Very Short Answers - Polynomials Class 9 Notes | EduRev

Question 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

Question 2.  then what is 

⇒

⇒ 

⇒ 

Question 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

Question 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). 

Question 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

Question 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)
 

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

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

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

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

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

Question 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

Question 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)

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

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

Question 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

Question 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

Question 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

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

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