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Class 10 Maths Chapter 2 Question Answers - Polynomials

Q.1. What is p(–2) for the polynomial p(t) = t2 – t + 1?

Solution. p(t) = t– t + 1
⇒ p(–2) = (–2)– (–2) + 1
= 4 + 2 + 1
= 7

Q.2. If x - 1/x = -1 then what is  Class 10 Maths Chapter 2 Question Answers - Polynomials

Solution. On squaring both sides we get;

Class 10 Maths Chapter 2 Question Answers - Polynomials

Class 10 Maths Chapter 2 Question Answers - Polynomials

⇒  Class 10 Maths Chapter 2 Question Answers - Polynomials

⇒  Class 10 Maths Chapter 2 Question Answers - Polynomials

Q.3. If x + y = –1, then what is the value of x3 + y3 – 3xy? 

Solution. We have x3 + y3 = (x + y)(x2 – xy + y2)
⇒ x3 + y3 = (–1)(x+ 2xy + y2) + 3xy
⇒ x+ y3 = –1(x + y)2 – 3xy
⇒ x+ y3 – 3xy = –1(–1)2 = –1(1)
⇒ x3 + y3 – 3xy = –1

Q.4. Show that p(x) is not a multiple of g(x), when p(x) = x3 + x – 1 g(x) = 3x – 1

Solution. g(x) = 3x – 1 = 0 ⇒ x = 1/3

∴ Remainder 

Class 10 Maths Chapter 2 Question Answers - Polynomials

Since remainder ≠ 0, so p(x) is not a multiple of g(x).

Q.5. (a) Find the value of ‘a’ if x – a is a factor of x3 – ax2 + 2x + a – 5.
(b) Find the value of ‘a’, if (x – a) is a factor of x– ax2 + 2x + a – 1 

[NCERT Exemplar]
(c) If x + 1 is a factor of ax3 + x2 – 2x + 4a – 9 find the value of ‘a’. [NCERT Exemplar]

Solution. (a) Let p(x) = x3 – ax+ 2x + a – 5
since x – a is a factor of p(x),
so p(a) = 0
⇒ (a)– a(a)2 + 2(a) + a – 5 = 0
⇒ (a)3 – a(a)2 + 2(a) + a – 5 = 0
⇒ 3a – 5 = 0 ⇒ a = Class 10 Maths Chapter 2 Question Answers - Polynomials

(b) Here, p(x) = x3 – ax2 + 2x + a – 1
∵ x – a is a factor of p(x)
∴ p(a) = 0
⇒ a3 – a(a)2 + 2(a) + a – 1 = 0
⇒ a3 – a3 + 2a + a – 1 = 0
⇒ 3a – 1 = 0
⇒ a = 1/3

(c) Here, x + 1 is a factor of p(x) = ax+ x2 – 2x + 4a – 9
∴ p(–1) = 0
⇒ a(–1)3 + (–1)2 – 2(–1) + 4a – 9 = 0
⇒ –a + 1 + 2 + 4a – 9 = 0
⇒ 3a – 6 = 0    
⇒ a = 2

Q.6. Without finding the cubes factorise (a – b)3 + (b – c)+ (c – a)3.

Solution. If x + y + z = 0
then x3 + y3 + z= 3xyz
Here, (a – b) + (b – c) + (c –a) = 0
∴ (a – b)+ (b – c)3 + (c – a)3
= 3(a – b)(b – c)(c – a)

Q.7. What is zero of a non-zero constant polynomial?
Solution. 
The zero of a non-zero constant polynomial is ‘0’.

Q.8. What is the coefficient of a zero polynomial?
Solution. 
The coefficient of a zero polynomial is 1.

Q.9. What is the degree of a biquadratic polynomial?
Solution.  
∵ The degree of a quadratic polynomial is 2.
∴ The degree of a biquadratic polynomial is 4.

Q.10. Is the statement: ‘0’ may be a zero of polynomial, true?
Solution. 
Yes, this statement is true.

Q.11. What is the value of (x + a) (x + b)?
Solution. 
The value of (x + a) (x + b)
= x2 + (a + b) x + ab.

