Polynomials

(i) y + 1/y is not a polynomial because in this given equation the second term becomes 1 × y^-1 where -1 is not a whole number. So the given algebraic expression is not a polynomial.

(ii) 2-5√x is not a polynomial because again the power of x is 1/2 which is not a whole number.

(iii) The given algebraic expression is a polynomial in one variable because all the powers of x are whole numbers.

(iv) This expression is not polynomial because the power of the first term is -2 which is not a whole number.

(v) The number 10 is a polynomial because 10 can be represented as follows: 10 × x⁰ which is equal to 10. So it is a polynomial.

i. The coefficient of any variable is the constant with which the variable is multiplied with. So the coefficient of m³ is one.

ii. The coefficient of m⁰ = -3/2

The coefficient of m = 1

The coefficient of m³ = -√3

iii. The coefficient of m⁰ = -1

The coefficient of m¹ = 7

The coefficient of m² = -5

The coefficient of m³ = -2/3

i. A polynomial is said to be monomial if it contains only one term in its entire expression. So the polynomial is as follows:

7x⁷

ii. A polynomial is said to be binomial if it contains only two term in its entire expression. So the polynomial is as follows:

2x³⁵ + 7

iii. A polynomial is said to be trinomial if it contains only three term in its entire expression. So the polynomial is as follows:

2x⁸ + 7x + 3

i. The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients. This algebraic expression has zero degree.

ii. The degree of the given polynomial is zero.

iii. The degree of the given polynomial is two.

iv. The degree of the given polynomial is ten.

v. The degree of the given polynomial is one.

vi. The degree of the given polynomial is five.

vii. The degree of the given polynomial is three.

viii. The degree of the given polynomial is ten.

i. A polynomial is said to be quadratic if the highest power of the variable in the polynomial is two.

So this polynomial is quadratic.

ii. The given polynomial is linear as the highest power of the variable is one.

iii. This polynomial is a linear one as the variable has the highest degree as one.

iv. This polynomial is a cubic polynomial because the highest power of the variable is three.

v. A polynomial is said to be quadratic if the highest power of the variable in the polynomial is two.

So this polynomial is quadratic.

vi. This polynomial is a cubic polynomial because the highest power of the variable is three.

i. In the standard form of any polynomial it is necessary for the power of the variable to go in descending order for each term.

So the standard form of this polynomial is as follows:

m³ + 5m + 3

ii. In the standard form of any polynomial it is necessary for the power of the variable to go in descending order for each term that is first term should consist of variable with highest power, second term should contain the variable with second highest power and so on.

So the standard form of this polynomial is as follows:

y⁵ + 2y⁴ + 3y³ –y² -7y – 1/2

i. In the coefficient form of the polynomial, the coefficient of each term of the variable present or absent is written inside the simple brackets. So the coefficient form is as follows:

⇒ x³ -2 = x³ + 0x² + 0x – 2

∴ The given polynomial in coefficient form is:

(1, 0, 0, -2)

ii. In the coefficient form of the polynomial, the coefficient of each term of the variable present or absent is written inside the simple brackets. So the coefficient form is as follows:

⇒ 5y = 5.y + 0

Therefore, the given polynomial in coefficient form is:

(5, 0)

iii. In the coefficient form of the polynomial, the coefficient of each term of the variable present or absent is written inside the simple brackets. So, the coefficient form is as follows

2m⁴ – 3m² + 7 = 2.m⁴ + 0.m³ – 3.m² + 0.m + 7

Therefore, the given polynomial in coefficient form is:

(2, 0, -3, 0, 7)

iv. In the coefficient form of the polynomial, the coefficient of each term of the variable present or absent is written inside the simple brackets.

Now in this case all the powers of the variable are zero and only the constant is present.

So the coefficient form is as follows:

(-2/3)

i. The given representation is the coefficient form of the polynomial. So, the first coefficient denotes the highest power of the variable.

Writing in the index form, the polynomial is:

⇒ 1.x³ + 2x² + 0.x + 3

⇒ x³ + 2x² + 3

∴ Index form of polynomial is x³ + 2x² + 3

ii. The given representation is the coefficient form of the polynomial. So the first coefficient denotes the highest power of the variable and the power of last term is always zero.

