Polynomial

(i) Since all the exponents (power) of variable(x) are whole no. (i.e., zero or positive integers)

⇒ 2x⁶ – 4x⁵ + 7x² + 3 is a polynomial.

(Note: The degree is the value of the greatest (highest) exponent of any expression (except the constant) in the polynomial. To find the degree all that you have to do is find the largest exponent in the polynomial).

And, since here the highest power is 6,

⇒ It is a polynomial of degree 6.

(ii) Since all the exponents (power) of variable(x) are not whole no. (i.e., zero or positive integers)

⇒ x^–2 + 2x^–1 + 4 is not a polynomial.

(iii) Since all the exponents (power) of variable(y) are whole no. (i.e., zero or positive integers)

⇒ is a polynomial.

And, since here the highest power is 3,

⇒ It is a polynomial of degree 3.

(iv)

Since all the exponents (power) of variable(x) are not whole no.(i.e., zero or positive integers)

is not a polynomial.

(v) Since all the exponents (power) of variable(x) are whole no. (i.e., zero or positive integers)

⇒ x⁵¹ – 1 is a polynomial.

And, since here the highest power is 51,

⇒ x⁵¹ – 1 is a polynomial of degree 51.

(vi)

Since all the exponents (power) of t are not whole no.(i.e., zero or positive integers)

is not a polynomial.

(vii) 15 is a constant polynomial as there is no variable present in that.

And, therefore the highest power is 0

⇒ 15 is a polynomial with degree 0.

(viii) 0 is a constant polynomial (zero polynomial) as there is no variable present in that.

⇒ 0 is a polynomial with degree 0.

(ix)

Since all the exponents (power) of variable(z) are not whole no.(i.e., zero or positive integers)

is not a polynomial.

(x) Since all the exponents (power) of variable(y) are whole no. (i.e., zero or positive integers)

⇒ y³ +4 is a polynomial.

And, since here the highest power is 3,

⇒ y³ +4 is a polynomial of degree 3.

(xi)

Since all the exponents (power) of variable(x) are whole no.(i.e., zero or positive integers)

And, since here the highest power is 2

is a polynomial of degree 2.

(i) (Note: The degree is the value of the greatest (highest) exponent of any expression (except the constant) in the polynomial. To find the degree all that you have to do is find the largest exponent in the polynomial).

Since, here the highest power of variable(x) is 1,

⇒ 2x + 17 is a first degree polynomial.

(ii) Since, here the highest power of variable(x) is 3,

⇒ x³ + x² + x+ 1 is a third degree polynomial.

(iii) Since, here the highest power of variable(x, y) is 2,

⇒ –3 + 2y² + 5xy is a second degree polynomial.

(Note: Whenever we have polynomial in two variable (i.e., x and y) and both are in multiplication we see for the degree by adding the powers of both the variable.)

(iv) Since, here the highest power of variable(x) is 3,

⇒ 5 – x – x³ is a third degree polynomial.

(v) Since, here the highest power of variable (t) is 2,

⇒ √2 +t+t² is a second degree polynomial.

(vi) Since, here the highest and only power of variable(x) is 1,

⇒ √ 5x is a first degree polynomial.

(i) We can see that with x, the constant written before that is 5,

⇒ The co-efficient of x³ in 5x³ – 13x² + 2 is 5.

(ii) We can see that with x, the constant written before that is -1,

⇒ The co-efficient of x in x² – x + 2 is -1.

(iii) We can see that with x², the constant written before that is 0, because there is no x² in polynomial.

⇒ The co-efficient of x² in 8x – 19 is 0.

(iv) We can see that with x⁰ (i.e., the constant) is √(11)

⇒ The co-efficient of x⁰ in √ (11) – 3√(11) x+x² is √(11).

(i) (Note: The degree is the value of the greatest (highest) exponent of any expression (except the constant) in the polynomial. To find the degree all that you have to do is find the largest exponent in the polynomial).

Since, here the highest power of variable is 4,

⇒ Degree of polynomial=4.

(ii) Since, here the highest power of variable is 1,

⇒ Degree of polynomial=1.

(iii) Since, here the highest power of variable is 0, because there exist no variable.

⇒ Degree of polynomial=0.

(iv) Since, here the highest power of variable is 3,

⇒ Degree of polynomial=3.

(v) Since, here the highest power of variable is 1,

⇒ Degree of polynomial=1.

(vi) Since, here the highest power of variable is 19,

⇒ Degree of polynomial=19.

(Note: The degree is the value of the greatest (highest) exponent of any expression (except the constant) in the polynomial. To find the degree all that you have to do is find the largest exponent in the polynomial).

(Binomial means the polynomial will have two terms and degree =17 means that highest power should be 17).

∴ First binomial: x¹⁷ – 4x⁵

Second Binomial= 7x¹⁷ + 27

(Note: The degree is the value of the greatest (highest) exponent of any expression (except the constant) in the polynomial. To find the degree all that you have to do is find the largest exponent in the polynomial).

(Monomial means the polynomial will have one term and degree =4 means that highest power should be 4).

∴ First binomial: xy³

Second Binomial=x⁴

(Note: The degree is the value of the greatest (highest) exponent of any expression (except the constant) in the polynomial. To find the degree all that you have to do is find the largest exponent in the polynomial).

(Trinomial means the polynomial will have three terms and degree 3 means that highest power should be 3).

∴ First binomial: 7 x³ – 4x -7

Second Binomial: x³+5x²-x

(i) Since, here the highest power of variable is 1, which is a whole no.

⇒ x² + 3x + 2is a polynomial.

But here only one variable, i.e., x is used,

⇒ x² + 3x + 2 is a polynomial in one variable.

(ii) Since, here the highest power of variable is 2, which is a whole no.

x² + y² + a² is a polynomial.

Since, here two variable, i.e., x and y are used,

⇒ x² + y² + a² is a polynomial in two variables.

(iii) Since, here the highest power of variable is 2, which is a whole no.

⇒ y² – 4ax is a polynomial.

Since, here two variable, i.e., x and y are used,

⇒ y² – 4ax is a polynomial in one variable.

(iv) Since, here the highest power of variable is 1, which is a whole no.

⇒ x + y + 2 is a polynomial.

But here only two variable, i.e., x and y and are used,

⇒ x + y + 2 is a polynomial in one variable.

(v) Since, here the highest power of variable is( 5+9=14 ), which is a whole no.

(Note: 5 for x and 9 for y)

⇒ x⁸ + y⁴ + x⁵y⁹ is a polynomial.

here only two variable, i.e., x and y is used,

⇒ x² + 3x + 2 is a polynomial in two variables.