Q.12. What is the value of (x + y + z)– 2[xy + yz + zx]?
Solution. 
∵ (x + y + z)
= x2 + y2 + z2 + 2 xy + 2 yz + 2 zx
= x2 + y+ z2 + 2[xy + yz + zx]
∴ (x + y + z)2 –  2[xy + yz + zx]
= x2 + y2 + z2

Q.13. What is the value of (x + y)3 – 3xy (x + y)?
Solution. 
∵ (x + y)3 = x3 + y3 + 3xy (x + y)
∴ [x3 + y3 + 3xy (x + y)] –  [3xy (x + y)] = x3 + y
⇒ (x + y)3 – 3xy(x + y) = x3 + y3
Thus, value of x+ y3 is (x + y)3 – 3xy (x + y)

Q.14. Write the value of x3 – y3.
Solution. 
The value of x– y3 is (x – y)3 + 3xy (x – y)

Q.15. Write the degree of the polynomial 4x4 + ox3 + ox5 + 5x + 7?
Solution. 
The degree of 4x4 + 0x3 + 0x5 + 5x + 7 is 4.

Q.16. What is the zero of the polynomial p(x) = 2x + 5?
Solution. 
∵ p(x) = 0 ⇒ 2x + 5 = 0

⇒ x = Class 10 Maths Chapter 2 Question Answers - Polynomials

∴ zero of 2x + 5 is Class 10 Maths Chapter 2 Question Answers - Polynomials

Q.17. Which of the following is one of the zero of the polynomial 2x2 + 7x – 4 ?
 2, 
Class 10 Maths Chapter 2 Question Answers - Polynomials -2 ? 

Solution. ∵ 2x2 + 7x – 4 = 2x2 + 8x – x – 4
⇒ 2x (x + 4) – 1 (x + 4) = 0
⇒ (x + 4) (2x – 1) = 0
⇒ x = – 4, x = 1/2,
∴ One of the zero of 2x2 + 7x – 4 is 1/2.

Q.18. If a + b + 2 = 0, then what is the value of a3 + b3 + 8.

Solution. ∵ x + y + z = 0
⇒ x+ y3 + z3 = 3xyz
∴ a + b + 2 = 0
⇒ (a)+ (b)3 + (2)3 = 3(a × b × 2) = 6ab
⇒ The value of a3 + b+ 8 is 6ab

Q.19. If 49x2 – p =Class 10 Maths Chapter 2 Question Answers - Polynomials, what is the value of p?
Solution. 
Class 10 Maths Chapter 2 Question Answers - Polynomials

Q.20. If Class 10 Maths Chapter 2 Question Answers - Polynomials = 1, then what is the value of Class 10 Maths Chapter 2 Question Answers - Polynomials ? 

Solution. On squaring both the sides we get;
Class 10 Maths Chapter 2 Question Answers - PolynomialsClass 10 Maths Chapter 2 Question Answers - Polynomials
Class 10 Maths Chapter 2 Question Answers - Polynomials

The document Class 10 Maths Chapter 2 Question Answers - Polynomials is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 10 Maths Chapter 2 Question Answers - Polynomials

1. What is a polynomial?
Ans. A polynomial is an algebraic expression consisting of variables, coefficients, and exponents that are combined using addition, subtraction, and multiplication operations. It can have one or more terms with different powers of the variables.
2. How to determine the degree of a polynomial?
Ans. The degree of a polynomial is determined by the highest power of the variable in the expression. To find the degree, examine the exponents of the variables in each term and select the highest exponent. That exponent represents the degree of the polynomial.
3. Can a polynomial have negative exponents?
Ans. No, a polynomial cannot have negative exponents. The exponents in a polynomial must be non-negative integers. Negative exponents would result in fractional or negative powers, which do not fit the definition of a polynomial.
4. What is the difference between a monomial and a polynomial?
Ans. A monomial is a polynomial with only one term, whereas a polynomial can have multiple terms. Monomials are simpler expressions with a single variable raised to a non-negative integer power, while polynomials can have multiple variables and different powers for each term.
5. How to perform polynomial division?
Ans. Polynomial division is performed similar to long division. Divide the highest degree term of the dividend (numerator) by the highest degree term of the divisor (denominator). Multiply the divisor by the result and subtract it from the dividend. Repeat this process until the degree of the remainder is smaller than the divisor's degree, or the remainder becomes zero. The final quotient is the result of the division.
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