Polynomials

(i)

Yes,

The given expression is a polynomial

This is because all the variables have integer exponents that are positive.

Since, the highest power of the variable is 5.

Hence, the degree of the polynomial is 5.

(ii)

Yes,

The given expression is a polynomial

This is because all of the variables have integer exponents that are positive.

Since, the highest power of the variable is 3

Hence, the degree of the polynomial is 3

(iii)

Yes,

The given expression is a polynomial

This is because all of the variables have integer exponents that are positive.

Since, the highest power of the variable is 2

Hence, the degree of the polynomial is 2

(iv)

No,

The given expression is not a polynomial

Since, the term has a fractional exponent.

(v)

No,

The given expression is not a polynomial

Since, the term has a negative exponent.

(vi)

Yes,

The given expression is a polynomial

This is because all of the variables have integer exponents that are positive.

Since, the highest power of the variable is 108

Hence, the degree of the polynomial is 108

(vii)

No,

The given expression is not a polynomial

Since, the term has a fractional exponent.

(viii)

Yes,

The given expression is a polynomial

This is because all of the variables have integer exponents that are positive.

Since, the highest power of the variable is 2

Hence, the degree of the polynomial is 2

(ix)

No,

The given expression is not a polynomial

Since, the term has a negative exponent.

(x) 1

Yes,

The given expression is a polynomial

This is because all of the variables have integer exponents that are positive.

Since, the highest power of the variable is 0

Hence, the degree of the polynomial is 0

(xi)

Yes,

The given expression is a polynomial

This is because all of the variables have integer exponents that are positive.

Since, the highest power of the variable is 0

Hence, the degree of the polynomial is 0

(xii)

Yes,

The given expression is a polynomial

This is because all of the variables have integer exponents that are positive.

Since, the highest power of the variable is 2

Hence, the degree of the polynomial is 2

(i)

Since,

In the given polynomial, the highest power of the variable is 1

Hence,

The degree of the polynomial is 1

(ii)

Since,

In the given polynomial, the highest power of the variable is 3

Hence,

The degree of the polynomial is 3

(iii) 9

Since,

In the given polynomial, the highest power of the variable is 0

Hence,

The degree of the polynomial is 0

(iv)

Since,

In the given polynomial, the highest power of the variable is 7

Hence,

The degree of the polynomial is 7

(v)

Since,

In the given polynomial, the highest power of the variable is 10

Hence,

The degree of the polynomial is 10

(vi)

Since,

In the given polynomial, the highest power of the variable is 2

Hence,

The degree of the polynomial is 2

(i)

Hence,

The Coefficient of in the given polynomial is -5

(ii)

Hence,

The Coefficient of x in the given polynomial is -22

(iii)

Hence,

The Coefficient of in the given polynomial is

_{(iv) 3x – 5}

_Since,

_{There isn’t any variable with exponent as 2}

Hence,

The Coefficient of in the given polynomial, is 0

An example of a binomial of degree 27 is a two-term polynomial with highest degree 27.

Hence,

The suitable example for the question can be y²⁷ – 29.

An example of a monomial of degree 16 is a single term polynomial with highest degree 16.

Hence,

The suitable example for the question can be y¹⁶

An example of a trinomial of degree is a three-term polynomial with highest degree 3.

Hence,

The suitable example for the question can be y³ –y² + 29

(i)

Since,

The degree of the given polynomial is 2

Hence,

The polynomial is a quadratic polynomial.

(ii)

Since,

The degree of the given polynomial is 3

Hence,

The polynomial is a cubic polynomial.

(iii)

Since,

The degree of the given polynomial is 2

Hence,

The polynomial is a quadratic polynomial.

(iv) -7+z

Since,

The degree of the given polynomial is 1

Hence,

The polynomial is a linear polynomial.

(v)

Since,

The degree of the given polynomial is 1

Hence,

The polynomial is a linear polynomial.

(vi)

Since,

The degree of the given polynomial is 3

Hence,

The polynomial is a cubic polynomial.

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= x³ + 3³

= (x + 3) (x² – 3x + 9)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (2x)³ + (3y)³

= (2x + 3y) [(2x)² – (2x) (3y) + (3y)²]

= (2x + 3y) (4x² – 6xy + 9y²)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (7)³ + (5b)³

= (7 + 5b) [(7)² – (7) (5b) + (5b)²]

= (7 + 5b) (49 – 35b + 25b²)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (1)³ + (4x)³

= (1 + 4x) [(1)² – 1 (4x) + (4x)²]

= (1 + 4x) (1 – 4x + 16x²)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (5a)³ + ()³

= (5a + ) [ (5a)² – 5a × + ()²]

= (5a + ) (25a² - + )

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (6x)³ + ()³

= (6x + ) [(6x)² – 6x × + ()²]

= (6x + ) (36x² - + )

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= 2x (8x³ + 27)

= 2x [(2x)³ + (3)³]

= 2x (2x + 3) [(2x)² – 2x(3) + 3²]

