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(C) (1, 0, 0, -1)

x³ – 1 = x³ + 0x³ + 0x – 1

(2y + 1) × (y² – 2y ³+ 3y) 

= 2y(y² – 2y³ + 3y) + 1(y² – 2y³ + 3y) 

= 2y³ – 4y⁴ + 6y² + y² – 2y³ + 3y 

= -4y⁴ + 2y³ – 2y³ + 6y² + y² + 3y 

= -4y⁴ + 7y²+ 3y

i. m³ + 5m + 3

 ii. y⁵ + 2y⁴ + 3y³– y²– 7y- \(\frac1{2}\)

i. p(x) = x³ 

∴ p(1) = 1³ = 1 

p(x) = x³ 

∴ p(0) = 0³ = 0 

p(x) = x³ 

∴ p(- 2) = (- 2)³ = - 8 

ii. p(y) = y² – 2y + 5 

∴ p(1) = 1² – 2(1) + 5 

= 1 – 2 + 5 

∴ P(1) = 4 

p(y) = y² – 2y + 5 

∴ p(0) = 0² – 2(0) + 5 

= 0 – 0 + 5 

∴ p(0) = 5

p(y) = y² – 2y + 5 

∴ p(- 2) = (- 2)² – 2(- 2) + 5 

= 4 + 4 + 5 

∴ p(- 2) = 13 

iii. p(x) = x⁴ – 2x² – x 

∴ p(1) = (1)⁴ – 2(1)² – 1 

= 1 – 2 – 1 

∴ p(1) = - 2 

∴ p(x) = x⁴ – 2x² – x 

∴ p(0) = (0)⁴ – 2(0)² – 0 

= 0 – 0 – 0 

∴ p(0) = 0 

p(x) = x⁴ – 2x² – x 

∴ p(-2) = (-2)⁴ – 2(-2)² – (-2) 

= 16 – 2(4) + 2 

= 16 – 8 + 2 

∴ p(-2) = 10

i. 5x⁷ 

ii. x³⁵ – 1 

iii. 3x⁸ + 2x⁶ + x⁵

Answer is (A) f(3) = 0

f(x) = x³ + x² – 17x + 15

f(3) = (3)³ + (3)² – 17(3) + 15 

= 27 + 9 – 51 + 15 = 0

f(3) = 0

(C) undefined

(D) the degree of the polynomial √7 is 0

(a) \(\frac{-5}{3}x+12\)

The function f(x) is not known. Here, 

a = 3, b = – 6 

A = 7, B = 22                  [Refer to Key Fact 8] 

∴ Required remainder = \(\frac{A-B}{a-b}x+\frac{Ba-Ab}{a-b}\)

= \(\frac{7-22}{3-(-6)}x+\frac{22\times3-7\times(-6)}{3-(-6)}\) = \(\frac{-5}{3}x+12.\)

i. x⁴ + 16 

Index form = x⁴ + 0x³ + 0x² + 0x + 16 

∴ Coefficient form of the polynomial = (1,0,0,0,16) 

ii. m⁵ + 2m² + 3m + 15 

Index form = m⁵ + 0m⁴ + 0m³ + 2m² + 3m + 15 

∴ Coefficient form of the polynomial = (1, 0, 0, 2, 3, 15)

i. √5 = √5 x° 

∴ Degree of the polynomial = 0 

ii. x° 

∴ Degree of the polynomial = 0 

iii. x²

∴ Degree of the polynomial = 2 

iv. √2m¹⁰ – 7 

Here, the highest power of m is 10. 

∴ Degree of the polynomial = 10 

v. 2p – √7 

Here, the highest power of p is 1. 

∴ Degree of the polynomial = 1

vi. 7y – y³ + y⁵ 

Here, the highest power of y is 5. 

∴ Degree of the polynomial = 5 

vii. xyz + xy – z 

Here, the sum of the powers of x, y and z in the term xyz is 1 + 1 + 1 = 3, which is the highest sum of powers in the given polynomial. 

∴ Degree of the polynomial = 3 

viii. m³n⁷ – 3m⁵n + mn 

Here, the sum of the powers of m and n in the term m³ n⁷ is 3 + 7 = 10, which is the highest sum of powers in the given polynomial. 

∴ Degree of the polynomial = 10

(D) √2x² + (1/2)

p(x) = x³ – mx² + 10x – 20 x – 2 is a factor of x³ – mx² + 10x – 20.

