 15) (x + 2)
= k (x3 - 8x2 + 15x + 2x2 - 16x + 30
= k (x3 + [- 8 + 2] x2 + [15 - 16] x + 30)
= k (x- 6x2 - x + 30)
Thus, the required polynomial is k (x3 - 6x2 - x + 30).

Q14: If α, β and γ be the zeroes of a polynomial p (x) such that (α + β +γ) = 3, (αβ + βγ + γα) = -10 and αβγ = - 24 then find p (x).

Here, α + β + γ = 3
αb + βγ + γα = - 10
αβγ = - 24
∵ A cubic polynomial having zeroes as α,β,γ is
p (x) = x3 - (a + b + γ) x+ (αβ + βγ + γα) x - (αβγ)
∴The required cubic polynomial is
= k {x3 - (3) x2 + (- 10) x - (- 24)}
= k(x- 3x2 - 10x + 24)

Note: If α, β and γ be the zeroes of a cubic polynomial p (x) then
p (x) = x3 - [Sum of the zeroes] x2 + [Product of the zeroes taken two at a time] x - [Product of zeroes]
i.e., p (x) = k {x3 - (α + β + γ) x2 + [αβ + βγ + γα] x - (αβγ).

Q15: Find all the zeroes of the polynomial 4x4 - 20x3 + 23x2 + 5x - 6 if two of its zeroes are 2 and 3.

Here, p (x) =4x4 - 20x3 + 23x2 + 5x - 6
Since, 2 and 3 are the zeroes of p (x),
∴ (x - 2) and (x - 3) are the factors of p(x)
⇒ (x - 2) (x - 3) is a factor of p (x)
⇒ x- 5x + 6 is a factor of p (x)
Now, using the division algorithm for x2 - 5x + 6 and the given polynomial p (x), we
get:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ We get (x2 - 5x + 6) (4x2 - 1) = p (x)
⇒ (x - 3) (x - 2) [(2x)2 - (1)2] = p (x)
⇒ (x - 3) (x - 2) (2x - 1) (2x + 1) = p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Thus, all the zeroes of p (x) are: 

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q16: If 1 is a zero of x3 - 3x2 - x + 3 then find all other zeroes.

Here,p (x) = x3 - 3x2 - x + 3
∵ 1 is a zero of p (x)
∴ (x - 1) is a factor of p (x).
Now, dividing p (x) by (x - 1), we have:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ p (x) = (x - 1) (x2 - 2x - 3)
⇒ p (x) = (x - 1) [(x2 - 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).
⇒ 1, 3, and - 1 are the zeroes of p (x).

Q17: Find all the zeroes of 2x4 - 3x3 - 3x2 + 6x - 2, if two of its zeroes are 1 and 1/2.

Here, p(x) = 2x4 - 3x3 - 3x2 + 6x - 2
∵ 1 and are the zeroes of p (x)
∴ (x - 1) and Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 are the factors of p (x)
⇒ (x - 1) (2x - 1) is a factor of p (x)
⇒ (2x2 - 3x + 1) is a factor of p (x).
Now, dividing p (x) by 2x2 - 3x + 1, we get

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ p (x) = (2x2 − 3x + 1) (x2 − 2)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 are the zeroes of p (x).

Q18: On dividing 4x3 - 8x2 + 8x + 1 by a polynomial g (x), the Quotient and remainder were (2x2 - 3x + 2) and (x + 3) respectively. Find g(x).

∵ Dividend = Divisor × Quotient + Remainder
i.e., p (x) = g (x) × Q (x) + r (x)
∴ g (x) = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Thus, the required polynomial g (x) = 2x - 1.

Q19: If α and β are the zeroes of the quadratic polynomial p (x) = kx2 + 4x + 4 such that α2 + β2 = 24, find the value of k.

Here, p (x) = kx2 + 4x + 4.
Comparing it with ax2 + bx + c, we have:
a = k; b = 4; c = 4
∴ Sum of the zeroes = -b/a
⇒ α + β = -4/k
and Product of the zeroes = c/a
⇒ αβ = 4/k
∵ α2 + β= 24
∴ (α + β)2 - 2αβ = 24
[∵ (x + y)2 = x2 + y2 + 2xy ⇒ (x + y)2 - 2xy = x2 + y2]
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒  Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒ 16 − 8k − 24k2 =0
⇒ 24k2 + 8k − 16 = 0
⇒ (3k − 2) (k + 1) = 0
⇒ 3k − 2= 0 or k + 1 = 0
⇒ k = 2/3 or k = -1

Q20: Find the zeroes of the quadratic polynomial 6x2 - 3 - 7x and verify the relationship between the zeroes and the coefficients of the polynomial.