(vi) Since, here the highest power of variable is 1, which is a whole no.

⇒

And, since the exponent of variable is not zero or whole number,

⇒ is not a polynomial.

Formula used/Theory.

Putting the value of ‘x’ in f(x) gives out the value of f(x) on ‘x’

We have,

f(x) = x² + 9x – 6

When we put

x = 0

Then,

f(0) = 0² + 9 × 0 – 6

f(0) = –6

when we put

x = 1

Then,

f(1) = 1² + 9 × 1 – 6

f(1) = 10 –6

f(1) = 4

When we put

x = 3

Then,

f(3) = 3² + 9 × 3 – 6

f(3) = 9+27 –6

f(3) = 30

Conclusion.

The values of f(0), f(1), f(3) are –6, 4, 30 respectively.

Formula used/Theory.

Putting the value of ‘x’ in f(x) gives out the value of f(x) on ‘x’

We have,

f(x) = 2x³ + x² + x + 4

When we put x = 1

Then,

f(1) = 2(1)³ + (1)² + 1 + 4

f(1) = 2+1+1+4

f(1) = 8

When we put x = –1

Then,

f(-1) = 2(-1)³ + (-1)² + (-1) + 4

f(-1) = (-2) +1–1+4

f(-1) = 2

Conclusion.

The value of f(1), f(-1) are 8, 2 respectively.

Formula used/Theory.

Putting the value of ‘x’ in f(x) gives out the value of f(x) on ‘x’

We have,

f(x) = 3x⁴ – 5x³ + x² + 8

When we put x = 1

Then,

f(1) = 3(1)⁴ – 5(1)³ + (1)² + 8

f(1) = 3 – 5 + 1 + 8

f(1) = 7

When we put x = -1

Then,

f(-1) = 3(-1)⁴ – 5(-1)³ + (-1)² + 8

f(-1) = 3 – (-5) + 1 + 8

f(-1) = 3 + 5 + 1 + 8

f(-1) = 17

Conclusion.

The value of f(1), f(-1) are 7, 17 respectively.

Formula used/Theory.

Putting the value of ‘x’ in f(x) gives out the value of f(x) on ‘x’

We have,

f(x) = 4 + 3x – x³ + 5x⁶

When we put x = 1

Then,

f(1) = 4 + 3 × 1 – 1 + 5 × 1

f(1) = 4 + 3 – 1 + 5

f(1) = 11

When we put x = -1

Then,

f(-1) = 4 + 3 × (-1) – (-1)³ + 5 × (-1)⁶

f(-1) = 4 + 3 × (-1) – (-1) + 5 × (-1)

f(-1) = 4 – 3 + 1 + 5

f(-1) = 7

Conclusion.

The value of f(1), f(-1) are 11, 7 respectively.

Formula used/Theory.

Putting the value of ‘x’ in f(x) gives out the value of f(x) on ‘x’

We have,

f(x) = 6 + 10x – 7x²

when we put x = 1

then,

f(1) = 6 + 10 × (1) – 7(1)²

f(1) = 6 + 10 × 1 – 7 × 1

f(1) = 6 + 10 – 7

f(1) = 9

when we put x = -1

then,

f(-1) = 6 + 10 × (-1) – 7(-1)²

f(-1) = 6 + 10 × (-1) – 7 × 1

f(-1) = 6 – 10 – 7

f(-1) = -11

Conclusion.

The value of f(1), f(-1) are 9, -11 respectively.

Formula used/Theory.

There is some value of ‘x’ for which the P(x) comes to 0, that value of ‘x’ is said to be zero of polynomial P(x)

We have,

P(x) = x-1

And zero of the polynomial P(x) is 1

Then,

Zero of the polynomial P(x)

Means P(x) = 0

x-1 = 0

x = 1

∴ Zero of polynomial P(x) is 1

Conclusion.

Hence, the statement is True

Formula used/Theory.

There is some value of ‘x’ for which the P(x) comes to 0, that value of ‘x’ is said to be zero of polynomial P(x)

We have,

P(x) = 3 – x

And zero of the polynomial P(x) is 3

Then,

Zero of the polynomial P(x)

Means P(x) = 0

3 – x = 0

-x = -3

x = 3

∴ Zero of polynomial P(x) is 3

Conclusion.

Hence, the statement is True

Formula used/Theory.

There is some value of ‘x’ for which the P(x) comes to 0, that value of ‘x’ is said to be zero of polynomial P(x)

We have,

P(x) = 5x+1

And zero of the polynomial P(x) is

Then,

Zero of the polynomial P(x)

Means P(x) = 0

5x+1 = 0

5x = -1

x =

∴ Zero of polynomial P(x) is

Conclusion.

Hence, the statement is True

Formula used/Theory.

There is some value of ‘x’ for which the P(x) comes to 0, that value of ‘x’ is said to be zero of polynomial P(x)

We have,

P(x) = x² – 9

And zero of the polynomial P(x) are 3 and -3

Then,

Zero of the polynomial P(x)

Means P(x) = 0

x² – 9 = 0

x² – (3)² = 0

⇒ a² – b² = (a+b)(a-b)

(x – 3)(x+3) = 0

x – 3 = 0 and x+3 = 0

x = 3 and x = -3

∴ Zero of polynomial P(x) is 3 and -3

Conclusion.

Hence, the statement is True

Formula used/Theory.

There is some value of ‘x’ for which the P(x) comes to 0, that value of ‘x’ is said to be zero of polynomial P(x)

We have,

P(x) = x² – 5x

And zero of the polynomial P(x) are 0 and 5

Then,

Zero of the polynomial P(x)

Means P(x) = 0

x² – 5x = 0

x(x– 5) = 0

x = 0 and (x-5) = 0

⇒ x = 0 and x = 5

∴ Zero of polynomial P(x) is 0 and 5

Conclusion.

Hence, the statement is True

Formula used/Theory.

There is some value of ‘x’ for which the P(x) comes to 0, that value of ‘x’ is said to be zero of polynomial P(x)

We have,

P(x) = x² – 2x – 8

And zero of the polynomial P(x) are 4 and -2

Then,

Zero of the polynomial P(x)

Means P(x) = 0

x² – 2x – 8 = 0

As in quadratic eq

We have to divide (-2x) in 2 parts such that product comes to (-8x²)

Hence the factors are :-

(-x) × (8x) addition gives 8x – x = 7x

(-2x) × (4x) addition gives 4x – 2x = 2x

(-4x) × (2x) addition gives 2x – 4x = -2x

Hence factors are -4x and 2x

x² – 4x + 2x – 8 = 0

x(x – 4)+2(x – 4) = 0

(x+2)(x – 4) = 0

x+2 = 0 and x – 4 = 0

x = (-2) and x = 4

∴ Zero of polynomial P(x) is 4 and -2

Conclusion.