= 5.x⁴ +0.x³ + 0.x² + 0.x – 1

= 5x⁴ – 1

Therefore, the index form of the polynomial is = 5x⁴ – 1

iii. The given representation is the coefficient form of the polynomial. So the first coefficient denotes the highest power of the variable and the power of last term is always zero.

= -2.x³ + 2.x² – 2.x + 2

= -2x³ + 2x² – 2x + 2

Therefore, the index form of the polynomial is = -2x³ + 2x² – 2x + 2

A polynomial is said to be quadratic if the highest power of the variable in the polynomial is two.

The polynomial is said to be linear if the variable has the highest degree as one.

The polynomial is said to be cubic if the variable has the highest degree as three.

A polynomial is said to be monomial if it contains only one term in its entire expression.

A polynomial is said to be binomial if it contains only two terms in its entire expression.

A polynomial is said to be trinomial if it contains only three terms in its entire expression.

i. Given:

Total number of trees in the village = a

Increase in the number of trees by = b

To find: Number of trees after x years

Number of trees in x years = (a + b) x

ii. Given:

Number of students in each row = y

Number of rows of Students = x

So total number of students = x × y

= xy

iii. For a number in tens place should be multiplied with ten and for the number in units place should be multiplied by one. So the number that can be formed as follows:

Since m is in tens place and n is in units place,

10 × m + 1 × n

10m + n

So the polynomial representing the two digit number is = 10m + n

i.

The result = 6x³ – 2x² + 2x

ii.

The result = 2m⁴ + 2m³ + 2m² + 3m – 6 +√2

iii.

The result = 5y² + 6y + 11

In the subtraction process, the sign of the subtrahend that is the second polynomial is inverted and then the operation is carried out.

The result = 6x² + 10x

ii. In the subtraction process, the sign of the subtrahend that is the second polynomial is inverted and then the operation is carried out.

The result = 10ab² – a²b – 7ab

The multiplication is as follows:

⇒ 2x × (x² – 2x – 1)

⇒ 2x.x² – 2x.2x – 2x.1

⇒ 2x²⁺¹ – 4x¹⁺¹ – 2x

⇒ 2x³ – 4x² – 2x

Therefore, the product = 2x³ – 4x² – 2x

“.” represents multiplication

ii. The multiplication is as follows:

⇒ (x⁵ – 1) × (x³ + 2x³ + 2)

⇒ x⁵.x³ + 2x³.x⁵ + 2.x⁵ – 1.x³ – 1.2x³ – 1.2

⇒ x⁵⁺³ + 2x^{3 + 5} +2x⁵ – x³ – 2x³ – 2

⇒ x⁸ + 2x⁸ + 5 + 2x⁵ – x³ – 2x³ – 2

⇒ 3x⁸ + 2x⁵ + 3x³ + 3

The product is = 3x⁸ + 2x⁵ + 3x³ + 3

“.” represents multiplication

iii. The multiplication is as follows:

⇒ (2y + 1) × (y² – 2y³ + 3y)

⇒ 2y.y² – 2y³.2y + 3y.2y + 1.y² – 1.2y³ + 1.3y

⇒ 2y²⁺¹ – 4y³⁺¹ + 6y¹⁺¹ + y² – 2y³ + 3y

⇒ 2y³ – 4y⁴ + 6y² + y² – 2y³ + 3y

⇒ -4y⁴ + 7y² + 3y

The product is = -4y⁴ + 7y² + 3y

“.” represents multiplication

i. The division is as follows:

x³-64 = [(x² + 4x + 16) × (x-4)] + 0

x³-64 = (x² + 4x + 16) × (x-4)

ii. The division is as follows:

5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = [(5x³ + 9x² + 6x + 8) × (x²-x)] + 8x +2

Quotient = 5x3 + 9x² + 6x + 8

Remainder = 8x + 2

Given:

Length of the farm, l = (2a² + 3b²) m

Breadth of the farm, b = (a² + b²) m

Side of the square plot, s = (a² - b²) m

To find: Area of remaining farm

Explain:

Area of Remaining Farm = Area of Rectangle – Area of Square

Area of Rectangle = l × b

= (2a² + 3b²) × (a² + b²)

= 2a².a² + 2a².b² + 3b².a² + 3b².b²

= 2a⁴ + 5a²b² + 3b⁴

Area of Square = s × s

= (a² - b²) × (a² - b²)