= 2x (2x + 3) (4x² – 6x + 9)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= 7 (a³ + 8b³)

= 7 (a + 2b) (a² – a × 2b + (2b)²]

= 7 (a + 2b) (a² – 2ab + 4b²)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= x² (x³ + 1)

= x² (x + 1) [(x²) – x (1) + (1)²]

= x² (x + 1) (x² – x + 1)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (a)³ + (0.2)³

= (a + 0.2) [(a)² – a (0.2) + (0.2)²]

= (a + 0.2) (a² – 0.2a + 0.04)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (x²)³ + (y²)³

= (x² + y²) [(x²)² – x²(y²) + (y²)²]

= (x² + y²) (x⁴ – x²y² + y⁴)

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= 2 (a³ + 8b³) – 5 (a + 2b)

= 2 [(a)³ + (2b)³] – 5 (a + 2b)

= 2 (a + 2b) [(a)² – a (2b) + (2b)²] – 5 (a + 2b)

= (a + 2b) [2 (a² – 2ab + 4b²) – 5]

We know that,

a³ + b³ = (a + b) (a² – a × b + b²)

Using this formula, we get

= (x)³ – (8)³

= (x - 8) [(x)² + x (8) + (8)²]

= (x – 8) (x² + 8x + 64)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= (4x)³ – (7)³

= (4x - 7) [(4x)² + 4x (7) + (7)²)

= (4x – 7) (16x² + 28x + 49)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= (1)³ – (3x)³

= (1 – 3x) [(1)² + 1 (3x) + (3x)²)

= (1 – 3x) (1 + 3x + 9x²)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= (x)³ – (5y)³

= (x – 5y) [(x)² + x (5y) + (5y)²

= (x – 5y) (x² + 5xy + 25y²)

We know that,

a³ - b³ = (a - b) (a²+ a × b + b²)

Using this formula, we get

= (2x)³ – ()³

= (2x - ) [(2x)² + 2x × + ()²]

=

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= (a)³ – (0.4)³

= (a – 0.4) [(a)² + a (0.4) + (0.4)²]

= (a – 0.4) (a² + 0.4a + 0.16)

(a + b)³ – (2)³

We know that,

a³ - b³ = (a - b) (a²+ a × b + b²)

Using this formula, we get

= (a + b – 2) [(a + b)² + (a + b) 2 + (2)²]

= (a + b – 2) [a² + b² + 2ab + 2 (a + b) + 4]

We know that,

a³ - b³ = (a - b) (a²+ a × b + b²)

Using this formula, we get

= (x²)³ – (9)³

= (x² – 9) [(x²)² + x²9 + (9)²]

= (x² – 9) (x⁴ + 9x² + 81)

= (x + 3) (x – 3) [(x² + 9)² – (3x)²]

= (x + 3) (x – 3) (x² + 3x + 9) (x² – 3x + 9)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= [a + b – (a – b)] [(a + b)² + (a + b) (a – b) + (a – b)²]

= (a + b – a + b) [a² + b² + 2ab + a² – b² + a² + b² – 2ab]

= 2b (3a² + b²)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= x (1 – 8y³)

= x [(1)³ – (2y)³]

= x (1 – 2y) [(1)² + 1 (2y) + (2y)²]

= x (1 – 2y) (1 + 2y + 4y²)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= 4x (8x³ – 125)

= 4x [(2x)³ – (5)³]

= 4x [(2x – 5) [(2x)² + 2x (5) + (5)²]

= 4x (2x – 5) (4x² + 10x + 25)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= 3a⁴b (a³ – 27b³)

= 3a⁴b [(a)³– (3b)³]

= 3a⁴b (a – 3b) [(a)² + a (3b) + (3b)²]

= 3a⁴b (a – 3b) (a² + 3ab + 9b²)

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= a³ - 1/a³ – 2 (a – 1/a)

= (a – 1/a) (a² + a × 1/a + 1/a²) – 2 (a – 1/a)

= (a – 1/a) (a² + 1 + 1/a² – 2)

=

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= 8a³ – b³ – 2x (2a – b)

= (2a)³ – (b)³ – 2x (2a – b)

= (2a – b) [(2a)² + 2a (b) + (b)²] – 2x (2a – b)

= (2a – b) (4a² + 2ab + b²) – 2x (2a – b)

= (2a – b) (4a² + 2ab + b² – 2x)

We know that,

(a + b)³ = a³ + b³ + 3ab (a + b)

Using this formula, we get

= (a + b)³ – 8

= (a + b)³ – (2)³

We know that,

a³ - b³ = (a - b) (a² + a × b + b²)

Using this formula, we get

= (a + b – 2) [(a+ b)² + 2 (a + b) + 4)]

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (5a)³ + (b)³ + (4c)³ – 3 (5a) (b) (4c)

= (5a + b + 4c) [(5a)² + b² + (4c)² – (5a) (b) – (b) (4c) – (5a) (4c)]

= (5a + b + 4c) (25a² + b² + 16c² – 5ab – 4bc – 20ac)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