∴By factor theorem,

Remainder = p(2) = 0

 p(x) = x³ – mx² + 10x – 20 

∴ p(2) = (2)³– m(2)² + 10(2) – 20 

∴ 0 = 8 – 4m + 20 – 20 

∴ 0 = 8 – 4m 

∴ 4m = 8 

∴ m = 2

p(x) = nx² – 5x + m 

(x – 2) is a factor of nx² – 5x + m. 

∴ By factor theorem, 

P(2) = 0 

∴ p(x) = nx² – 5x + m 

∴ p(2) = n(2)²– 5(2) + m 

∴ 0 = n(4) – 10 + m 

∴ 4n – 10 + m = 0 …(i) 

Also, ( x = \(\frac1{2} \)) is a factor of nx² – 5x + m.

∴ By factor theorem,

p( \(\frac1{2} \)) = 0 

p(x) = nx²– 5x + m 

∴ p(\(\frac1{2} \) ) = n(\(\frac1{2} \))² – 5\(\frac1{2} \) + m 

0 = \(\frac{n}{4} \)–\(\frac{5}{2} \) + m 

∴ 0 = n- 10 +4m … [Multiplying both sides by 4] 

∴ n = 10 – 4m ……(ii)

 Substituting n = 10 – 4m in equation (i)

 4(10 – 4m) – 10 + m = 0 

∴ 40 – 16m – 10 + m = 0 

∴ -15m+ 30 = 0 

∴ -15m = -30 

∴ m = 2 Substituting m = 2 in equation (ii), 

n = 10 – 4(2)

 = 10 – 8 

∴ n = 2 

∴ m = n = 2

i. (5x²– 2y + 9) – (3x²+ 5y – 7) 

= 5x² – 2y+ 9 – 3x² – 5y + 1 

= 5x² – 3x² – 2y – 5y + 9 + 7 

= 2x² – 1y + 16 

ii. (2x² + 3x + 5) – (x² – 2x + 3)

 = 2x² + 3x + 5 – x² + 2x – 3 

= 2x² – x² + 3x + 2x + 5 – 3 

= x²+ 5x + 2

i. 5 + 3x⁴ 

Here, the highest power of x is 4. 

∴ Degree of the polynomial = 4 

ii. 7 = 7x° 

∴ Degree of the polynomial = 0 

iii. ax⁷ + bx⁹ 

Here, the highest power of x is 9. 

∴ Degree of the polynomial = 9

(d) 2

Let f(x) = x³ + 2x² – 5ax – 7

∴ R₁ = f(–1) = (–1)³ + 2. (–1)² – 5.a. (–1) –7 

= –1 + 2 + 5a –7 = 5a – 6 

g(x) = x³ + ax² – 12x + 6 

R₂ = g(2) = (2)³ + a.(2)² – 12.(2) + 6 

= 8 + 4a – 24 + 6 = 4a – 10 

Given, 2R₁ + R₂ = 6 ⇒ 2(5a – 6) + (4a – 10) = 6 

⇒ 10a – 12 + 4a – 10 = 6 

⇒ 14a – 22 = 6 ⇒ 14a = 28 ⇒ a = 2.

i. x² – 9x + √3 -(- 19x + √3 + 7x²) 

= x² – 9x + √3 + 19x – √ 3 – 7x² 

= x² – 7x² – 9x + 19x + √3 – √3 

= – 6x² + 10x 

ii. (2ab² + 3a²b – 4ab) – (3ab – 8ab² + 2a²b) 

= 2ab² + 3a²b – 4ab – 3ab + 8ab² – 2a²b

= 2ab² + 8ab² + 3a²b – 2a²b – 4ab – 3ab 

= 10ab² + a²b – 7ab

Linear polynomials: ii, iii 

Quadratic polynomials: i, v 

Cubic polynomials: iv, vi

(A) a linear polynomials is x + 5

Let the required polynomial be A. 

