Class 9 Exam  >  Class 9 Notes  >  Advance Learner Course: Mathematics (Maths) Class 9  >  Short Answer Questions: Polynomials - 2

Class 10 Maths Chapter 2 Question Answers - Polynomials - 2

Ques 21: Find the zeroes of the quadratic polynomial 5x2 - 4 - 8x and verify the relationship between the zeroes and the coefficients of the polynomial.
Sol:
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 - 2
∴ zeroes of p (x) are 2 and Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Relationship Verification
Sum of the zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
⇒ 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 - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
i.e., L.H.S. = R.H.S.
⇒ The relationship is verified.

Ques 22: Find the quadratic polynomial, the sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.
Sol:
∵ 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.

Ques 23: If one zero of the polynomial (a2 - 9) x2 + 13x + 6a is reciprocal of the other, find the value of ‘a’.
Sol:
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 - 2

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

Ques 24: If the product of zeroes of the polynomial ax2 - 6x - 6 is 4, find the value of ‘a’
Sol: Here, p (x) = ax2 - 6x - 6
∵ Product of zeroes = Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
but product of zeroes is given as 4
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 ⇒ − 6 = 4 × a
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 ⇒ Class 10 Maths Chapter 2 Question Answers - Polynomials - 2

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

Ques 25: Find all the zeroes of the polynomial x4 + x3 - 34x2 - 4x + 120, if two of its zeroes are 2 and - 2.
Sol: 
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 - 2

∵ 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

Ques 26: Find all the zeroes of the polynomial 2x4 + 7x3 - 19x- 14x + 30, if two of its zeroes are √2 and -√2.  
Sol: 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 - 2are the factors of p (x).
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2i.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 - 2

∴ 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 - 2
∴ 3/2 and - 5 are zeroes of p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 are the zeroes of p (x).

Ques 27: Find the quadratic polynomial whose zeroes are 1 and - 3. Verify the relation between the coefficients and the zeroes of the polynomial.
Sol: 
∵ 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 - 2

Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
⇒− 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 - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
⇒− 3= − 3

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

Ques 28: Find the zeroes of the quadratic polynomial 4x2 - 4x - 3 and verify the relation between the zeroes and its coefficients.
Sol: 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 - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 are zeroes of p (x).

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

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

Class 10 Maths Chapter 2 Question Answers - Polynomials - 2

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

Ques 29: Obtain all other zeroes of the polynomial 2x3 - 4x - x2 + 2, if two of its zeroes are √2 and -√2.
Sol: p (x) = 2x- 4x - x2 + 2
√2 and -√2 are the zeroes of p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 and are the factors of p (x)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 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 - 2

⇒ 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 - 2is another factor of px.
⇒ 1/2 is another zero of p (x)

Ques 30: Find all the zeroes of x4 - 3x3 + 6x - 4, if two of its zeroes are √2 and - √2.
Sol: 
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 - 2 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 - 2

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.

Ques 31: Find a quadratic polynomial whose zeroes are - 4 and 3 and verify the relationship between the zeroes and the coefficients.
Sol: 
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.

Ques 32: 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
Sol: 
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 - 2

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

Ques 33: 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’.
Sol: 
We have:

Class 10 Maths Chapter 2 Question Answers - Polynomials - 2

∴ 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.

Ques 34: 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.
Sol: 
We have:
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2

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

Ques 35: Find all the zeroes of the polynomial x3 + 3x2 - 2x - 6, if two of its zeroes are - √2 and √2.
Sol: 
p (x) = x3 + 3x2 - 2x - 6
∵ Two of its zeroes are -√2 and √2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 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 - 2

∴ 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.

Ques 36: Find all the zeroes of the polynomial 2x3 + x2 - 6x - 3, if two of its zeroes are -√3 and √3.
Sol:
p (x) = 2x3 + x2 - 6x - 3
Two of its zeroes are -√3 and √3
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 and are factors of p (x)
i.e., Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 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 - 2

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

Ques 37: Find the zeroes of the polynomial Class 10 Maths Chapter 2 Question Answers - Polynomials - 2and verify the relation between the coefficients and the zeroes of the above polynomial.
Sol: The given polynomial is

Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
∴ zeroes of the given polynomial are Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Now in, Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
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 - 2

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

Ques 38: Find the quadratic polynomial, the sum and product of whose zeroes are Class 10 Maths Chapter 2 Question Answers - Polynomials - 2 respectively. Also find its zeroes.  
Sol: Sum of zeroes = √2
Product of zeroes Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
∵ 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 - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Since = Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2

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

Ques 39: 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.
Sol: 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 - 2

∴ 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

Ques 40: If a and b are zeroes of the quadratic polynomial x2 – 6x + a; find the value of ‘a’ if 3α + 2β = 20.
Sol: 
We have quadratic polynomial = x2 – 6x + a ...(1)
∵ a and b are zeroes of (1)
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
Class 10 Maths Chapter 2 Question Answers - Polynomials - 2
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 - 2
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 - 2 is a part of the Class 9 Course Advance Learner Course: Mathematics (Maths) Class 9.
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FAQs on Class 10 Maths Chapter 2 Question Answers - Polynomials - 2

1. What is a polynomial?
Ans. A polynomial is a mathematical expression that consists of variables, coefficients, and exponents, combined using addition, subtraction, multiplication, and non-negative integer exponents. It can have one or more terms, and each term consists of a coefficient multiplied by one or more variables raised to non-negative integer exponents.
2. How do you identify the degree of a polynomial?
Ans. The degree of a polynomial is determined by the highest exponent of the variable in the polynomial. For example, if the highest exponent of the variable in a polynomial is 3, then the degree of the polynomial is 3. The degree helps classify polynomials into different categories, such as linear, quadratic, cubic, etc.
3. Can a polynomial have negative exponents?
Ans. No, a polynomial cannot have negative exponents. In a polynomial, the exponents of the variables must be non-negative integers. Negative exponents would violate this rule and would make the expression not a polynomial. However, negative exponents can appear in the denominator of a polynomial fraction.
4. How do you add or subtract polynomials?
Ans. To add or subtract polynomials, you combine like terms. Like terms are terms with the same variables raised to the same exponents. You simply add or subtract the coefficients of these like terms while keeping the variables and exponents unchanged. For example, to add 2x^2 + 3x + 1 and 4x^2 - 2x + 5, you add the coefficients of the like terms to get 6x^2 + x + 6.
5. Can a polynomial have more than one variable?
Ans. Yes, a polynomial can have more than one variable. Polynomials can have multiple variables, each raised to different non-negative integer exponents. For example, a polynomial like 3x^2y^3 + 2xy + 1 is a polynomial with two variables, x and y, raised to different exponents. The variables can be combined using addition, subtraction, and multiplication operations according to the rules of polynomial algebra.
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