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

= (a + 2b + 4c) (a² + 4b² + 16c² – 2ab – 8bc – 4ac)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (1)³ + (b)³ + (2c)³ – 3 (1) (b) (2c)

= (1 + b + 2c) [(1)² + (b)² + (4c)² – (1) (b) – (2b) (c) – (2c) (1)]

= (1 + b + 2c) (1 + b² + 4c² – b – 2bc – 2c)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (6)³ + (3b)³ + (2c)³ – 3 (6) (3b) (2c)

= (6 + 3b + 2c) [(6)² + (3b)² + (2c)² – (6) (3b) – (3b) (2c) – (2c) (6)]

= (6 + 3b + 2c) (36 + 9b² + 4c² – 18ab – 6bc – 12ac)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (3a)³ + (-b)³ + (2c)³ – 3 (3a) (-b) (2c)

= [3a + (-b) + 2c] [(3a)² + (-b)² + (2c)² – (3a) (-b) – (-b) (2c) – (2c) (3a)]

= (3a - b + 2c) (9a² + b² + 4c²+ 3ab + 2bc – 6ca)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

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

= (2a + 5b - 4c) [(2a)² + (5b)² + (-4c)² – (2a) (5b) – (5b) (-4c) – (-4c) (2a)]

= (2a + 5b - 4c) (4a² + 25b² + 16c² – 10ab + 20bc + 8ca)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (2)³ + (-3b)³ + (-7c)³ – 3 (2) (-3b) (-7c)

= (2 - 3b - 7c) [(2)² + (-3b)² + (-7c)² – (2) (-3b) – (-3b) (-7c) – (-7c) (2)]

= (2 - 3b - 7c) (4 + 9b² + 49c² + 6b – 21bc + 14c)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (5)³ + (-2x)³ + (-3y)³ – 3 (5) (-2x) (-3y)

= (5 – 2x – 3y) [(5)² + (-2x)² + (-3y)² – (5) (-2x) – (-2x) (-3y) – (-3y) (5)]

= (5 – 2x – 3y) (25 + 4x² + 9y² + 10x – 6xy + 15y)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

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

= ( + 2b + c) [()² + (2b)² + (c)² – () (2b) – (2b) (c) – (c) (a)]

= ( + 2b + c) (2a² + 8b² + c² – 4ab – 2bc – ac)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (x)³ + (y)³ + (4)³ – 3 (x) (y) (4)

= (x + y + 4) [(x)² + (y)² + (4)² – (x) (y) – (y) (4) – (4) (x)]

= (x + y + 4) (x² + y² + 16 –xy – 4y – 4x)

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

Putting (a – b) = x, (b – c) = y and (c – a) = z, we get

(a - b)³ + (b - c)³ + (c - a)³= x³ + y³ + z³

Where (x + y + z) = (a – b) + (b – c) + (c – a) = 0

= 3xyz [Since, (x + y + z) = 0 so (x³ + y³ + z³) = 3xyz]

= 3 (a – b) (b – c) (c – a)

We know that,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

We have,

(3a – 2b) (2b – 5c) + (5c – 3a) = 0

So,

(3a – 2b)³ + (2b – 5c)³ + (5c – 3a)³ = 3 (3a – 2b) (2b – 5c) (5c – 3a)

We have,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Also,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

= [a (b – c)]³ + [b (c – a)]³ + [c (a – b)]³

Since,

a (b – c) + b (c – a) + c (a – b) = ab – ac + bc – ba + ca – bc = 0

So,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

= 3a (b – c) b (c – a) c (a – b)

= 3abc (a – b) (b – c) (c – a)

We have,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Also,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

Since,

(5a – 7b) + (9c – 5a) + (7b – 9c) = 5a – 7b + 9c – 5a + 7b – 9c = 0

Therefore,

(5a – 7b)³ + (9c – 5a)³ + (7b – 9c)³ = 3 (5a – 7b) (9c – 5a) (7b – 9c)

We have,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Also,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

Using this, we get

= [x + y + (-z)] [(x)² + (y)² + (-z)² – (x) (y) – (y) (-z) – (-z) (x)]

= x³ + y³ – z³ + 3xyz

We have,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Also,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

Using this, we get

= [x + (-2y) + 3] [(x)² + (-2y)² + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]

= (a + b + c) (a² + b² + c² – ab – bc – ca)

= a^# + b³ + c³ – 3abc

Where,

x = a, b = -2y and c = 3

(x – 2y + 3) (x² + 4y² + 2xy – 3x + 6y + 9)

= (x)³ + (-2y)³ + (3)³ – 3 (x) (-2y) (3)

= x³ – 8y³ + 27 + 18xy

We have,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Also,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

Using this, we get

= [x + (-2y) + (-z)] [(x)² + (-2y)² + (-z)² – (x) (-2y) – (-2y) (-z) – (-z) (x)

= (a + b + c) (a² + b² + c² – ab – bc – ca)

