Previous Year Questions : Polynomials | Mathematics (Maths) Class 9 PDF Download

Sol:

g(x) = 3x – 1 = 0 ⇒ x = 1/3

∴ Remainder 

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

Sol:

 ∵ (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

Sol:

∵ (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)

Sol:

The value of x³ – y³ is (x – y)³ + 3xy (x – y)

Sol:

The degree of 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is 4.

Sol:

∵ p(x) = 0 ⇒ 2x + 5 = 0

⇒ x =

∴ zero of 2x + 5 is

Sol:

 ∵ 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.

Sol:

∵ 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

Sol:

Sol:

On squaring both the sides we get;

Sol:

p(x) = x³ – ax² + 4x – a
(x – a) = 0
⇒ x = a
∴ p(a) = (a)³ – a(a)² + 4(a) – a = a³ – a³ + 4a – a = 4a – a = 3a
∴ The required remainder = 3a.

Sol:

Here, p(x) = kx³ + 9x² + 4x – 8
Since, Divisor = x + 3
∴ x + 3 = 0
⇒ x = –3
∴ p(–3) = 7
⇒ k(–3)³ + 9(–3)² + 4(–3) – 8 = 7
⇒ –27k + 81 – 12 – 8 = 7
⇒ –27k = 7 – 81 + 12 + 8
⇒ –27k = 27 – 81
⇒ –27k = –54
⇒ k= -(54/27) = 2
Thus, the required value of k = 2.

Sol:

(a) Here p(x) = x² + 4x – kx + 2
If p(x) is exactly divisible by x – 2, then p(2) = 0
i.e. (2)² + 4(2) – k(2) + 2 = 0
⇒ 4 + 8 – 2k + 2 = 0
⇒ 14 – 2k = 0
⇒ 2k = 14
⇒ k= 14/2 = 7
Thus, the required value of k is 7.

(b) Here, p(x) = x³ – 2mx² + 16
∴ p(–2) = (–2)³ –2(–2)²m + 16
⇒ –8 –8m + 16
⇒ –8m + 8
Since, p(x) is divisible by x + 2
∴ p(–2) = 0
or –8m + 8 = 0
⇒ m = 1

Sol:

We have  x² – x – 12
⇒ x² – 4x + 3x – 12
⇒ x(x – 4) + 3(x – 4)
⇒ (x – 4)(x + 3)
Thus, x² – x – 12 = (x – 4)(x + 3)

Sol:

We have x + (1/2x)  = 5
Squaring both sides, we get:
⇒ ⇒ ⇒ Thus, the required value of

Sol:

Here, p(x) = x³ – 27x² + 8x + 18, (x – 1) will be a factor of p(x) only if (x – 1) divides p(x) leaving a remainder 0.
For  x – 1 = 0
⇒ x = 1
∴ p(1) = (1)³ – 27(1)² + 8(1) + 18
⇒ 1 – 27 + 8 + 18
⇒ 27 – 27
⇒ 0
Since, p(1) = 0
∴ (x – 1) is a factor of p(x).
Thus, (x – 1) is a factor of x³ – 27x² + 8x + 18.

Sol:

x³ + y³ – 12xy + 64
⇒ (x)³ + (y)³ + (4)³ – 3(x)(y)(4)
⇒ [x² + y² + 42 – xy – y. 4 – 4 .  x](x + y + 4)
⇒ [x² + y² + 16 – xy – 4y – 4x][x + y + 4]                    ...(1)
Since, x + y = –4 ∴ x + y + 4 = 0                                 ...(2)
From (1) and (2), we have x³ + y³ – 12xy + 64
⇒ [x² + y² + 16 – xy – 4y – 4x][0] = 0
Thus, x³ + y³ – 12xy + 64 = 0.

Sol:

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

Sol:

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, 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.

Sol:

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

Sol: 

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