= a².a² – a².b² – a².b² + b².b²

= a⁴ – 2a²b² + b⁴

Area of Remaining Part = (2a⁴ + 5a²b² + 3b⁴) – (a⁴ – 2a²b² + b⁴)

= 2a⁴ + 5a²b² + 3b⁴ - a⁴ + 2a²b² - b⁴

= a⁴ + 7a²b² + 2b⁴

Therefore, the area of remaining portion is = a⁴ + 7a²b² + 2b⁴

The linear division method is as follows:

Representing the polynomial in first polynomial in coefficient form:

⟹ 2.m² – 3.m + 10

⟹ (2, – 3, 10)

Bring the first term as it is. Then multiply 5 with the value written in the bottom result area. Continue this until done.

Quotient = 2m + 7

Remainder = 45

The linear division is as follows:

Representing the polynomial in first polynomial in coefficient form:

⟹ 1.x⁴ + 2.x³ + 3x² + 4x + 5

⟹ (1 2 3 4 5)

So the final answer is written in the following form:

Quotient = x³ + 3x – 2

Remainder = 9

The linear division method is as follows:

Representing the polynomial in first polynomial in coefficient form:

⟹ 1.y³ + 0.y² + 0.y – 216

⟹ (1, 0, 0, -216)

So the final answer is written in the following form:

Quotient = y² + 6y + 36

Remainder = 0

The linear division is as follows:

Representing the polynomial in first polynomial in coefficient form:

⟹ 2.x⁴ + 3.x³ – 2.x² + 4.x

⟹ (2, 3, -2, 4, 0)

So, the final answer is written in the following form:

Quotient = 2x³ – 3x² + 7x – 17

Remainder = 51

The linear division is as follows:

Representing the polynomial in first polynomial in coefficient form:

⟹ 1.x⁴ + 0.x³ – 3.x² + 0.x – 8

⟹ (1, 0, -3, 0, -8)

So the final answer is written in the following form:

Quotient = x³ – 4x² + 13x - 52

Remainder = 200

The linear division is as follows:

Representing the polynomial in first polynomial in coefficient form:

⟹ 1.y³ – 3.y² + 5.y – 1

⟹ (1, -3, 5, -1)

So, the final answer is written in the following form:

Quotient = y² – 2y + 3

Remainder = 2

Put x = 0 throughout the polynomial

⇒ 0² – 5 × 0 + 5

x (0) = 5

Put x = 3√2 throughout the polynomial

⇒ (3√2)² – (3√2 × 3√2) + 1

⇒ (9 × 2) – (9 × 2) + 1

⇒ 18 – 18 + 1

p(3√2) = 1

Put x = a throughout the polynomial

(a)³ + 2 × (a)² –a + 10

p (a) = a³ + 2a² – a + 10

Put x = -a throughout the polynomial

(-a)³ + 2 × (-a)² –(-a) + 10

p (-a) = -a³ + 2a² + a + 10

p(a) + p(-a):

a³ + 2a² – a + 10 + -a³ + 2a² + a + 10

∴ p(a) + p(-a) = 4a² + 20

Put x = 2 throughout the polynomial

= 2 × (2)² – 6 × (2)² – 5 × 2 + 7

= 8 – 24 – 10 + 7

= -19

∴ p(2) = -19

The given polynomial is as follows:

2x – 2x³ + 7

(i) Put x = 3 in the given polynomial

= 2 × 3 – 2 × (3)³ + 7

= 6 – 2 × 27 + 7

= 6 -54 + 7

= -41

(ii) Put x = -1 in the given polynomial

= 2 × (-1) – 2 × (-1)³ + 7

= -2 + 2 + 7

= 7

(iii) Put x = 0 in the given polynomial

= 2 × 0 – 2 × (0)³ + 7

= 0 + 0 + 7

= 7

To find: p (1)

Put x = 1 in the given polynomial

p (1) = (1)³

p (1) = 1

To find: p (0)

Put x = 0 in the given polynomial

p (0) = 0

To find: p (-2)

Put x = -2 in the given polynomial

p (-2) = (-2)³

p (-2) = -8

To find: p (1)

Put x = 1 in the given polynomial

p (1) = (1)² – 2 × 1 + 5

= 1 – 2 + 5

= -1 + 5

= 4

To find: p (0)