Here, p (x) = 6x2 - 3 - 7x = 6x2 - 7x - 3
= 6x- 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (2x - 3) (3x + 1)
= Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
∴ Zeroes of p (x) are 3/2 and Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 
To verify the relationship:
Sum of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒ 7/6 = 7/6
L.H.S = R.H.S ⇒ Relationship is verified.
Product of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
i.e., L.H.S = R.H.S ⇒ Relationship is verified.

Q21: Find the zeroes of the quadratic polynomial 5x2 - 4 - 8x and verify the relationship between the zeroes and the coefficients of the polynomial.

p (x) = 5x2 - 4 - 8x
= 5x2 - 8x - 4
= 5x2 - 10x + 2x - 4
= 5x (x - 2) + 2 (x - 2)
= (x - 2) (5x + 2)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
∴ zeroes of p (x) are 2 and Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Relationship Verification
Sum of the zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒ 8/5 = 8/5

i.e., L.H.S. = R.H.S. ⇒ relationship is verified.
Product of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
i.e., L.H.S. = R.H.S.
⇒ The relationship is verified.

Q22: Find the quadratic polynomial, the sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.

The quadratic polynomial p (x) is given by
x2 - (Sum of the zeroes) x + (Product of the zeroes)
∴ The required polynomial is
= x2 - [8] x + [12]
= x2 - 8x + 12
To find zeroes:
∵ x2 - 8x + 12 = x2 - 6x - 2x + 12
= x (x - 6) - 2 (x - 6)
= (x - 6) (x - 2)
∴ The zeroes of p (x) are 6 and 2.

Q23: If one zero of the polynomial (a2 - 9) x2 + 13x + 6a is reciprocal of the other, find the value of ‘a’.

Here, p (x) = (a2 - 9) x2 + 13x + 6a
Comparing it with Ax2 + Bx + C, we have:
A = (a2 - 9); B = 13; C = 6
Let one of the zeroes = a
∴ The other zero = 1/α
Now, Product of the zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒ 6a = a2 − 9    ⇒ a2 − 6a + 9 = 0
⇒ (a − 3)2 =0   ⇒ a − 3=0
⇒ a = 3
Thus, the required value of a is 3.

Q24: If the product of zeroes of the polynomial ax2 - 6x - 6 is 4, find the value of ‘a’

Here, p (x) = ax2 - 6x - 6
∵ Product of zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
but product of zeroes is given as 4
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 ⇒ − 6 = 4 × a
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 ⇒ Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Thus, the required value of a is -3/2.

Q25: Find all the zeroes of the polynomial x4 + x3 - 34x2 - 4x + 120, if two of its zeroes are 2 and - 2.

Here p (x) = x4 + x3 - 34x2 - 4x + 120
∵ The two zeroes of p (x) are 2 and - 2
∴ (x - 2) and (x + 2) are factors of p (x)
⇒ (x - 2) (x + 2) is a factor of p (x)
⇒ x2 - 4 is a factor of p (x).
Now, dividing p(x) by x2 - 4, we have:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∵ Remainder = 0
∴ p (x) = (x2 - 4) (x2 + x - 30)
i.e., x2 + x - 30 is also a factor of p (x).
∵ x2 + x - 30 = x2 + 6x - 5x - 30 = x (x + 6) - 5 (x + 6)
= (x + 6) (x - 5) = [x - (- 6)] [x - 5]
- 6 and 5 are also zeroes of p (x).
⇒ All the zeroes of the given polynomial are : 2, - 2, 5 and - 6

Q26: Find all the zeroes of the polynomial 2x4 + 7x3 - 19x- 14x + 30, if two of its zeroes are √2 and -√2.  

P(x) = 2x4 + 7x3 - 19x2 - 14x + 30
∵ √2 and -√2 are the two zeroes of p (x).
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1are the factors of p (x).
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1i.e., x2 - 2 is a factor of p (x).
Now, dividing p (x) by x2 - 2, we have:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ p (x) = (2x2 + 7x - 15) (x2 - 2)
[∵ Remainder = 0]
⇒ 2x2 + 7x - 15 is a factor of p (x)
∵ 2x2 + 7x - 15 = 2x2 + 10x - 3x - 15
= 2x (x + 5) - 3 (x + 5)
= (2x - 3) (x + 5)
= Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
∴ 3/2 and - 5 are zeroes of p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 are the zeroes of p (x).

Q27: Find the quadratic polynomial whose zeroes are 1 and - 3. Verify the relation between the coefficients and the zeroes of the polynomial.

