Class 9 Maths Chapter 2 Question Answers - Polynomials

Question 1. Factorise: 

Solution:   

Thus,    

Question 2. Factorise: (x⁶ – y⁶)
Solution: x⁶ – y⁶ = (x³)² – (y³)² 
= (x³ – y³)(x³ + y³)  [∵ a² – b² = (a + b)(a – b)]

= [(x – y)(x² + xy + y²)][(x + y)(x² – xy + y²)]  

[∵ a³ + b³ = (a² + b² – ab)(a + b) and a³ – b³ = (a² + ab + b²)(a – b)]

= (x – y)(x + y)(x² + xy + y²)(x² – xy + y²)

Thus, x⁶ – y⁶ = (x – y)(x + y)(x² + xy + y²)(x² – xy + y²)

Question 3. If the polynomials 2x³ + 3x² – a and ax³ – 5x + 2 leave the same remainder when each is divided by x – 2, find the value of ‘a’
Solution: Let p(x) = 2x³ + 3x² – a and f(x) = ax³ – 5x + 2

 28 – a = 8a – 8
⇒ 8a + a = 28 + 8
⇒ 9a = 36

Thus , a = 4

Question 4. Find the values of ‘p’ and ‘q’, so that (x – 1) and (x + 2) are the factors of x³ + 10x² + px + q.
Solution: Here f(x) = x³ + 10x² + px + q
Since, x + 2 = 0 [∵ x + 2 is a factor of f(x)]
⇒ x= –2 If x + 2 is a factor f(x),
then f(–2) = 0 i.e. (–2)³ + 10(–2)² + p(–2) + q = 0 [∵Factor theorem]
⇒ –8 + 40 + (–2p) + q = 0 ⇒ 32 – 2p + q = 0 ...(1)
⇒ 2p – q = 32 Also x – 1 = 0 ⇒ x = 1
If (x – 1) is a factor of f(x), then f(1) must be equal to 0. [∵Factor theorem]
i.e. (1)³ + 10(1)² + p(1) + q = 0
⇒ 1 + 10 + p + q = 0
⇒ 11 + p + q = 0
⇒ p + q = –11                                 ...(2)
Now, by adding (1) and (2), we get

Now we put p = 7 in (2), we have  7 + q = –11
⇒ q = –11 – 7 = –18
Thus, the required value of p and q are 7 and –18 respectively.

Question 5. If (x² – 1) is a factor of the polynomial px⁴ + qx³ + rx² + sx + t, then prove that p + r + t = q + s = 0.
Solution: We have f(x) = px⁴ + qx³ + rx² + sx + t
Since, (x² – 1) is a factor of f(x), [∵ x² – 1 = (x + 1)(x – 1)]
then (x + 1) and (x – 1) are also factors of f(x).
∴ By factor theorem, we have   f(1) = 0 and f(–1) = 0
For f(1) = 0, p(1)4 + q(1)³ + r(1)² + s(1) + t = 0
⇒ p + q + r + s + t = 0                       ...(1)
For f(–1) = 0, p(–1)⁴ + q(–1)³ + r(–1)² + s(–1) + t = 0
⇒ p – q + r – s + t = 0                     ...(2)

From (4) and (3), we get p + r + t = q + s = 0

Question 6. If x + y = 12 and xy = 27, find the value of x³ + y³.
Solution: Since, (x + y)³ = x³ + y³ + 3xy (x + y)
∴ Substituting x + y = 12 and xy = 27,
we have: (12)³ =x³ + y³ + 3 (27) (12)
⇒ x³ + y³ = 81(12) – 123
= [92 – 122] × 12
= [(9 + 12) (9 – 12)] × 12
= 21 × 3 × 12 = 756

Question 7. If a + b + c = 5 and ab + bc + ca = 10, Then prove that a³ + b³ + c³ – 3abc = –25.
Solution: Since, a³ + b³ + c³ – 3abc
= (a + b + c) (a² + b² + c² – ab – bc – ca)
∴ a³ + b³ + c³ – 3 abc = (a + b + c) [(a² + b² + c² + 2ab+ 2bc+ 2cb) − 3ab − 3bc − 3ca]
= (a + b + c) [(a + b + c)2 − 3 (ab + bc + ca)]
= 5 [ 52 − 3(10)]
= 5[25 – 30]
= 5[–5] = –25

Question 8. If a, b, c are all non-zero and a + b + c = 0, prove that
Solution: Since, a + b + c = 0
∴ a³ + b³ + c³ = 3abc                         ..... (1)
Now, in   = 3, we have

                   [Multiplying and dividing by ‘abc’]

                   ..... (2)

From (1) and (2), we have