Hence, the statement is True

Formula used/Theory.

There is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)

We have,

f(x) = 2 – x

Then,

Zero of the polynomial f(x)

Means f(x) = 0

2 – x = 0

-x = -2

x = 2

∴ Zero of polynomial f(x) is 2

Conclusion.

Hence, the zero of the polynomial f(x) is 2

Formula used/Theory.

There is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)

We have,

f(x) = 7x+2

Then,

Zero of the polynomial f(x)

Means f(x) = 0

7x+2 = 0

7x = -2

x =

∴ Zero of polynomial f(x) is

Conclusion.

Hence, the zero of the polynomial f(x) is

Formula used/Theory.

There is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)

We have,

f(x) = x+9

Then,

Zero of the polynomial f(x)

Means f(x) = 0

x+9 = 0

x = -9

∴ Zero of polynomial f(x) is –9

Conclusion.

Hence, the zero of the polynomial f(x) is –9

Formula used/Theory.

There is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)

We have,

f(x) = 6 – 2x

Then,

Zero of the polynomial f(x)

Means f(x) = 0

6 – 2x = 0

-2x = -6

x = = 3

∴ Zero of polynomial f(x) is 3

Conclusion.

Hence, the zero of the polynomial f(x) is 3

Formula used/Theory.

There is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)

We have,

f(x) = 2x

Then,

Zero of the polynomial f(x)

Means f(x) = 0

2x = 0

x = 0

∴ Zero of polynomial f(x) is 0

Conclusion.

Hence, the zero of the polynomial f(x) is 0

Formula used/Theory.

There is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)

We have,

f(x) = ax+b

Then,

Zero of the polynomial f(x)

Means f(x) = 0

ax+b = 0

ax = -b

x =

∴ Zero of polynomial f(x) is

Conclusion.

Hence, the zero of the polynomial f(x) is

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

When x³ – 3x² + 2x + 5 is divided by (x – 2).

Let f(x) = x³ – 3x² + 2x + 5 …(1)

Now, let’s find out the zero of the linear polynomial, (x – 2).

To find zero,

x – 2 = 0

⇒ x = 2

This means that by remainder theorem, when x³ – 3x² + 2x + 5 is divided by (x – 2), the remainder comes out to be f(2).

From equation (1), remainder can be calculated as,

Remainder = f(2)

⇒ Remainder = (2)³ – 3(2)² + 2(2) + 5

⇒ Remainder = 8 – 12 + 4 + 5

⇒ Remainder = 5

∴ the required remainder = 5.

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

When x³ – 3x² + 2x + 5 is divided by (x + 2).

Let f(x) = x³ – 3x² + 2x + 5 …(1)

Now, let’s find out the zero of the linear polynomial, (x + 2).

To find zero,

x + 2 = 0

⇒ x = -2

This means that by remainder theorem, when x³ – 3x² + 2x + 5 is divided by (x + 2), the remainder comes out to be f(-2).

From equation (1), remainder can be calculated as,

Remainder = f(-2)

⇒ Remainder = (-2)³ – 3(-2)² + 2(-2) + 5

⇒ Remainder = -8 – 12 – 4 + 5

⇒ Remainder = -19

∴ the required remainder = -19

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

When x³ – 3x² + 2x + 5 is divided by (2x – 1).

Let f(x) = x³ – 3x² + 2x + 5 …(1)

Now, let’s find out the zero of the linear polynomial, (2x – 1).

To find zero,

2x – 1 = 0

⇒ 2x = 1

⇒ x = 1/2

This means that by remainder theorem, when x³ – 3x² + 2x + 5 is divided by (2x – 1), the remainder comes out to be f(1/2).

From equation (1), remainder can be calculated as,

Remainder = f( 1/2 )

⇒ Remainder

⇒ Remainder

⇒ Remainder = -19

∴ the required remainder = -19

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

When x³ – 3x² + 2x + 5 is divided by (2x + 1).

Let f(x) = x³ – 3x² + 2x + 5 …(1)

Now, let’s find out the zero of the linear polynomial, (2x + 1).

To find zero,

2x + 1 = 0

⇒ 2x = -1

⇒ x = -1/2

This means that by remainder theorem, when x³ – 3x² + 2x + 5 is divided by (2x + 1), the remainder comes out to be f(-1/2).

From equation (1), remainder can be calculated as,

Remainder = f(-1/2)

⇒ Remainder

⇒ Remainder

⇒ Remainder

⇒ Remainder

⇒ Remainder

⇒ Remainder

∴ the required remainder = 25/8

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

Let f(x) = x³ – 6x² + 13x + 60 …(1)

When x³ – 6x² + 13x + 60 is divided by (x – 1).

Now, let’s find out the zero of the linear polynomial, (x – 1).

To find zero,

x – 1 = 0

⇒ x = 1

This means that by remainder theorem, when x³ – 6x² + 13x + 60 is divided by (x – 1), the remainder comes out to be f(1).

From equation (1), remainder can be calculated as,

Remainder = f(1)

⇒ Remainder = (1)³ – 6(1)² + 13(1) + 60

⇒ Remainder = 1 – 6 + 13 + 60

⇒ Remainder = -5 + 73

⇒ Remainder = 68

∴ the required remainder = 68

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

Let f(x) = x³ – 3x² + 4x + 50 …(1)

When x³ – 3x² + 4x + 50 is divided by (x – 1).

Now, let’s find out the zero of the linear polynomial, (x – 1).

To find zero,

x – 1 = 0

⇒ x = 1

This means that by remainder theorem, when x³ – 3x² + 4x + 50 is divided by (x – 1), the remainder comes out to be f(1).

From equation (1), remainder can be calculated as,

Remainder = f(1)

⇒ Remainder = (1)³ – 3(1)² + 4(1) + 50

⇒ Remainder = 1 – 3 + 4 + 50

⇒ Remainder = 1 + 1 + 50

⇒ Remainder = 52

∴ the required remainder = 52

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

Let f(x) = 4x³ + 4x² – x – 1 …(1)

When 4x³ + 4x² – x – 1 is divided by (x – 1).

Now, let’s find out the zero of the linear polynomial, (x – 1).

To find zero,

x – 1 = 0

⇒ x = 1

This means that by remainder theorem, when 4x³ + 4x² – x – 1 is divided by (x – 1), the remainder comes out to be f(1).