The given zeroes are 1 and - 3.
∴ Sum of the zeroes = 1 + (- 3) = - 2
Product of the zeroes = 1 × (- 3) = - 3
A quadratic polynomial p (x) is given by
x2 - (sum of the zeroes) x + (product of the zeroes)
∴ The required polynomial is
x- (- 2) x + (- 3)
⇒ x2 + 2x - 3
Verification of relationship
∵ Sum of the zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒− 2= − 2

i.e., L.H.S = R.H.S ⇒ The sum of zeroes is verified
∵ Product of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒− 3= − 3

i.e., L.H.S = R.H.S ⇒ The product of zeroes is verified.

Q28: Find the zeroes of the quadratic polynomial 4x2 - 4x - 3 and verify the relation between the zeroes and its coefficients.

Here, p (x) = 4x2 - 4x - 3 = 4x2 - 6x + 2x - 3
= 2x (2x - 3) + 1 (2x - 3)
= (2x - 3) (2x + 1)
= Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 are zeroes of p (x).

Verification of relationship
∵ Sum of the zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒ 2/2 = 1 ⇒ 1= 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
2/2 = 1 ⇒ 1 = 1
i.e., L.H.S = R.H.S ⇒ Sum of zeroes is verified
Now, Product of zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

⇒  Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
i.e., L.H.S = R.H.S ⇒ Product of zeroes is verified.

Q29: Obtain all other zeroes of the polynomial 2x3 - 4x - x2 + 2, if two of its zeroes are √2 and -√2.

p (x) = 2x- 4x - x2 + 2
√2 and -√2 are the zeroes of p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 and are the factors of p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 is a factor of p (x)
⇒ x2 - 2 is a factor of p (x)
Now, Dividing p (x) by (x2 - 2), we have:
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

⇒ p (x) = (x2 - 2) (2x - 1)
∴ 2x - 1 is also a factor of p (x)
i.e.,Class 10 Maths Chapter 2 Question Answers - Polynomials - 1is another factor of px.
⇒ 1/2 is another zero of p (x)

Q30: Find all the zeroes of x4 - 3x3 + 6x - 4, if two of its zeroes are √2 and - √2.

p (x) = x- 3x3 + 6x - 4
2 and (- √2) are the zeroes of p(x)
∴ x-2 and x-(- 2) are factors of p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 is a factor of p (x).
⇒ x2 - 2 is a factor of p (x)
On Dividing p (x) by x- 2, we have:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Since, remainder = 0
∴ (x2 - 2) (x2 - 3x + 2) = p (x)
Now, x2 - 3x + 2 = x2 - 2x - x + 2
= x (x - 2) - 1 (x - 2) = (x - 1) (x - 2)
i.e., (x - 1) (x - 2) is a factor of p (x)
∴ 1 and 2 are zeroes of p (x).
∴ All the zeroes of p (x) are ,2 , - √2, 1 and 2.

Q31: Find a quadratic polynomial whose zeroes are - 4 and 3 and verify the relationship between the zeroes and the coefficients.

We know that:
P (x) = x2 - [Sum of the zeroes] x + [Product of the zeroes] ...(1)
∵ The given zeroes are - 4 and 3
∴ Sum of the zeroes = (- 4) + 3 = - 1
Product of the zeroes = (- 4) × 3 = - 12
From (1), we have
x2 - (- 1) x + (- 12)
= x2 + x - 12    ...(2)
Comparing (2) with ax2 + bx + c, we have
a = 1, b = 1, c = - 12
∴ Sum of the zeroes = -b/a
⇒ (+ 3) + (- 4) = -1/1
i.e., L.H.S = R.H.S ⇒ Sum of zeroes is verified.
Product of zeroes = c/a
⇒ 3 × (- 4) = -12/1
⇒ - 12 = - 12
i.e., L.H.S = R.H.S ⇒ Product of roots is verified.

Q32: Using division algorithm, find the quotient and remainder on dividing f (x) by g (x), where f (x) = 6x3 + 13x2 + x - 2 and g (x) = 2x + 1

Here, f (x) = 6x3 + 13x2 + x - 2
g (x) = 2x + 1
Now, dividing f (x) by g (x), we have:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Thus, The quotient = 3x2 + 5x - 2
remainder = 0

Q33: If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1 then the remainder comes out to be ax + b, find ‘a’ and ‘b’.

We have:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ Remainder = x + 2
Comparing x + 2 with ax + b, we have
a = 1 and b = 2
Thus, the required value of a = 1 and b = 2.

Q34: If the polynomial x4 + 2x3 + 8x2 + 12x + 18 is divided by another polynomial x2 + 5, the remainder comes out to be px + q. Find the values of p and q.