= a³ + b³ + c³ – 3abc

Where,

x = a, b = -2y and c = -z

(x – 2y – z) (x² + 4y² + z² + 2xy + zx – 2yz)

= (x)³ + (-2y)³ + (-z)³ – 3 (x) (-2y) (-z)

= x³ – 8y³ – z³ – 6xyz

We have,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Also,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

Given,

x + y + 4 = 0

We have,

(x³ + y³ – 12xy + 64)

= (x)³ + (y)³ + (4)³ – 3 (x) (y) (4) = 0

We have,

a³ (b – c)³ + b³ (c – a)³ + c³ (a – b)³

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Also,

If, (x + y + z) = 0

Then (x³ + y³ + z³) = 3xyz

Given, x = 2y + 6

Or, x – 2y – 6 = 0

We have,

(x³ – 8y³ – 36xy – 216)

= (x³ – 8y³ – 216 – 36xy)

= (x)³ + (-2y)³ + (-6)³ – 3 (x) (-2y) (-6)

= (x – 2y – 6) [(x)² + (-2y)² + (-6)² – (x) (-2y) – (-2y) (-6) – (-6) (x)]

= (x – 2y – 6) (x² + 4y² + 36 + 2xy – 12y + 8x)

= 0 (x² + 4y² + 36 + 2xy – 12y + 6x)

= 0

Polynomials in one variable are algebraic expressions that consist of terms having same variable all through

∴ since in the expression the only polynomial used is x.

Hence, option C is correct

*Note: A polynomial is an expression having variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

Now, and is not a polynomial because it does not contain integer power of the variable “x”.

And, is not a polynomial because it contains a negative power of the variable, which is not the criteria for a polynomial.

∴ is a polynomial it contains only integral powers of the variable (x) i.e. x²

Hence, option D is correct

*Note: A polynomial is an expression having variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables

and is not a polynomial because it does not contain integer power of the variable “y”

And, is not a polynomial because it contains a negative power of the variable, which is not the criteria for a polynomial.

∴ y is a polynomial as it follows the criteria of polynomial and all the other do not follow these criteria.

Hence, option C is correct

As we know that,

A polynomial is an expression having variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables

Out of the 4 given options, we have:

and are not polynomials as they have negative exponent.

is not a polynomial as it has non integral exponent.

And -4 can be written as -4x⁰, thus it is a polynomial with integer power.

Thus, -4 is the correct option.

Out of the 4 given options, D is the correct option because all the other has negative exponential​ value which shows that they aren't a polynomial

A quadratic polynomial is a polynomial of degree 2 which means that it must have a variable with degree 2.

∴ x² + 5x + 4 is a quadratic polynomial

Hence, option D is correct

x + 1 is a linear polynomial because it has degree one which means that highest power of the variable must be 1

Hence, option B is correct

In algebra, a binomial is a polynomial which is the sum of two terms, each of which is a monomial. It is the simplest kind of polynomial after the monomials

∴ x² + 4 is a binomial

Hence, option B is correct

is a polynomial of degree 0 this is because it do not have any variable

Hence, option D is correct

The degree of the zero polynomial is left undefined

Hence, option C is correct

We have,

p (x) = 2x + 3

So, zero of the given polynomial can be calculated as follows:

0 = 2x + 3

2x = - 3

x =

Hence, option B is correct

We have,

p (x) = 2 – 5x

So, zero of the given polynomial can be calculated as follows:

0 = 2 – 5x

5x = 2

x =

Hence, option A is correct

The zero of the zero polynomial is not defined

Hence, option D is correct

We have,

p (x) = x + 4

p (-x) = - x + 4

Then, the value of p (x) + (-x) and it can be calculated as follows:

p (x) + p (-x) = x + 4 – x + 4

= 4 + 4

= 8

Hence, option D is correct

We have,

p (x) = x² - 2x + 1

= 8 – 8 + 1

= 1

Hence, option B is correct

We have,

p (x) = x² + x - 6

So, zero of the given polynomial can be calculated as follows:

x² + x – 6 = 0

x² + 3x – 2x – 6 = 0

x (x + 3) – 2 (x + 3) = 0

(x + 3) (x – 2) = 0

Now, x + 3 = 0

x = - 3

And, x – 2 = 0

x = 2

Hence, the zeros of the given polynomial are -3 and 2

∴ Option C is correct

We have,

p (x) = 2x² + 5x - 3

So, zero of the given polynomial can be calculated as follows:

2x² + 5x – 3 = 0

2x² + 6x – x – 3 = 0

2x (x + 3) – 1 (x + 3) = 0

(2x – 1) (x + 3) = 0

Now, (2x – 1) = 0

2x = 1

Also, (x + 3) = 0

x = -3

Hence, zeros of the given polynomial are and – 3

∴ Option B is correct

We have, (x² + kx – 3) = (x – 3) (x + 1)

So, the value of k can be calculated as follows:

(x² + kx – 3) = x² – 3x + x – 3

x² + kx – 3 = x² – 2x – 3

On comparing the coefficients, we get,

kx = - 2x

k = - 2

Thus, the value of k = - 2

Hence, option B is correct

Let p (x) = 2x² + kx

It is given that, (x + 1) is a factor of (2x² + kx)