From equation (1), remainder can be calculated as,

Remainder = f(1)

⇒ Remainder = 4(1)³ + 4(1)² – (1) – 1

⇒ Remainder = 4 + 4 – 1 – 1

⇒ Remainder = 8 – 2

⇒ Remainder = 6

∴ the required remainder = 6

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

Let f(x) = 11x³ – 12x² – x + 7 …(1)

When 11x³ – 12x² – x + 7 is divided by (x – 1).

Now, let’s find out the zero of the linear polynomial, (x – 1).

To find zero,

x – 1 = 0

⇒ x = 1

This means that by remainder theorem, when 11x³ – 12x² – x + 7 is divided by (x – 1), the remainder comes out to be f(1).

From equation (1), remainder can be calculated as,

Remainder = f(1)

⇒ Remainder = 11(1)³ – 12(1)² – (1) + 7

⇒ Remainder = 11 – 12 – 1 + 7

⇒ Remainder = -1 – 1 + 7

⇒ Remainder = -2 + 7

⇒ Remainder = 5

∴ the required remainder = 5

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

Let f(x) = x³ – 6x² + 9x – 8 …(1)

When x³ – 6x² + 9x – 8 is divided by (x – 3).

Now, let’s find out the zero of the linear polynomial, (x – 3).

To find zero,

x – 3 = 0

⇒ x = 3

This means that by remainder theorem, when x³ – 6x² + 9x – 8 is divided by (x – 3), the remainder comes out to be f(3).

From equation (1), remainder can be calculated as,

Remainder = f(3)

⇒ Remainder = (3)³ – 6(3)² + 9(3) – 8

⇒ Remainder = 27 – 54 + 27 – 8

⇒ Remainder = -27 + 27 – 8

⇒ Remainder = 0 – 8

⇒ Remainder = -8

∴ the required remainder = -8

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the following questions on the basis of this remainder theorem.

Let f(x) = x³ – ax² + 2x – a …(1)

When x³ – ax² + 2x – a is divided by (x – a).

Now, let’s find out the zero of the linear polynomial, (x – a).

To find zero,

x – a = 0

⇒ x = a

This means that by remainder theorem, when x³ – ax² + 2x – a is divided by (x – a), the remainder comes out to be f(a).

From equation (1), remainder can be calculated as,

Remainder = f(a)

⇒ Remainder = (a)³ – a(a)² + 2(a) – a

⇒ Remainder = a³ – a³ + 2a – a

⇒ Remainder = 2a – a

⇒ Remainder = a

∴ the required remainder = a

Remainder theorem says that,

f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Let us solve the questions on the basis of this theorem.

Here, let f(x) = 4x³ + 4x² – x – 1 …(i)

First, we need to find zero of the linear polynomial, (2x + 1).

To find zero,

2x + 1 = 0

⇒ 2x = -1

⇒ x = - 1/2

f(x) will be multiple of (2x + 1) if f(-1/2) = 0.

⇒

⇒

⇒

⇒

⇒ P(x) = 4x³ + 4x² – x – 1 is a multiple of (2x + 1).

Let the two polynomials be,

P(x) = ax³ + 3x² – 3 …(i)

Q(x) = 2x³ – 5x + a …(ii)

Now, we understand by the question that,

P(x) and Q(x) divided by (x – 4) gives the same remainder.

We need to find zero of the linear polynomial, (x – 4).

To find zero, put (x – 4) = 0

⇒ x – 4 = 0

⇒ x = 4

By Remainder theorem that says, f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Here, a = 4.

This means, remainder when P(x) is divided by (x – 4) is P(4).

⇒ Remainder = P(4)

⇒ Remainder = a(4)³ + 3(4)² – 3

⇒ Remainder = 64a + 48 – 3

⇒ Remainder = 64a + 45 …(iii)

And remainder when Q(x) is divided by (x – 4) is Q(4).

⇒ Remainder = Q(4)

⇒ Remainder = 2(4)³ – 5(4) + a

⇒ Remainder = 128 – 20 + a

⇒ Remainder = 108 + a …(iv)

When P(x) and Q(x) are divided (x – 4) , they leave same remainder.

Comparing equations (iii) and (iv), we have

64a + 45 = 108 + a

⇒ 64a – a = 108 – 45

⇒ 63a = 63

⇒

⇒ a = 1

Thus, a = 1.

Let the polynomials be:

A(x) = x³ + 2x² – px – 7

B(x) = x³ + px² – 12x + 6

We will use remainder theorem here.

By Remainder theorem that says, f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

When A(x) is divided by (x + 1), it will leave remainder R₁.

Let’s also find zero of linear polynomial, (x + 1).

To find zero,

⇒ x + 1 = 0

⇒ x = -1

The required remainder, R₁ = A(-1).

⇒ R₁ = (-1)³ + 2(-1)² – p(-1) – 7

⇒ R₁ = -1 + 2 + p – 7

⇒ R₁ = 1 – 7 + p

⇒ R₁ = p – 6 …(i)

When B(x) is divided by (x – 2), it will leave remainder R₂.

Let’s also find zero of linear polynomial, (x – 2).

To find zero,

⇒ x – 2 = 0

⇒ x = 2

The required remainder, R₂ = B(2)

⇒ R₂ = (2)³ + p(2)² – 12(2) + 6

⇒ R₂ = 8 + 4p – 24 + 6

⇒ R₂ = 4p + 14 – 24

⇒ R₂ = 4p – 10 …(ii)

We have, 2R₁ + R₂ = 6

Substituting values from (i) and (ii), we get

2(p – 6) + (4p – 10) = 6

⇒ 2p – 12 + 4p – 10 = 6

⇒ 6p – 22 = 6

⇒ 6p = 6 + 22

⇒ 6p = 28

⇒

⇒

Thus, p = 14/3.

We have the polynomial x⁴ – 2x³ + 3x² – ax + b.

Let it be P(x), such that

P(x) = x⁴ – 2x³ + 3x² – ax + b …(i)

According to the question,

P(x) is divided by (x – 1) and (x + 1), and leaves the remainder 5 and 19 respectively.

We will use remainder theorem here.

By Remainder theorem that says, f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

At first, let’s find the zero of the linear polynomial, (x – 1).

To find zero,

x – 1 = 0

⇒ x = 1

Now, let’s find the zero of the linear polynomial, (x + 1).

To find zero,

x + 1 = 0

⇒ x = -1

From Remainder theorem, we can say

When P(x) is divided by (x – 1), the remainder comes out to be 5.

We can also say that,

The remainder comes out to be P(1).

⇒ P(1) = 5

⇒ (1)⁴ – 2(1)³ + 3(1)² – a(1) + b = 5

⇒ 1 – 2 + 3 – a + b = 5

⇒ b – a + 2 = 5

⇒ b – a = 5 – 2

⇒ b – a = 3 …(ii)

And when P(x) is divided by (x + 1), the remainder comes out to be 19.