∴ (x² + 13x + 7) – A = 3x² + 5x – 4 

∴ A = (x² + 13x + 7) – (3x² + 5x – 4) 

= x² + 13x + 7 – 3x² – 5x + 4 

= x² – 3x² + 13x – 5x + 7+4 

= -2x²+ 8x + 11 

∴ – 2x² + 8x + 11 must be subtracted from x² + 13x + 7 to get 3x² + 5x – 4

i. (x³ – 2x² – 9) + (5x³ + 2x + 9) 

= x³ – 2x² – 9 + 5x³ + 2x + 9 

= x³ + 5x³ – 2x² + 2x – 9 + 9 

= 6x³ – 2x² + 2x

ii. (-7m⁴ + 5m³ + √2 ) + (5m⁴ – 3m³ + 2m² + 3m – 6) 

= -7m⁴ + 5m³ + √2 + 5m⁴ – 3m³ + 2m² + 3m – 6 

= -7m⁴ + 5m⁴ + 5m³ – 3m³ + 2m² + 3m +√2 – 6 

= -2m⁴ + 2m³ + 2m² + 3m + √2 – 6

iii. (2y² + 7y + 5) + (3y + 9) + (3y² – 4y – 3) 

= 2y² + 7y + 5 + 3y + 9 + 3y² – 4y – 3 

= 2y² + 3y² + 7y + 3y – 4y + 5 + 9 – 3 

= 5y² + 6y + 11

i. x³ – 2 = x³ + 0x² + 0x – 2 

∴ Coefficient form of the given polynomial = (1, 0, 0, -2) 

ii. 5y = 5y + 0 

∴ Coefficient form of the given polynomial = (5,0)

iii. 2m⁴ – 3m² + 7 

= 2m⁴ + 0m³ – 3m² + 0m + 7 

∴ Coefficient form of the given polynomial = (2, 0, -3, 0, 7) 

iv. – (2/3)

∴Coefficient form of the given polynomial = - (2/3)

i. m³ + 5m + 3 

ii. y⁵ + 2y⁴ + 3y³ – y² – 7y – (1/2)

p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7.

Divisor = x + 1

x + 1 = 0

x = -1

So, substituting the value of x = – 1 in p(x), we get,

p(-1) = (-1)⁴ – 2(-1)³ + 3(-1)² – a(-1) + 3a – 7.

19 = 1 + 2 + 3 + a + 3a – 7

19 = 6 – 7 + 4a

4a – 1 = 19

4a = 20

a = 5

Since, a = 5.

We get the polynomial,

p(x) = x⁴ – 2x³ + 3x² – (5)x + 3(5) – 7

p(x) = x⁴ – 2x³ + 3x² – 5x + 15 – 7

p(x) = x⁴ – 2x³ + 3x² – 5x + 8

As per the question,

When the polynomial obtained is divided by (x + 2),

We get,

x + 2 = 0

x = – 2

So, substituting the value of x = – 2 in p(x), we get,

p(-2) = (-2)⁴ – 2(-2)³ + 3(-2)² – 5(-2) + 8

⇒ p(-2) = 16 + 16  + 12 + 10 + 8

⇒ p(-2) = 62

Therefore, the remainder = 62.

f(- 1) = (- 1)³ – (- 1)² – (- 1) + 1 

= -1 – 1 + 1 + 1 = 0 

∴ (x + 1) is a factor.

i. (5x² – 2y + 9) - (3x² + 5y – 7)

= 5x² – 2y + 9 – 3x² – 5y + 1 

= 5x² – 3x² – 2y – 5y + 9 + 7 

= 2x² – 1y + 16

ii. (2x² + 3x + 5) - (x² – 2x + 3)

= 2x² + 3x + 5 – x² + 2x – 3 

= 2x² – x² + 3x + 2x + 5 – 3 

= x² + 5x + 2

(A) 4

p(-1) = 2(-1)³ + 2(-1) 

= -2 – 2 = -4

p(x) = nx² – 5x + m 

(x – 2) is a factor of nx² – 5x + m. 

∴ By factor theorem, 

P(2) = 0 

∴ p(x) = nx² – 5x + m 

∴ p(2) = n(2)² – 5(2) + m 

∴ 0 = n(4) – 10 + m 

∴ 4n – 10 + m = 0 …(i) 

Also, (x = 1/2) is a factor of nx² – 5x + m. 

∴ By factor theorem, p(1/2) = 0 

p(x) = nx² – 5x + m 

∴ p(1/2) = n(1/2)² – 5 + m 

0 = (n/4) – (5/2) + m 

∴ 0 = n - 10 + 4m … [Multiplying both sides by 4] 

∴ n = 10 – 4m ……(ii) 

Substituting n = 10 – 4m in equation (i), 

4(10 – 4m) – 10 + m = 0 

∴ 40 – 16m – 10 + m = 0 

∴ -15m + 30 = 0 

∴ -15m = -30 

∴ m = 2 

Substituting m = 2 in equation (ii), 

n = 10 – 4(2) 