Thus, x = -1 is a factor of (2x² + kx)

∴ p (-1) = 0

2 (-1)² + k (-1) = 0

2 – k = 0

k = 2

Thus, the value of k = 2

Hence, option C is correct

We have,

2x² – 4x⁴ + 5x² – x⁵ + 3

From the given polynomial,

The highest power of x = 5

Coefficient of x⁵ = - 1

Hence, option D is correct

Let, p (x) = (x³¹ + 31)

And, x + 1 = 0

x = - 1

It is given that, (x + 1) is a factor of p (x) so the remainder is equal to p (-1)

∴ p (-1) = (-1)³¹ + 31

= - 1 + 31

= 30

Hence, option C is correct

We have,

x³ – ax² + x

Let, p (x) = x³ – ax² + x

And, x - a = 0

x = a

It is given that, (x - a) is a factor of p (x) so the remainder is equal to p (a)

∴ p (a) = (a)³ – a (a)² + a

= a³ – a³ + a

= a

Hence, option B is correct

We have,

(x³ + ax² + 2x + a)

Let, p (x) = (x³ + ax² + 2x + a)

And, x + a = 0

x = - a

It is given that, (x + a) is a factor of p (x) so the remainder is equal to p (-a)

∴ p (-a) = (-a)³ + a (-a)² + 2 (-a) + a

= - a³ + a³ – 2a + a

= - a

Hence, option C is correct

We have,

x⁴ + 2x³ – 3x² + x - 1

Let, p (x) = x⁴ + 2x³ – 3x² + x - 1

And, x - 2 = 0

x = 2

It is given that, (x - 2) is a factor of p (x) so the remainder is equal to p (2)

∴ p (2) = (2)⁴ + 2 (2)³ – 3 (2)² + 2 - 1

= 16 + 16 – 12 + 2 – 1

= 34 – 13

= 21

Hence, option D is correct

We have,

x³ - 3x² + 4x + 32

Let, p (x) = x³ - 3x² + 4x + 32

And, x + 2 = 0

x = - 2

It is given that, (x + 2) is a factor of p (x) so the remainder is equal to p (- 2)

∴ p (- 2) = (- 2)³ - 3 (- 2)² + 4 (- 2) + 32

= - 8 - 12 – 8 + 32

= - 28 + 32

= 4

Hence, option D is correct

We have,

4x³ - 12x² + 11x - 5

Let, p (x) = 4x³ - 12x² + 11x - 5

And, (2x – 1) = 0

2x = 1

x =

It is given that, (2x - 1) is a factor of p (x) so the remainder is equal to p()

∴ p (2) = 4 ()³ - 12 ()² + 11 () - 5

= 4 × - 12 × + 11 × - 5

= - 3 + - 5

= – 8

= 6 – 8

= - 2

Hence, option C is correct

We have, (x + 1) = 0

x = - 1

Firstly, putting (x = - 1) in x³ – 2x² + x + 2 we get:

= (- 1)³ – 2 (-1)² + (-1) + 2

= - 1 – 2 – 1 + 2

= - 2

∴ (x + 1) is not a factor of x³ – 2x² + x + 2

Secondly, putting (x = - 1) in x³ + 2x² + x - 2 we get:

= (-1)³ + 2 (-1)² + (-1) – 2

= - 1 + 2 – 1 – 2

= - 2

∴ (x + 1) is not a factor of x³ + 2x² + x – 2

Thirdly, putting (x = - 1) in x³ + 2x² – x – 2 we get:

= (-1)³ + 2 (-1)² – (-1) – 2

= - 1 + 2 + 1 – 2

= 0

Hence, (x + 1) is a factor of x³ + 2x² + x – 2

Thus, option C is correct

We have,

4x² + 4x – 3

= 4x² – 2x + 6x – 3

= 2x (2x – 1) + 3 (2x – 1)

= (2x – 1) (2x + 3)

Hence, option D is correct

We have,

6x² + 17x + 5

= 6x² + 2x + 15x + 5

= 2x (3x + 1) + 5 (3x + 1)

= (2x + 5) (3x + 1)

Hence, option D is correct

We have,

x² - 4x – 21

= x² + 3x - 7x – 21

= x (x + 3) - 7 (x + 3)

= (x + 3) (x - 7)

Hence, option C is correct

We have,

p (x) = x³ – 20x + 5k

It is given in the question that, (x + 5) is a factor of p (x) so:

p (- 5) = 0

(- 5)³ – 20 (- 5) + 5 k = 0

- 125 + 100 + 5k = 0

- 25 + 5k = 0

5k = 25

k =

k = 5

Hence, option B is correct

We have,

3x³ + 2x² + 3x + 2

= x² (3x + 2) + 1 (3x + 2)

= (x² + 1) (3x + 2)