We can also say that,

The remainder comes out to be P(-1)

⇒ P(-1) = 19

⇒ (-1)⁴ – 2(-1)³ + 3(-1)² – a(-1) + b = 5

⇒ 1 + 2 + 3 + a + b = 5

⇒ a + b + 6 = 5

⇒ a + b = 5 – 6

⇒ a + b = -1 …(iii)

Solving equations (ii) and (iii), we get

(b – a) + (a + b) = 3 + (-1)

⇒ 2b = 3 – 1

⇒ 2b = 2

⇒

⇒ b = 1

Putting b = 1 in equation (ii),

b – a = 3

⇒ 1 – a = 3

⇒ a = 1 – 3

⇒ a = -2

The values of a = -2 and b = 1.

∵ The polynomial = x⁴ – 2x³ + 3x² – ax + b

∴ The polynomial = x⁴ – 2x³ + 3x² – (-2)x + 1

⇒ The polynomial = x⁴ – 2x³ + 3x² + 2x + 1

So, when polynomial x⁴ – 2x³ + 3x² + 2x + 1 is divided by (x + 2), the remainder can be calculated by using Remainder theorem.

First, we need to find zero of the linear polynomial, (x + 2).

To find zero,

Put (x + 2) = 0

⇒ x = -2

So, Required remainder = P(-2)

⇒ Required remainder = (-2)⁴ – 2(-2)³ + 3(-2)² + 2(-2) + 1

⇒ Required remainder = 16 + 16 + 12 – 4 + 1

⇒ Required remainder = 32 + 12 – 3

⇒ Required remainder = 44 – 3

⇒ Required remainder = 41

Thus, remainder is 41.

We have

Let us simplify it,

⇒

⇒

⇒ f(x) = x …(i)

To show, f(a) + f(b) = f(a + b)

Take LHS: f(a) + f(b)

Just replace x by a in equation (i),

f(a) = a

Now, replace x by b in equation (i),

f(b) = b

LHS: f(a) + f(b)

⇒ f(a) + f(b) = a + b …(ii)

Now, Put (a + b) in equation (i),

f(a + b) = a + b

Replace (a + b) in equation (ii) from f(a + b), we get

f(a) + f(b) = f(a + b)

Thus, shown that f(a) + f(b) = f(a + b).

We have,

f(x) = ax + b …(i)

f(0) = 3

f(2) = 5

Replace x by 0 in equation (i), we get

f(0) = a(0) + b

⇒ 3 = 0 + b [∵ f(0) = 3]

⇒ 3 = b

⇒ b = 3

Replace x by 2 in equation (i), we get

f(2) = a(2) + b

⇒ 5 = 2a + b [∵ f(2) = 5]

⇒ 5 = 2a + 3 [∵ b = 3]

⇒ 2a = 5 – 3

⇒ 2a = 2

⇒ a = 1

Thus, a = 1 and b = 3.

We have,

f(x) = ax² + bx + c …(i)

f(0) = 2

f(1) = 1

f(4) = 6

Replace x by 0 in equation (i), we get

f(0) = a(0)² + b(0) + c

⇒ 2 = 0 + 0 + c [∵ f(0) = 2]

⇒ 2 = c

⇒ c = 2 …(ii)

Now, replace x by 1 in equation (i), we get

f(1) = a(1)² + b(1) + c

⇒ 1 = a + b + 2 [∵ f(1) = 1 & c = 2]

⇒ a + b = 1 – 2

⇒ a + b = -1 …(iii)

Finally, replace x by 4 in equation (i), we get

f(4) = a(4)² + b(4) + c

⇒ 6 = 16a + 4b + 2 [∵ f(4) = 6 & c = 2]

⇒ 16a + 4b = 6 – 2

⇒ 16a + 4b = 4

⇒ 4 (4a + b) = 4

⇒ 4a + b = 1 …(iv)

Solving equations (iii) & (iv), we get

⇒ -3a = -2

⇒

Put a = 2/3 in equation (iii), we get

⇒

⇒

⇒

Thus, a = 2/3, b = -5/3 and c = 2.

Let us check for the option (a),

We have

We can write it as, x + 2x^-1 + 3.

Here, x^-1 is a polynomial.

⇒ x + 2x^-1 + 3 is not a polynomial.

⇒ is not a polynomial.

For option (b),

We have

We can write it as, 3√x + 2x^-1/2 + 5.

Here, x^{- 1/2} is not a polynomial.

⇒ 3√x + 2x^{- 1/2} + 5 is not a polynomial.

⇒ is not a polynomial.

For option (c),

We have √2x² - √3x + 6

Term-wise, √2 x² is a polynomial

√3 x is a polynomial

6 is a polynomial

⇒ √2 x² - √3 x + 6 is a polynomial.

For option (d),

We have x¹⁰ + y⁵ + 8.

Here, we have two variables, x and y.

Hence, it is clearly not a polynomial of single variable.

Thus, option (d) is correct.

For option (a),

We have x – 1

Look at each term individually, there’s no violation of the definition of polynomials.

The exponent in the algebraic expression is non-negative integers, that’s why we can say that (x – 1) is a polynomial.

For option (b),

We have

Rearranging the expression, we get

Even after re-arranging the expression, notice that the variable in both terms is appearing in denominator, which does not satisfy the definition of polynomials as variables cannot appear in denominator in the expression of a polynomial.

For option (c),

We have

Simplifying the expression, we get

x² – 2x^-2 + 5

Look at the term 2x^-2.

In polynomials, the exponent in the algebraic expression should be non-negative integers, but in this expression the exponent is negative.

⇒ is not a polynomial.

For option (d),

We have

Simplifying the expression, we get

⇒

⇒

⇒

Observe the rational root in the second term of the expression, but in polynomials, the exponent should be non-negative integers.

⇒ is not a polynomial.

Thus, option (a) is correct.

For option (a),

We have x + x²

In this expression, we have 2 terms.

The degree of this expression is 2 (largest exponent of x).

But linear polynomial has a degree of 1, so that it can give exactly one root.

⇒ x + x² is not a linear polynomial.

For option (b),

We have x + 1

In this expression, we have 2 terms.

And the degree of this expression is 1 (largest exponent of x).

This compliments the definition of a polynomial.

⇒ x + 1 is a linear polynomial.

For option (c),

We have 5x² – x + 3.

In this expression, we have 3 terms.

And the degree of this expression is 2 (largest exponent of x).

This violates the definition of linear polynomial, as the degree of linear polynomial is always 1.