= 10 – 8 

∴ n = 2 

∴ m = n = 2

p(y) = y³ – 5y² + 7y + m

 Divisor = y + 2 

∴ take y = – 2

 ∴ By remainder theorem, 

Remainder = p(- 2) = 50

 P(y) = y³ – 5y ²+ 7y + m 

∴ P(-2) = (- 2)³ – 5(- 2)² + 7(- 2) + m 

∴ 50 = -8 – 5(4) – 14 + m 

∴ 50 = -8 – 20 – 14 + m 

∴ 50 = – 42 + m 

∴ m = 50 + 42 

∴ m = 92

p(x) = x³ – mx² + 10x – 20 x – 2 is a factor of x³ – mx² + 0x – 20. 

∴By factor theorem, 

Remainder = p(2) = 0 

p(x) = x³ – mx² + 10x – 20 

∴ p(2) = (2)³ – m(2)² + 10(2) – 20 

∴ 0 = 8 – 4m + 20 – 20 

∴ 0 = 8 – 4m 

∴ 4m = 8 

∴ m = 2

(A) 5

Here, degree of first polynomial = 2 and degree of second polynomial 3 

∴ Degree of polynomial obtained by multiplication = 2 + 3 = 5

p(x) = x³¹ + 31 

Divisor = x + 1 

∴ take x = – 1 

∴ By remainder theorem, 

Remainder = p(-1) 

p(x) = x³¹ + 31 … 

∴ p(-1) = (-1)³¹ + 31 

= -1 + 31 = 30 

∴ Remainder = 30

When polynomial bx² + x + 5 is divided by (x – 3), the remainder is m. 

∴ By remainder theorem,

Remainder = p(3) = m 

p(x) = bx² + x + 5 

∴ p(3) = b(3)² + 3 + 5 

∴ m = b(9) + 8 

m = 9b + 8 …(i) 

When polynomial bx³ – 2x + 5 is divided by x – 3 the remainder is n 

∴ remainder = p(3) = n 

p(x) = bx³ – 2x + 5 

∴ P(3) = b(3)³ – 2(3) + 5 

∴ n = b(27) – 6 + 5 

∴ n = 27b – 1 …(ii) 

Now, m – n = 0 …[Given] 

∴ m = n 

∴ 9b + 8 = 27b – 1 …[From (i) and (ii)] 

∴ 8 + 1 = 27b – 9b 

∴ 9 = 18b 

∴ b = 1/2

p(y) = y³ – 5y² + 7y + m 

Divisor = y + 2 

∴ take y = – 2 

∴ By remainder theorem, 

Remainder = p(- 2) = 50 

P(y) = y³ – 5y² + 7y + m 

∴ P(-2) = (- 2)³ – 5(- 2)² + 7(- 2) + m 

∴ 50 = -8 – 5(4) – 14 + m 

∴ 50 = -8 – 20 – 14 + m 

∴ 50 = – 42 + m 

∴ m = 50 + 42 

∴ m = 92

i. p(m) = m – 1 

Divisor = m – 1 

∴ take m = 1 

Remainder = p(1) 

p(m) = m²¹ – 1 

∴ P(1) = 1²¹ – 1 = 1 – 1 = 0 

∴ By factor theorem, m - 1 is a factor of m²¹ -1.

ii. p(m) = m²² – 1 

Divisor = m – 1 

∴ take m = 1 

Remainder = p(1) 

p(m) = m²² – 1 

∴ P(1) = 1²² – 1 = 1 – 1 = 0 

∴ By factor theorem, m -1 is a factor of m²² – 1.

(C) -3

p(1) = 0 

∴ 3(1)² + m(1) = 0 

∴ 3 + m = 0 

∴ m = -3

Here, p(x) = x³ + 4x – 5 

Substituting x = 1 in p(x), we get 

p(1) = (1)³ + 4(1) – 5 

= 1 + 4 – 5 

P(1) = 0 

∴ By remainder theorem, 

Remainder = 0 

∴ (x -1) is the factor of x³ + 4x – 5.

Let the required polynomial be A. 

∴ (4m + 2n + 3) + A = 6m + 3n + 10 

∴ A = 6m + 3n + 10 – (4m + 2n + 3) 

= 6m + 3n + 10 – 4m – 2n – 3 

= 6m – 4m + 3n – 2n + 10 – 3 

= 2m + n + 7 

∴ 2m + n + 7 must be added to 4m + 2n + 3 to get 6m + 3n + 10