Class 9 Maths Question Answers - Polynomials

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

Solution. p(t) = t² – t + 1
⇒ p(–2) = (–2)² – (–2) + 1
= 4 + 2 + 1
= 7

Q.2. If x - 1/x = -1 then what is  

Solution. On squaring both sides we get;

⇒

⇒  

⇒  

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

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 x³ – 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 – ax² + 2x + a – 5
since x – a is a factor of p(x),
so p(a) = 0
⇒ (a)³ – a(a)² + 2(a) + a – 5 = 0
⇒ (a)³ – a(a)² + 2(a) + a – 5 = 0
⇒ 3a – 5 = 0 ⇒ a =

(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) = ax³ + 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)³ + (c – a)³.

Solution. If x + y + z = 0
then x³ + y³ + z³ = 3xyz
Here, (a – b) + (b – c) + (c –a) = 0
∴ (a – b)³ + (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 – 2[xy + yz + zx]?
Solution. ∵ (x + y + z)² 
= x² + y² + z² + 2 xy + 2 yz + 2 zx
= x² + y² + 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 + y³ 
⇒ (x + y)³ – 3xy(x + y) = x³ + y³
Thus, value of x³ + y³ is (x + y)³ – 3xy (x + y)

Q.14. Write the value of x³ – y³.
Solution. The value of x³ – 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 =

∴ zero of 2x + 5 is

Q.17. Which of the following is one of the zero of the polynomial 2x² + 7x – 4 ?
 2,  -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
⇒ x³ + y³ + z³ = 3xyz
∴ a + b + 2 = 0
⇒ (a)³ + (b)³ + (2)³ = 3(a × b × 2) = 6ab
⇒ The value of a³ + b³ + 8 is 6ab

Q.19. If 49x² – p =, what is the value of p?
Solution. 


Q.20. If  = 1, then what is the value of  ? 

Solution. On squaring both the sides we get;