⇒ 5x² – x + 3 is not a linear polynomial.

For option (d),

We have

Let us simplify the expression.

⇒

⇒ x² = 1

⇒ x² – 1 = 0

This now is an algebraic expression, having 2 terms.

The degree of this expression is 2 (largest exponent of x).

This again doesn’t satisfy the criteria of linear polynomial, which says that degree should be 1.

⇒ is not a linear polynomial.

Thus, option (b) is correct.

For option (a),

We have √x – 4.

Observe the two terms, x has an exponent of 1/2.

But for second degree polynomials, the degree should be 2 (that is, largest exponent of x should be 2)

⇒ √x – 4 is not a second degree polynomial.

For option (b),

We have x³ + x.

The degree of this expression is 3 (largest exponent of x is 3).

But for second degree polynomials, the degree should be 2 (that is, largest exponent of x should be 2)

⇒ x³ + x is not a second degree polynomial.

For option (c),

We have x³ + 2x + 6.

The degree of this expression is also 3 (largest exponent of x is 3).

But for second degree polynomials, the degree should be 2 (that is, largest exponent of x should be 2)

⇒ x³ + 2x + 6 is not a second degree polynomial.

For option (d),

We have x² + 5x + 6.

The degree of this expression is 2 (largest exponent of x is 2).

This satisfies the criteria of second degree polynomial.

⇒ x² + 5x + 6 is second degree polynomial.

Thus, option (d) is correct.

We have got polynomial, √3.

We can re-write it as, √3 x⁰ [∵ x⁰ = 1; any variable or number with exponent 0 gives 1 every time]

So, notice the exponent of x. Here, it is 0.

Since, degree of a term is sum of exponents of the variables that appears in it, and thus is a non-negative integer.

⇒ Degree of the polynomial = 0

Thus, option (d) is correct.

A zero is root of a polynomial function which is a number that, when plugged in for the variable, makes the function equal to zero.

To find the zero of a polynomial, just equate the polynomial equals to 0 and find out the value of the variable.

Here, p(x) = 0

⇒ 2x – 3 = 0

⇒ 2x = 3

⇒

Thus, zero of the polynomial is 3/2.

We know,

p(x) = x + 4 …(i)

Now, to find p(-x), just replace x by –x in the above expression. Let us do it.

p(-x) = (-x) + 4

⇒ p(-x) = -x + 4 …(ii)

Now, add equations (i) and (ii), we get

p(x) + p(-x) = (x + 4) + (-x + 4)

⇒ p(x) + p(-x) = x + 4 – x + 4

⇒ p(x) + p(-x) = 8

Thus, the values of p(x) + p(-x) = 8

If the polynomial x³ + 4x² + 4x – 3 is divided by x, we need to find the remainder.

Let P(x) = x³ + 4x² + 4x – 3 …(i)

Next, we need to find the zero of the linear polynomial, x.

To find zero, equate x to 0.

⇒ x = 0

So, when P(x) is divided by x, it leaves a remainder P(0).

Replace x by 0 in equation (i), we get

P(0) = (0)³ + 4(0)² + 4(0) – 3

⇒ P(0) = 0 + 0 + 0 – 3

⇒ P(0) = -3

⇒ Required remainder = -3

Thus, -3 is the answer.

We have been given that,

(3x – 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ + … + a₁x + a₀

If we simply put x = 1 in the above equation,

(3(1) – 1)⁷ = a₇(1)⁷ + a₆(1)⁶ + a₅(1)⁵ + …+ a₁(1) + a₀

⇒ a₇ + a₆ + a₅ + … + a₁ + a₀ = (3 – 1)⁷

⇒ a₇ + a₆ + a₅ + … + a₁ + a₀ = 2⁷

⇒ a₇ + a₆ + a₅ + … + a₁ + a₀ = 128

Thus, value of a₇ + a₆ + a₅ + … + a₁ + a₀ = 128.

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial x + 1

x + 1 = 0

x = – 1

if f(x) is 2x³ + 3x² – 1

then putting x = – 1;

f( – 1) = 2( – 1)³ + 3( – 1)² – 1

= 2 × ( – 1) + 3 × (1) – 1

= – 2 + 3 – 1

= 0

Conclusion.

∴ (x + 1) is a factor of 2x³ + 3x² – 1

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial x + 1

x + 1 = 0

x = – 1

if f(x) is x⁴ + x³ – x² + 4x + 5

then putting x = – 1;

f( – 1) = ( – 1)⁴ + ( – 1)³ – ( – 1)² + 4( – 1) + 5

= 1 – 1 – 1 – 4 + 5

= 0

Conclusion.

∴ (x + 1) is a factor of x⁴ + x³ – x² + 4x + 5

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial x + 1

x + 1 = 0

x = – 1

if f(x) is 7x³ + x² + 7x + 1

then putting x = – 1;

f( – 1) = 7( – 1)³ + ( – 1)² + 7( – 1) + 1

= 7 × ( – 1) + 1 + 7 × ( – 1) + 1

= – 7 + 1 – 7 + 1

= – 12

Conclusion.

∴ (x + 1) is not a factor of 7x³ + x² + 7x + 1

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial x + 1

x + 1 = 0

x = – 1

if f(x) is 3 + 3x – 5x³ – 5x⁴

then putting x = – 1;

f( – 1) = 3 + 3( – 1) – 5( – 1)³ – 5( – 1)⁴

= 3 + 3 × ( – 1) – 5 × ( – 1) – 5 × 1

= 3 – 3 + 5 – 5

= 0

Conclusion.

∴ (x + 1) is a factor of 3 + 3x – 5x³ – 5x⁴

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial x + 1

x + 1 = 0

x = – 1

if f(x) is x⁴ + x² + x + 1

then putting x = – 1;

f( – 1) = ( – 1)⁴ + ( – 1)² + ( – 1) + 1

= 1 + 1 – 1 + 1

= 2

Conclusion.

∴ (x + 1) is not a factor of x⁴ + x² + x + 1

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial x + 1

x + 1 = 0

x = – 1

if f(x) is x³ + x² + x + 1

then putting x = – 1;

f( – 1) = ( – 1)³ + ( – 1)² + ( – 1) + 1

= – 1 + 1 – 1 + 1

= 0

Conclusion.

∴ (x + 1) is a factor of x³ + x² + x + 1

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

x + 2 = 0

x = – 2

if f(x) is x⁴ – x²

then putting x = – 2 ;

f( – 2) = ( – 2)⁴ + ( – 2)²

= 16 + 4

= 20

Conclusion.