Put x = 0 in the given polynomial

p (0) = 0 – 0 + 5

= 5

To find: p (-2)

Put x = -2 in the given polynomial

p (-2) = (-2)² – 2 × (-2) + 5

= 4 + 4 + 5

p (-2) = 13

To find: p (1)

Put x = 1 in the given polynomial

p (1) = (1)⁴ – 2 × (1)² – 1

= 1 – 2 – 1

p (1) = -2

To find: p (0)

Put x = 0 in the given polynomial

p (0) = 0 – 0 – 1

p (0) = -1

To find: p (-2)

Put x = -2 in the given polynomial

p (-2) = (-2)⁴ – 2 × (-2)² – 1

= 16 – 2 × 4 – 1

= 16 -8 -1

= 7

Put m = 2 throughout the polynomial

(2)³ + 2 × 2 + a = 12

8 + 4 + a = 12

12 + a = 12

a = 0

Put x = -1 throughout the polynomial

m × (-1)² – 2 × (-1) + 3 = 7

m + 2 + 3 = 7

m + 5 = 7

m = 7-5

m = 2

The coefficient form of the first polynomial is as follows:

⟹ 1.x² – 7.x + 9

⟹ (1, -7, 9)

Therefore, the remainder is = 17

The coefficient form of the first polynomial is as follows:

⟹ 2.x³ – 2.x² + a.x – a

(2, -2, a, -a)

The remainder is = 2a³-a² –a

The coefficient form of the first polynomial is as follows:

⟹ 54.m³ + 18.m² – 27.m + 5

(54, 18, -27, 5)

The last term in the synthetic division is the remainder of the given division method.

The remainder is =1544.

Put y = -2 throughout the given polynomial

(-2)³ – 5 × (-2)² + 7 × (-2) + m = 50

-8 – 20 -14 + m = 50

-28 + 14 + m = 50

-14 + m = 50

m = 50 + 14

m = 64

The coefficient form of the polynomial:

⟹ 1.x² + 2.x – 3

(1, 2, -3)

Since the last term is zero which indicates that the polynomial on dividing by x + 3 leaves no remainder and hence x + 3 is the factor of x² + 2x – 3

Put x = 2 throughout the polynomial and equate it to zero as x -2 is the factor of the given polynomial.

(2)³ – m × (2)² + 10 × 2 – 20 = 0

8 – 4m + 20 – 20 = 0

4m = 8

m = 2

So for x-2 to be the factor of the given polynomial, the value of m = 2.

The coefficient form of p(x) is:

p(x) = 1.x³ – 1.x² – 1.x – 1

(2, -1, 0, -45)

Since the last term is not term is not zero, q(x) is not the factor of p(x).

The coefficient form of p(x) is:

p(x) = 2.x³ – 1.x² + 0.x – 45

(2, -1, 0, -45)

Since the last term is not term is not zero, q(x) is not the factor of p(x).

Put x= -1 throughout the polynomial

x(-1) = (-1)³¹ + 31

= -1 + 31

= 30

So the remainder is 30.

Put m = 1 in the first polynomial

m (1) = (1)²¹ – 1

= 1 – 1

= 0

So m-1 is the factor of m²¹ -1

Put m = 1 in the second polynomial

m (1) = (1)²² – 1

= 1 – 1

= 0

So m-1 is the factor of m²² -1

Put x = 2 in the given polynomial:

n× 4 – 5× 2 + m = 0

4n + m = 10 ……………………………. (i)

Put x = 1/2 in the given polynomial

n × 1/4 - 5 × 1/2 + m = 0

n + 4m – 10 = 0

n + 4m = 10 ……………………………….. (ii)

Since the RHS of both the equations is same,

4n + m = n + 4m

3n = 3m

n = m

Put n = m in the equation (i)

4m + m = 10

5m = 10

m = 2 = n

Put x = 2 in the polynomial

p(2) = 2 + 5× 2

= 2 + 10

= 12

Put x = -2 in the polynomial

p(-2) = 2 + (5× -2)

= 2 – 10

= -8

Put x = 1 in the given polynomial

P(1) = 2 + 5× 1

= 2 + 5

= 7

p(2) + p(-2) – p(1) = 12 – 8 - 7

= -3

Put x = 5√3 in the given polynomial

p(5√3) = 2×(5√3)² – 5√3 × 5√3 + 5

= 2 × 25 × 3 – 25 × 3 + 5

= 150 – 75 + 5

p (5√3) = 80

2x² + x – 1

⟹ 2x² + 2x - x – 1

⟹ 2x (x + 1) – 1 (x + 1)