Hence, option D is correct

We have,

= - 1

x² + y² = - xy

x² + y² + xy = 0

∴ Value of (x³ – y³) = (x – y) (x² + y² + xy)

= (x - y) × 0

= 0

Hence, option C is correct

The correct answer is D

It is given that: a + b + c = 0

We know,

(a + b + c)³ = a³ + b³ + c³ + 2(ab + bc + ac)(a + b + c)

Put a + b + c = 0 in the above equation we get,

Then, a³ + b³ + c³ = 3 abc

Hence, option D is correct

We have,

(x³ + 10x²mx + n)

Let, p (x) = x³ + 10x² + mx + n

It is given in the question that (x + 2) and (x – 1) are the factors of p (x)

∴ p (-2) = 0

(-2)³ + 10 (-2)² + m (-2) + n = 0

- 8 + 40 – 2m + n = 0

32 – 2m + n = 0

2m – n – 32 = 0 ……….(i)

And,

p (1) = 0

(1)³+ 10 (1)² + m (1) n

1 + 10 + m + n = 0

11 + m + n = 0

m + n + 11 = 0 ………. (ii)

Now, adding equation (i) and (ii) we get

2m – n - 32 + m + n + 11 = 0

3m - 21 = 0

3m = 21

m =

m = 7

Now, putting the value of m in (ii) we get:

7 + n + 11 = 0

18 + n = 0

n = - 18

∴ the value of m is 7 and that of n is – 18

Hence, option B is correct

We have,

(369)² – (368)²

We know that,

(a² – b²) = (a+ b) (a – b)

Using this identity, we get:

(369)² – (368)² = (369 + 368) (369 – 368)

= 737 × 1

= 737

Hence, option D is correct

We have,

104 × 96 = (100 + 4) (100 – 4)

= (100)² – (4)²

= 10000 – 16

= 9984

Hence, option B is correct

We have,

4a² + b² + 4ab + 8a + 4b + 4

We know that,

x² + y² + z² + 2xy + 2yz + 2xz = (x + y + z)²

= (2a)² + (b)² + (2)² + 2 (2a) (b) + 2 (b) (2) + 2 × 2 (2a)

= (2a + b + 2)²

Hence, option A is correct

The coefficient of x in the expansion of (x+3)² can be calculated as follows:

(x + 3)³ = x³ + (3)³ + 3 (x) (3) (x + 3)

= x³ + 27 + 9x (x + 3)

= x³ + 27 + 9x² + 27x

∴ Coefficient of x is 27

Hence, option D is correct

It is given in the question that,

a + b + c = 0

So,

= 3

Hence, option D is correct

It is given that,

x + y + z = 9

And, xy + yz + zx = 23

As we know that,

(x + y + z)² = (x² + y² + z² + 2xy + 2yz + 2zx)

∴ (9)² = [x² + y² + z² + 2 (xy + yz + zx)]

x² + y² + z² = 81 – 2 × 23

x² + y² + z² = 81 – 46

x² + y² + z² = 35

We also know that:

(x³ + y³ + z³ – 3xyz) = (x + y + z) [x² + y² + z² – (xy + yz + zx)]

= 9 (35 – 23)

= 9 × 12

= 108

Hence, option A is correct

Let p (x) = x¹⁰⁰ + 2x⁹⁹ + k

It is given in the question that, (x + 1) is divisible by (x + 1)

So, p (-1) = 0

(-1)¹⁰⁰ + 2 (-1)⁹⁹ + k = 0

1 + 2 (-1) + k = 0

1 – 2 + k = 0

- 1 + k = 0

k = 1

Thus, the value of k = 1

Hence, option A is correct

We know that,

In any polynomial in x, the indices of x must be a non-negative integer

Hence, option C is correct

We have,

p (x) = 2x³ – kx² + 3x + 10

It is given in the question that (x + 2) is exactly divisible by p (x)

∴ p (-2) = 0

2 (-2)³ – k (-2)² + 3 (-2) + 10 = 0

2 × (-8) – k × (4) – 6 + 10 = 0

- 16 – 4k – 6 + 10 = 0

- 22 – 4k + 10 = 0

- 12 – 4k = 0

- 12 = 4k

k =

k = - 3

Hence, option D is correct

We have,

207 × 193 = (200 + 7) (200 – 7)

= (200)² – (7)²

= 40000 – 49

= 39951

Hence, option B is correct

We have,

305 × 308 = 305 × (300 + 8)

= 305 × 300 + 305 × 8

= 91500 + 2440

= 93940

Hence, option C is correct

We have,

p (x) = x² – 3x

∴ Zeros of p (x) are:

p (x) = 0

x² – 3x = 0

x (x – 3) = 0

So, x = 0

And, x – 3 = 0

x = 3

Thus, zeros of the given polynomial are 0 and 3

Hence, option B is correct

We have,

p (x) = 3x² – 1

∴ Zeros of p (x) are:

p (x) = 0

3x² – 1 = 0

3x² = 1

So, x² =

And, x =

Thus, zeros of the given polynomial are and -

Hence, option D is correct

The correct answer is: (a)