∴ g(x) is not a factor of f(x)

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

x + 5 = 0

x = – 5

if f(x) is 2x³ + 9x² – 11x – 30

then putting x = – 5 ;

f( – 5) = 2( – 5)³ + 9( – 5)² – 11( – 5) – 30

= 2 × ( – 125) + 9 × (25) – 11x( – 5) – 30

= – 250 + 225 + 55 – 30

= 0

Conclusion.

∴ g(x) is a factor of f(x)

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

x – 3 = 0

x = 3

if f(x) is 2x³ + 7x² – 24x – 45

then putting x = 3 ;

f(3) = 2(3)³ + 7(3)² – 24(3) – 45

= 2 × (27) + 7 × (9) – 24x(3) – 45

= 54 + 63 – 72 – 45

= 0

Conclusion.

∴ g(x) is a factor of f(x)

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

3x – 2 = 0

x =

if f(x) is 3x³ + x² – 20x + 12

then putting x = ;

f(3) = 3()³ + ()² – 20() + 12

=

=

= = 0

Conclusion.

∴ g(x) is a factor of f(x)

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

x + 2 = 0

x = – 2

if x + 2 is factor of f(x) = 2x⁴ + 3x³ + 2kx² + 3x + 6

then f( – 2) = 0 ;

f( – 2) = 2( – 2)⁴ + 3( – 2)³ + 2k( – 2)² + 3( – 2) + 6 = 0

= 2 × 16 – 3 × 8 + 8 × k – 6 + 6

= 32 – 24 + 8k – 6 + 6

8 + 8k = 0

8k = – 8

k = = – 1

Conclusion.

∴ if (x + 2) is factor of f(x) then k = – 1

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

x – 1 = 0

x = 1

if x – 1 is factor of f(x) = 2x³ + 9x² + x + k

then f(1) = 0 ;

f(1) = 2(1)³ + 9(1)² + 1 + k = 0

= 2 + 9 + 1 + k

12 + k = 0

k = – 12

Conclusion.

∴ if (x – 1) is factor of f(x) then k = – 12

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

x – 1 = 0

x = 1

if x – 1 is factor of f(x) = kx² – 3x + k

then f(1) = 0 ;

f(1) = k(1)² – 3(1) + k = 0

= k – 3 + k

2k – 3 = 0

k =

Conclusion.

∴ if (x – 1) is factor of f(x) then k = –

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

2x – 3 = 0

x =

if 2x – 3 is factor of f(x) = 2x⁴ + x³ – kx² – x + 6

then f() = 0 ;

f() = 2()⁴ + ³ – k()² – + 6 = 0

=

=

=

= = 0

144 – 18k = 0

k = = 8

Conclusion.

∴ if (2x – 3) is factor of f(x) then k = 8

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of polynomial g(x)

2x – 1 = 0

x =

if 2x – 1 is factor of f(x) = 2x³ + kx² + 11x + k + 3

then f() = 0 ;

f() = 2³ + k² + 11() + k + 3 = 0

=

=

=

= = 0

70 + 10k = 0

k = = – 7

Conclusion.

∴ if (2x – 1) is factor of f(x) then k = – 7

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

⇒ a² – b² = (a + b)(a – b)

1^st we find out zero of polynomial g(x)

x² – 4 = 0

x² – 2² = (x + 2)(x – 2) = 0

x + 2 = 0 and x – 2 = 0

x = – 2 and x = 2

if x + 2 is factor of f(x) = ax⁴ + 2x³ – 3x² + bx – 4

then f( – 2) = 0 ;

f( – 2) = a( – 2)⁴ + 2( – 2)³ – 3( – 2)² + b( – 2) – 4

= 16a – 16 – 12 – 2b – 4

16a – 2b – 32 = 0

16a = 32 + 2b ………eq 1

if x – 2 is factor of f(x) = ax⁴ + 2x³ – 3x² + bx – 4

then f(2) = 0 ;

f(2) = a(2)⁴ + 2(2)³ – 3(2)² + b(2) – 4

= 16a + 16 – 12 + 2b – 4

16a + 2b = 0 ………eq 2

Putting value of 16a from eq 1 into eq 2

(32 + 2b) + 2b = 0

32 + 4b = 0

4b = – 32

b = = – 8

Putting value of b in eq 1

16a = 32 + 2 × ( – 8)

16a = 32 – 16

16a = 16

a = = 1

Conclusion.

∴ if (x² – 4) is factor of f(x) then b = – 8 and a = 1

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

1^st we find out zero of both polynomial (x + 1)(x + 2)

x + 1 = 0 and x + 2 = 0

x = – 1 and x = – 2

if x + 2 is factor of f(x) = x³ + 3x² + 2ax + b

then f( – 2) = 0 ;

f( – 2) = ( – 2)³ + 3( – 2)² + 2a( – 2) + b

= – 8 + 12 – 4a + b

b – 4a + 4 = 0

b = 4a – 4 ………eq 1

if x + 1 is factor of f(x) = x³ + 3x² + 2ax + b

then f( – 1) = 0 ;

f( – 1) = ( – 1)³ + 3( – 1)² + 2( – 1)a + b

= – 1 + 3 – 2a + b

b – 2a + 2 = 0 ………eq 2

Putting value of b from eq 1 into eq 2

(4a – 4) – 2a + 2 = 0

2a – 2 = 0

2a = 2

a = = 1

Putting value of a in eq 1

b = 4 × (1) – 4

= 0

Conclusion.

∴ if (x + 1) and (x + 2) is factor of f(x) then b = 0 and a = 1

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

⇒ Dividend = Divisor × Quotient + Remainder

When x + 2 is factor of polynomial

Then

x + 2 = 0

x = – 2

if x + 2 is factor of polynomial f(x) = ax³ + bx² + x – 6

then f( – 2) is 0

f( – 2) = a( – 2)³ + b( – 2)² + ( – 2) – 6 = 0

– 8a + 4b – 2 – 6 = 0

4b – 8a = 8

4b = 8 + 8a ……… eq 1

When x – 2 divides polynomial gives remainder 4

Then

Dividend = Divisor × Quotient + Remainder

ax³ + bx² + x – 6 = (x – 2) × Quotient + 4

ax³ + bx² + x – 6 – 4 = (x – 2) × Quotient

ax³ + bx² + x – 10 = (x – 2) × Quotient

(x – 2) = 0

x = 2

if (x – 2) is factor of polynomial ax³ + bx² + x – 10

then f(2) = 0

f(2) = a(2)³ + b(2)² + 2 – 10

8a + 4b – 8 = 0

8a + 4b = 8 ………eq 2

Putting value of 4b from eq 1 into eq 2

8a + (8 + 8a) = 8

16a + 8 = 8

16a = 8 – 8 = 0

a = 0

Putting value of ‘a’ in eq 1

4b = 8 + 8a

4b = 8 + 8 × 0

4b = 8

b = = 2

Conclusion.