⟹ (x + 1) (2x – 1)

Therefore, the factors of the given polynomial = (x + 1) (2x – 1)

2m² + 5m – 3

⟹ 2m² + 6m - m – 3

⟹ 2m (x + 3) – 1 (m + 3)

⟹ (m+ 3) (2m – 1)

Therefore, the factors of the given polynomial = (m+ 3) (2m – 1)

12x² + 61x + 77

⟹ 12x² + 28x + 33x + 77

⟹ 4x (3x + 7) + 11 (3x + 7)

⟹ (4x + 11) (3x + 7)

Therefore, the factors of the given polynomial = (4x + 11) (3x + 7)

3y² – 2y – 1

⟹ 3y² – 3y + y – 1

⟹ 3y (y - 1) + 1 (y - 1)

⟹ (3y + 1) (y – 1)

Therefore, the factors of the given polynomial = (3y + 1) (y – 1)

√3x² + 4x + √3

⟹ √3x² + 3x + x + √3

⟹ √3x (x + √3) + 1 (x + √3)

⟹ (x + √3) (√3x + 1)

Therefore, the factors of the given polynomial = (x + √3) (√3x + 1)

1/2 x² - 3x + 1

⟹ 1/2x² - 2x - x + 4

⟹ 1/2x (x - 4) – 1 (x - 4)

⟹ (x - 4) (1/2x – 1)

Therefore, the factors of the given polynomial = (x - 4) (1/2x – 1)

Put (x² – x) = a

⟹ a² – 8a + 12

⟹ a² – 2a – 6a + 12

⟹ a (a-2) – 6(a-2)

⟹ (a-6) × (a-2)

⟹ but a = (x² – x)

⟹ ((x² – x)-6) × ((x² – x) – 2

⟹ (x² – x -6) × (x² – x -2)

⟹ (x² –3x + 2x – 6) × (x² – 2x + x -2)

⟹ (x (x-3) + 2(x – 3)) × (x(x-2) + 1(x-2))

⟹ (x + 2)(x-3)(x-2)(x+1)

Therefore, the factorized form = (x + 2)(x-3)(x-2)(x+1)

(x-5)² – 5(x-5) -24

Put (x – 5) = a

⟹ a² – 5a - 24

⟹ a² – 8a + 3a -24

⟹ a (a-8) + 3(a-8)

⟹ (a-8) × (a+3)

⟹ But a = (x – 5)

⟹ (x – 5 - 8) × (x-5 +3)

⟹ (x – 13) × (x-2)

Therefore, the factorized form of the polynomial = (x – 13) × (x-2)

(x² – 6x) ² – 8(x² – 6x + 8) – 64

⟹ (x² – 6x) ² – 8(x² -6x) - 64 – 64

⟹ (x² – 6x) ² – 8(x² -6x) – 128

Put (x² -6x) = a

⟹ (a) ² – 8(a) – 128

⟹ a² – 8a – 128

⟹ a² – 16a + 8a - 128

⟹ a (a-16) + 8(a – 16)

⟹ (a + 8) × (a-16)

⟹ But a = (x² -6x)

⟹ ((x² -6x) + 8) × ((x² -6x) – 16)

⟹ (x² -6x + 8) × (x² -6x – 16)

⟹ (x² -4x – 2x + 8) × (x² -8x + 2x – 16)

⟹ (x(x-4) – 2(x-4)) × (x(x-8) + 2(x-8)

⟹ (x-2)(x-4)(x-8)(x+2)

Therefore, the factorized form = (x-2) (x-4) (x-8) (x+2)

(x² – 2x + 3) (x² – 2x + 5) – 35

Put (x² – 2x) = a

⟹ (a + 3) (a + 5) – 35

⟹ (a² + 5a + 3a + 15) – 35

⟹ a² + 8a + 15 – 35

⟹ a² + 8a – 20

⟹ a² + 10a – 2a – 20

⟹ a (a + 10) – 2 (a + 10)

⟹ (a – 2) (a + 10)

⟹ But a = (x² – 2x)