We have:
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ Remainder = 2x + 3
Comparing 2x + 3 with px + q, we have
p = 2 and q = 3

Q35: Find all the zeroes of the polynomial x3 + 3x2 - 2x - 6, if two of its zeroes are - √2 and √2.

p (x) = x3 + 3x2 - 2x - 6
∵ Two of its zeroes are -√2 and √2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 is a factor of p (x)
⇒ x2 - 2 is a factor of p (x).
Now, dividing p (x) by x2 - 2 we have:
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ p (x) = (x2 - 2) (x + 3)
i.e., (x + 3) is a factor of p (x),
⇒ (- 3) is a zero of p (x)
∴All the zeroes of p (x) are - √2, √2 and - 3.

Q36: Find all the zeroes of the polynomial 2x3 + x2 - 6x - 3, if two of its zeroes are -√3 and √3.

p (x) = 2x3 + x2 - 6x - 3
Two of its zeroes are -√3 and √3
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 and are factors of p (x)
i.e., Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 is a factor of p (x)
⇒ x2 - 3 is a factor of p (x)
Now, Dividing p (x) by x2 - 3, we have:
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ p (x) = (x2 - 3) (2x + 1)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 is a factor of p (x)
⇒ -1/2 is a zero of p (x)
∴ All the zeroes of p (x) are -√3 , √3 and -1/2.

Q37: Find the zeroes of the polynomial Class 10 Maths Chapter 2 Question Answers - Polynomials - 1and verify the relation between the coefficients and the zeroes of the above polynomial.

The given polynomial is

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
∴ zeroes of the given polynomial are Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Now in, Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
co-efficient of x2 = 1
co-efficient of x = 1/6
constant term = –2
∴ Sum of zeroes  Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Product of zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q38: Find the quadratic polynomial, the sum and product of whose zeroes are Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 respectively. Also find its zeroes.  

Sum of zeroes = √2
Product of zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
∵ A quadratic polynomial is given by
x2 – [sum of roots] x + [Product of roots]
∴ The required polynomial is Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Since = Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
⇒ zeroes are Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

Q39: If the remainder on division of x3 + 2x2 + kx + 3 by x - 3 is 21, then find  the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x3 + 2x2 + kx - 18.

Let x3 + 2x2 + kx + 3 = p(x)
∵ The divisor = x – 3
∴ p(3) = 33 + 2 × 32 + 3k + 3
21 = 27 + 18 + 3k + 3
[∵ Remainder = 21]
⇒ 21 – 18 – 3 – 27 = 3k
⇒ –27 = 3k ⇒ k = – 9
Now, the given cubic polynomial
= x3 + 2x2 – 9x + 3
since, 

Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

∴ The required quotient = x2 + 5x + 6
Now, x3 + 2x2 – 9x – 18 = (x – 3) (x2 + 5x + 6)
= (x – 3) (x + 3) (x + 2)
⇒ The zeroes of x3 + 2x2 – 9x – 18 are 3, –3 and – 2

Q40: If a and b are zeroes of the quadratic polynomial x2 – 6x + a; find the value of ‘a’ if 3α + 2β = 20.

We have quadratic polynomial = x2 – 6x + a ...(1)
∵ a and b are zeroes of (1)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
It is given that: 3α + 2β = 20      ...(2)
Now, α +β = 6    ⇒ 2 (α+ β) = 2(6)
2α + 2β = 12      ...(3)
Subtracting (3) from (2), we have
Class 10 Maths Chapter 2 Question Answers - Polynomials - 1
Substituting a = 8 in α + β= 6, we get
8 +β = 6 ⇒ β = –2
Since, αβ = a
8(–2) = α ⇒ α = –16

The document Class 10 Maths Chapter 2 Question Answers - Polynomials - 1 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 2 Question Answers - Polynomials - 1

1. What is a polynomial?
Ans. A polynomial is an algebraic expression consisting of variables and coefficients, combined using addition, subtraction, multiplication, and non-negative integer exponents.
2. What are the different types of polynomials?
Ans. The different types of polynomials include monomials (one term), binomials (two terms), trinomials (three terms), and polynomials with four or more terms.
3. How do you add and subtract polynomials?
Ans. To add or subtract polynomials, you combine like terms by adding or subtracting the coefficients of the same variables and exponents.
4. How do you multiply polynomials?
Ans. To multiply polynomials, you can use the distributive property or the FOIL method, where you multiply each term of one polynomial by each term of the other polynomial and then combine like terms.
5. What is the degree of a polynomial?
Ans. The degree of a polynomial is the highest exponent of the variable in the polynomial. It determines the overall behavior and complexity of the polynomial.
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