∴ The value of a and b comes to be 0 and 2 respectively

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

⇒ Dividend = Divisor × Quotient + Remainder

If (x – y) is a factor of xⁿ – yⁿ

Then

We have to prove

On dividing xⁿ – yⁿ by (x – y) Remainder gets 0

When n = 1

(x – y) becomes factor of (x¹ – y¹)

Hence,

x – y = 0

x = y

Suppose on dividing xⁿ – yⁿ with x – y we get Remainder R

xⁿ – yⁿ = (x – y) × Quotient + R

Putting x = y

yⁿ – yⁿ = (y – y) × Quotient + R

0 = 0 + R

R = 0

∴ For every value n (x – y) is a factor of xⁿ – yⁿ

Conclusion.

⇒ For every value n (x – y) is a factor of xⁿ – yⁿ

Let us suppose, if xⁿ + yⁿ is divided by x + y, the quotient is Q and remainder without x is R.

Dividend = Divisor × Quotient + Remainder

⸫ xⁿ + yⁿ = (x + y) × Q + R ……. [This is an identity]

Since x does not belong to the remainder R, the value of R will not change for any value of x.

So, in the above identity, putting (-y) for x, we get:

(-y)ⁿ + yⁿ = (-y + y) × Q + R

Now, as n is odd, we get that (-y)ⁿ = - yⁿ

so, we get,

- yⁿ + yⁿ = 0 × Q + R

0 = R

⸫ R = 0

⸫ (x + y) is a factor of the polynomial xⁿ + yⁿ, when n is an odd positive integer.

Let us suppose, if xⁿ + yⁿ is divided by x - y, the quotient is Q and remainder without x is R.

Dividend = Divisor × Quotient + Remainder

⸫ xⁿ + yⁿ = (x - y) × Q + R ……. [This is an identity]

Since x does not belong to the remainder R, the value of R will not change for any value of x.

So, in the above identity, putting (y) for x, we get:

(y)ⁿ + yⁿ = (y - y) × Q + R

2(yⁿ ) = 0 + R

2(yⁿ ) = R

Since, value of R is not “0”, we can say that x – y is not a factor of the polynomial xⁿ + yⁿ

⸫ we can say that (x – y) can never be a factor of the polynomial xⁿ + yⁿ.

As it is given that, (x + 2) is a factor of the given polynomial, then -2 is the root of the polynomial.

So, as -2 is root of the polynomial and when we put -2 in the polynomial then it will give us “0” as the answer.

Now, we have,

(-2)³ + 6(-2)² + 4(-2) + k = 0

-8 + 6×4 -8 + k = 0

-16 + 24 + k = 0

k + 8 = 0

k = -8

⸫ value of k is -8.

Hence, the correct option is (c).

In the question, it is given that which means that is a root of the polynomial f(x).

Which means that, [x – ()] =0 is a factor of the polynomial f(x).

[x + ()] =0 is a factor.

2x + 1 = 0 is a factor.

Hence the correct option is (b).

We know that (x – 1) is a factor of the polynomial f(x) but it is not the factor of g(x).

Which means that, f(1) = 0 but g(1) ≠ 0.

But for (x - 1) to be a factor of the given options, when we put x=1 in the given functions we should get an answer as “0”.

So, for option (a), let F(x) = f(x) g(x)

To check whether (x – 1) is a factor of the polynomial F(x) we have to put x=1 in F(x) and see whether it gives us the answer as “0” or not.

F(1) = f(1) × g(1)

F(1) = 0 × g(1)

F(1) = 0

So, we can say that F(x) = f(x) g(x) is the polynomial whose factor is (x – 1).

⸫ (x – 1) will be the factor of f(x) g(x).

Hence the correct option is (a).

As we know that (x + y) is a factor of the polynomial xⁿ + yⁿ, when n is an odd positive integer, so here we can see that y = 1, so, we can say that (x + 1) is a factor of the polynomial xⁿ + 1 when n is a positive odd integer.

Hence the correct option is (a).

For the polynomial an⁴ + bn³ + cn² + dn + e, we have n² – 1 is a factor.

So, n² – 1 = 0 is the root of the polynomial.

(n – 1)(n + 1) = 0

n = -1 or 1 are the roots of the given polynomial.

Now, when n = 1, we have:

a(1)⁴ + b(1)³ + c(1)² + d(1) + e = 0

a + b + c + d + e = 0 ……... (1)

Now, when n = -1, we have:

a(-1)⁴ + b(-1)³ + c(-1)² + d(-1) + e = 0

a – b + c – d + e = 0

a + c + e = d + b

Hence the correct option is (a).

We have the polynomial as x³ + ax² – 2x + a – 12.

Given that x + a is the factor of the polynomial, which means that x = -a is the root of the polynomial.

So f(-a) = 0.

(-a)³ + a(-a)² - 2(-a) + a – 12 = 0

-a³ + a³ + 2a + a – 12 = 0

3a – 12 = 0

3a = 12

a = 4.

⸫ value of a is 4.

We have the polynomial as k²x³ – kx² + 3kx – k.

Given that x - 3 is the factor of the polynomial, which means that x = 3 is the root of the polynomial.

So f(3) = 0.

k²(3)³ – k(3)² + 3k(3) – k = 0

27k² – 9k + 9k – k = 0

27k² – k = 0

k(27k – 1) = 0

k= 0 or

⸫ value of k is 0 or .

We have f(x) = 2x + 5

So, f(-x) = 2(-x) + 5

Now, f(x) + f(–x) = 2x + 5 + 2(-x) + 5

f(x) + f(–x) = 2x + 5 - 2x + 5

f(x) + f(–x) = 10.

⸫ value of f(x) + f(–x) when f(x) = 2x + 5 is 10.

The given polynomial is px² + 5x + r.

Given that, (x – 2) is the factor of the polynomial, we have f(2) = 0.

So, f(2) = p(2)² + 5(2) + r = 0

4p + 10 + r = 0 ……… (1)

Now, given that, is the factor of the polynomial, we have

So,

From (1), we get r = -4p – 10

So, we can say that the relation between p and r is r = -4p – 10.

⸫ the relation between p and r is r = -4p – 10.

The polynomial is f(x) = 2x + 3.

The zero of the polynomial means value of x for which f(x) = 0.

So, f(x) = 2x + 3 = 0

2x + 3 = 0

2x = -3

⸫ zero of the polynomial f(x) = 2x + 3 is .