⟹ (x² – 2x) + 10) ((x² – 2x) – 2)

⟹ (x² – 2x + 10) (x² – 2x – 2)

Therefore, the factorized form = (x² – 2x + 10) (x² – 2x – 2)

(y+2) (y+3) (y-3) (y + 8) + 56

⟹ (y² + 3y + 2y + 6) (y² + 8y - 3y - 24) + 5

⟹ (y² + 5y + 6) (y² + 5y - 24) + 56

Put (y² + 5y) = a

⟹ (a + 6) (a – 24) + 56

⟹ a² -24a + 6a – 144 + 56

⟹ a² – 18a – 88

⟹ a² -22a + 4a – 88

⟹ a (a-22) + 4 (a-22)

⟹ (a+4) (a-22)

⟹ But a = (y² + 5y)

⟹ ((y² + 5y) + 4) ((y² + 5y)-22)

⟹ (y² + 5y + 4) (y² + 5y -22)

⟹ (y² + 4y + y + 4) (y² + 5y -22)

⟹ (y (y+4) + 1(y+4)) (y² + 5y -22)

⟹ (y+1) (y+ 4) (y² + 5y -22)

Therefore, the factorized form = (y+1) (y+ 4) (y² + 5y -22)

Put (y² + 5y) = a

⟹ a (a -2) – 24

⟹ a² -2a – 24

⟹ a² -6a + 4a – 24

⟹ a (a – 6) + 4 (a – 6)

⟹ (a + 4) (a – 6)

⟹ But a = (y² + 5y)

⟹ ((y² + 5y) + 4) ((y² + 5y) – 6)

⟹ (y² + 5y + 4) (y² + 5y – 6)

⟹ (y² + 4y + y + 4) (y² + 6y - y – 6)

⟹ (y (y + 4) + 1 (y + 4)) (y (y + 6) – 1 (y + 6))

⟹ (y + 4) (y + 1) (y + 6) (y – 1)

Therefore, the factorized form = (y + 4) (y + 1) (y + 6) (y – 1)

(x – 3) (x- 5) (x – 4)² – 6

⟹ (x² – 8x + 15) (x² – 8x + 16) – 6

⟹ Put (x² – 8x) = a

⟹ (a + 15) (a + 16) – 6

⟹ a² + 15a + 16a + 240 – 6

⟹ a² + 31a + 234

⟹ a² + 13a + 18a + 234

⟹ a (a + 13) + 18 (a + 13)

⟹ (a + 18) (a + 13)

⟹ But a = (x² – 8x)

⟹ ((x² – 8x) + 18) ((x² – 8x) + 13)

⟹ (x² – 8x + 18) (x² – 8x + 13)

Therefore, the factorized form =(x² – 8x + 18) (x² – 8x + 13)

Option (A) is not a polynomial. Hence, incorrect option.

(B) is x^√2 – 3x is not a polynomial because it does not have natural number power. Hence, incorrect.

(C) x^-2 + 7 is not a polynomial. Because it has negative power. Hence, incorrect.

(D) is the quadratic polynomial.

The option D is the only valid polynomial because in this polynomial only the power of the variable is a whole number.

The degree of this polynomial is zero as √7x⁰.

Option (A) is not the correct polynomial as a degree of polynomial cannot be fractional.

Option (B) is also not the correct answer as √7 is not multiplied with a variable with degree 5.

Option (C) is also not the correct answer as √7 is not multiplied with a variable with degree.

Option (D) is the correct answer as x⁰ is multiplied with √7

The degree of this polynomial is zero as √7x⁰.

Option (A) is not the degree of zero polynomial as zero polynomial does not have any variable and so its degree cannot be zero.

Option (B) is not the degree of zero polynomial as zero polynomial does not have any variable and so its degree cannot be one.

Option (C) is the degree of zero polynomial as zero polynomial does not have any variable and so its degree is not defined.

Option (D) is not the degree of zero polynomial as zero polynomial does not have any variable and so its degree is not defined and hence the degree cannot be any real number.

So the correct answer is C.

The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients.

Therefore, the degree of this polynomial is 3

Option (B) is not correct as the degree does not match to the highest degree of the variable.

Option (C) is not correct as the degree does not match to the highest degree of the variable.

Option (D) is not correct as the degree does not match to the highest degree of the variable.

So option A is correct.