We have,

p (x) = x² + kx + 1

(x – 1) is a factor of p (x)

∴ p (1) = 0

(1)² + k (1) + 1 = 0

1 + k + 1 = 0

2 + k = 0

k = - 2

Hence, both Assertion (A) and Reason (R) are true also Reason (R) is the correct explanation of Assertion (A)

∴ Option A is correct

We have,

p (x) = x³ – ax² + 6x - a

(x – a) is divided by p (x)

∴ p (a) = (a)³ – a (a)² + 6 (a) - a

= a³ – a³ + 6a - a

= 5a

Hence, both Assertion (A) and Reason (R) are true also Reason (R) is the correct explanation of Assertion (A)

∴ Option A is correct

The correct answer is: (B)

We have,

p (x) = x³ – 2x + 3k

(x – 2) is a factor of p (x)

∴ p (2) = 0

(2)³ – 2 (2) + 3k = 0

8 – 4 + 3k = 0

4 + 3k = 0

3k = - 4

k =

Hence, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A)

∴ Option B is correct

The correct answer is: (a)

We have,

(25)³ + (-16)³ + (-9)³

Since, 25 – 16 – 9 = 25 – 25

= 0

∴ 3 (-25) (-16) (-9)

= 75 × 144

= 10800

Hence, both Assertion (A) and Reason (R) are true also Reason (R) is the correct explanation of Assertion (A)

∴ Option A is correct

The correct match for the above is as follows:

Hence, the correct answer is:

(a) – (r)

(b) – (p)

(c) – (s)

(d) – (q)

The correct match for the above is as follows:

Hence, the correct answer is:

(a) – (r)

(b) – (p)

(c) – (s)

(d) – (q)

We have,

p (x) = 3x³ + 4x² – 5x + 8

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

= 3 × -8 + 4 × 4 + 10 + 8

= - 24 + 16 + 10 + 8

= - 24 + 34

= 10

We have,

p (x) = 4x³ + 8x² – 17x + 10

Also, 2x – 1 = 0

2x = 1

x =

∴ p () = 4 ()³ + 8 ()² – 17 () + 10

= 4 × + 8 × - 17 × + 10

= + 2 – + 10

= + 12

= + 12

= - 8 + 12

= 4

Hence, remainder = 4

It is given in the question that (x – 2) is a factor of 2x³ – 7x² + 11x + 5a

Let f (x) = 2x³ – 7x² + 11x + 5a

Now, x – 2 = 0

x = 2

∴ f (2) = 0

2 (2)³ – 7 (2)² + 11 (2) + 5a = 0

16 – 28 + 22 + 5a = 0

38 – 28 + 5a = 0

10 + 5a = 0

5a = - 10

a =

a = - 2

Hence, the value of a = - 2

It is given in the question that (x + 2) is a factor of x³ – 2mx² + 16

p (x) = x³ – 2mx² + 16

Now, x + 2 = 0

x = - 2

∴ f (-2) = 0

(- 2)³ – 2m (- 2)² + 16 = 0

- 8 – 8m + 16 = 0

- 8 + 8m = 0

- 8 = - 8m

m =

m = 1

Hence, the value of m = 1

It is given in the question that,

(a + b + c) = 8

And, (ab + bc + ca) = 19

We know that,

(a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)

∴ (8)² = a² + b² + c² + 2 × 19

64 = a² + b² + c² + 38

64 – 38 = a² + b² + c²

a² + b² + c² = 26

We have,

(3a + 4b + 5c)²

We know that,

(a + b + c)² = a² + y² + c² + 2 (xy + yz + zx)

∴ (3a + 4b + 5c)² = (9a² + 16b² + 25c² + 2 × 3a × 4b + 2 × 4b × 5c + 2 × 5c × 3a)

= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

We have,

(3x + 2)²

We know that,

(a + b)³ = a³ + b³ + 3ab (a + b)

∴ (3x + 2)² = (3x)³ + (2)³ + 3 (3x) (2) (3x + 2)

= 27x³ + 8 + 18x (3x + 2)

= 27x³ + 8 + 54x² + 36x

= 27x³ + 54x² + 36x + 8

We have,

[(28)³ + (-15)³ + (-13)³]

Now putting a = 28, b = - 15 and c = - 13

Now, a + b + c = 28 – 15 – 13

= 28 – 28

= 0

So, x³ + y³ + z³ = 3xyz

(28)³ + (-15)³ + (-13)³ = 3 × (28) × (-15) × (-13)

= 84 × 195

= 16380

It is given in the question that (x + 1) is divided by (x⁶⁰ + 60)

Let f (x) = x⁶⁰ + 60

And, (x + 1) = 0

x = - 1

∴ f (x) = x⁶⁰ + 60

f (-1) = (-1)⁶⁰ + 60

= 1 + 60

= 61

Hence, remainder = 61

Thus, option C is correct

We have,

(36x² – 1) + (1 + 6x)²

= [(6x)² – (1)²] + (1 + 6x)²

= (6x – 1) (6x + 1) + (6x + 1) (6x + 1)