The coefficient form is as follows:

x³ – 1 = x³ + 0x² + 0x – 1

The coefficient form is:

(1, 0, 0, -1)

Option (A) does not match to our solution.

Option (B) does not match to our solution.

Option (C) does match to our solution.

Option (D) also does not match to our solution.

So option (C) is correct answer.

Put p =7√7 throughout the polynomial

= (7√7)² – (7√7 × 7√7) + 3

=49 × 7 – (49 × 7) + 3

=343 – 343 + 3

=3

Option (A) does match to our solution.

Option (B) does not match to our solution.

Option (C) does not match to our solution.

Option (D) also does not match to our solution.

So option (A) is correct answer.

Put x = -1 throughout the polynomial

= 2 × (-1)3 + (2 × -1)

= -2 -2

= -4

Option (A) does not match to our solution.

Option (B) does not match to our solution.

Option (C) does not match to our solution.

Option (D) does match to our solution.

So option (D) is correct answer.

Put x = 1 in the given polynomial

3 × 1 + m = 0

m = -3

Option (A) does not match to our solution.

Option (B) does not match to our solution.

Option (C) does match to our solution.

Option (D) also does not match to our solution.

So option (C) is correct answer.

The multiplication is as follows:

= (x² -3) × (2x – 7x³ + 4)

= x².2x – 7x³.x² + 4x² -3.2x + 3.7x³ -3.4

= 2.x²⁺¹ – 7x³⁺² + 4x² -6x + 21x³-12

= 2x³ – 7x⁵ + 4x² – 6x + 21x³ – 12

“.” represents Multiplication

Therefore the degree of the product is 5.

Option (B) is not correct as the degree does not match to the highest degree of the variable.

Option (C) is not correct as the degree does not match to the highest degree of the variable.

Option (D) is not correct as the degree does not match to the highest degree of the variable.

So option A is correct.

A polynomial is said to be linear if the highest power of the variable is one. So among the options x + 5 only has highest power as one.

Option (B) is not correct as the highest degree of the variable is 2 and the equation is quadratic.

Option (C) is not correct as the highest degree of the variable is 3 and the equation is cubic.

Option (D) is not correct as the highest degree of the variable is 4.

So option A is correct.

i. The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients.
Here,

3x⁴ + 0.x³ + 0.x² + 0.x + 5

Therefore, the degree of this polynomial is 4

ii. The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients.

Here,

7x⁰.

Therefore, the degree of this polynomial is 0

iii. The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients.

Here,

b.x⁹ + 0.x⁸ + a.x⁷ + 0.x⁶ + 0.x⁵ + 0.x⁴ + 0.x³ + 0.x² + 0.x¹ + 1

Therefore, the degree of this polynomial is 9.

i. In standard form the terms are arranged from highest power of the variable to the lowest power.

So the standard form is as follows:

7x⁴ – x³ + 4x² – x + 9.

ii. In standard form the terms are arranged from highest power of the variable to the lowest power.

So the standard form is as follows:

5p⁴ + 2p³ + 10p² + p – 8.

i. In the coefficient form of the polynomial, the coefficients of the terms are written from highest power of the variable to lowest power.

⟹ x⁴ + 16 = 1.x⁴ + 0.x³ + 0.x² + 0.x + 16

So the coefficient form is as follows:

(1, 0, 0, 0, 16).

ii. In the coefficient form of the polynomial, the coefficients of the terms are written from highest power of the variable to lowest power.

⟹ m⁵ + 2m² + 3m + 15 = 1.m⁵ + 0.m⁴ +0.m³ + 2.m² + 3.m + 15

So the coefficient form is as follows:

(1, 0, 0, 2, 3, 15).

i. The given representation is the coefficient form of the polynomial. So, the first coefficient denotes the highest power of the variable and the power of last term is always zero.

So the index form is as follows:

3x⁴ – 2x³ + 0x² + 7x + 18

= 3x⁴ – 2x³ + 7x + 18

ii. The given representation is the coefficient form of the polynomial. So, the first coefficient denotes the highest power of the variable and the power of last term is always zero.

So the index form is as follows:

6x³ + x² + 0x + 7

=6x³ + x² + 7

iii. The given representation is the coefficient form of the polynomial. So, the first coefficient denotes the highest power of the variable and the power of last term is always zero.

So the index form is as follows:

4x³ + 5x² -3x + 0

= 4x³ + 5x² -3x

The addition is as follows:

The result is as follows: 10x⁴ + 13x³ + 9x² + 7x + 12

The addition is as follows:

The result is as follows: p³q + 4p²q + 4pq + 7

The subtraction is as follows:

The result is as follows: 2x² – 7y + 16

The subtraction is as follows:

The result is as follows: x² + 5x + 2

The multiplication is as follows:

= (m³ – 2m + 3) × (m⁴ -2m² + 3m + 2)

= m³ × m⁴ - 2m² × m³ + 3m × m³ + 2 × m³ - 2m × m⁴ + 4m² × m – 3 × 2m - 4m + 3m⁴ – 6m² + 6m + 12

= m⁷ – 2m⁵ + 3m⁴ + 2m³ – 2m⁵ + 4m³ – 6m - 4m + 3m⁴ - 6m² + 6m + 12

= m⁷ - 4m⁵ + 6m⁴ + 6m³ – 16m + 12

The product = m⁷ - 4m⁵ + 6m⁴ + 6m³ – 16m + 12

The multiplication is as follows:

= (5m³ – 2) × (m² –m +3)

= 5m³.m² – 5m³.m + 3.5m³ – 2m² + 2m – 6

= 5m⁵ – 5m⁴ + 15m³ – 2m² + 2m -6

The product is = 5m⁵ – 5m⁴ + 15m³ – 2m² + 2m -6

The coefficient forms the first polynomial is as follows:

⟹ 3x³ – 8x² + x + 7

⟹ (3, -8, 1, 7)

The quotient is as follows = 3x³ + x² + 4x

The remainder is = 19

Put x = -3 in the given polynomial and equate it to zero as x+3 is to be a factor of the polynomial.

(-3)³ – 2 × m × (-3) + 21 = 0

-27 + 6m + 21 = 0

6m -6 = 0

6m = 6

m = 1

So for x+3 to be a factor of the given polynomial, the value of m = 1.

Older Population of Kovad, Pk = 5x² – 3y²

Older Population of Varud, Pv = 7y² + 2xy

Older Population of Chikhali, Pc = 9x² + 4xy

Number of people migrated from Kovad, Mk = x² + xy – y²

Number of people migrated from Varud, Mv = 5xy

Number of people migrated from Chikhali, Mc = 3x² + xy

Present Population of Kovad = Pk – Mk

= (5x² – 3y²) – (x² + xy – y²)

= 4x² – 2y² – xy

Present population of Varud = Pv – Mv

= (7y² + 2xy) – 5xy

= 7y² – 3xy

Present Population of Chikhali = Pc – Mc

= (9x² + 4xy) – (3x² + xy)

= 6x² + 3xy

Put x = 3 in the first polynomial:

b× (3)² + 3 + 5 = m

9b + 8 = m ………………………………….. (i)

Put x = 3 in the second polynomial

b× (3)³ -2× 3 + 5 = n

27b -6 + 5 = n

27b -1 = n …………………………………. (ii)

Equation (i) – (ii)

m –n = 9b + 8 – (27b -1)

but m-n = 0

9b + 8 -27b + 1 = 0

-18b + 9 = 0

18b = 9

b = 9/18

b = 1/2

(8m² + 3m – 6) - (9m-7) + (3m² – 2m + 4)

= 8m² + 3m – 6 – 9m + 7 + 3m² – 2m + 4

= 11m² – 8m + 5

Let the polynomial to be subtracted be ax² + bx + c

3x² + 5x – 4 = (x² + 13x + 7) - (ax² + bx + c)

(ax² + bx + c) = (x² + 13x + 7) – (3x² + 5x – 4)

= (x² – 3x²) + (13x – 5x) + (7 + 4)

= -2x² – 8x + 11

Comparing with coefficients we get:

a= -2

b = -8

c = 11

Therefore, the polynomial is = -2x² -8x + 11

Let the polynomial to be added be am + bn + c

6m + 3n + 10 = (4m + 2n + 3) + (am + bn + c)

(am + bn + c) = (6m + 3n + 10) - (4m + 2n + 3)

= (6-4) m + (3-2) n + (10-3)

= 2m + n + 7

Therefore, the polynomial to be added to 4m + 2n + 3 is 2m + n + 7.