= (6x + 1) (6x – 1 + 1 + 6x)

= (6x + 1) (12x)

∴ (6x + 1) is a factor of (36x² – 1) + (6x + 1)²

Hence, option B is correct

It is given in the question that,

a² + b² = - ab

a² + b² + ab = 0 (i)

We know that,

a³ – b³ = (a – b) (a² + b² + ab)

From (i), we have a² + b² + ab = 0

∴ a³ – b³ = (a – b) × 0

a³ – b³ = 0

Hence, option D is correct

We have,

(x + 5)³ = x³ + (5)³ + 3 × x × 5 (x + 5)

= x³ + 125 + 15x (x + 5)

= x³ + 125 + 15x² + 75x

∴ Coefficient of x = 75

Hence, option D is correct

We know that,

is a polynomial of degree 0

Hence, option C is correct

Let f(x) = 2x² + 7x – 4

= 2x² + 8x – x – 4

= 2x (x + 4) – 1 (x + 4)

= (2x – 1) (x + 4)

So, 2x – 1 = 0

2x = 1

x =

And, x + 4 = 0

x = - 4

Thus, one of zero of the given polynomial is

Hence, option B is correct

We know that,

The zero of the zero polynomial is not defined

Hence, option D is correct

We have,

p (x) = ax³ + x² – 2x + b

It is given in the question that,

(x + 1) and (x – 1) are the factors of p (x)

∴ p (-1) = p (1)

So, p (-1) = a (-1)³ + (-1)² – 2 (-1) + b

0 = - a + 1 + 2 + b

0 = - a + 3 + b (i)

Also, p (1) = a (1)³ + (1)² – 2 (1) + b

0 = a + 1 – 2 + b

0 = a + b – 1 (ii)

As, p (-1) = p (1)

- a + 3 + b = a + b - 1

- 2a = - 4

a = 2

Now, putting the value of a in (ii), we get:

2 + b – 1 = 0

1 + b = 0

b = - 1

Hence, the value of a is 2 and that of b is -1

We have,

p (x) = ax³ + bx² + x – 6

It is given in the question that, (x + 1) is a factor of p (x)

x + 2 = 0

x = - 2

∴ f (-2) = 0

a (-2)³ + b (-2)² + (-2) – 6 = 0

- 8a + 4b – 2 – 6 = 0

- 8a + 4b – 8 = 0

- 4 (2a – b + 2) = 0

2a – b + 2 = 0 (i)

Also, it is given in the question that when p (x) is divided by (x – 2) then it leaves a remainder 4

∴ p (2) = 4

a (2)³ + b (2)² + (2) – 6 = 4

8a + 4b + 2 – 6 = 4

8a + 4b – 4 - 4= 0

4 (2a + b – 2) = 0

2a + b – 2 = 0 (ii)

Now, adding (i) and (ii) we get:

2a – b + 2 + 2a + b – 2 = 0

4a = 0

a = 0

Putting the value of a in (ii), we get:

2 (0) + b – 2 = 0

b – 2 = 0

b = 2

Hence, it is proved that the value of a is 0 and that of b is 2

We have,

(3x – 5)³ = (3x)³ – (5)³ – 3 (3x) (5) (3x – 5)

= 27x³ – 125 – 45x (3x – 5)

= 27x³ - 125 – 135x² + 225x

= 27x³ – 135x² + 225x - 125

Hence, option D is correct

It is given in the question that,

a + b + c = 5

And, ab + bc + ca = 0

We know that,

(a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)

Putting the given values, we get:

(5)² = a² + b² + c² + 20

25 – 20 = a² + b² + c²

a² + b² + c² = 5 (i)

Also, we know that

(a³ + b³ + c³ – 3abc) = (a + b + c) (a² + b² + c² – ab – bc – ca)

= (a + b + c) [(a² + b² + c²) – (ab + bc + ca)]

Now, putting the values we get:

= (5) × (5 – 10)

= 5 × (-5)

= - 25

Hence, it is proved that

(a³ + b³ + c³ – 3abc) = - 25

We have,

p (x) = 2x³ + ax² + 3x – 5

q (x) = x³ + x² – 4x + a

It is given in the question that, when p (x) and q (x) is divided by (x – 2) it leaves same remainder

∴ p (2) = q (2)

2 (2)³ + a (2)² + 3 (2) – 5 = (2)³ + (2)² – 4 (2) + a

2 × 8 + a × 4 + 3 × 2 – 5 = 8 + 4 – 4 × 2 + a

16 + 4a + 6 – 5 = 12 – 8 + a

16 + 4a + 1 = 4 + a

4a – a = 4 – 17

3a = - 13

a =

Hence proved

(i) We have,

Now,

Put x = 0

p(0) = 5 - 4(0) + 2(0)²

= 5 - 4(0) + 2(0)

= 5 – 0 + 0

= 5

Hence,

P(0) = 5

(ii) We have,

Now,

Put x = 3

p(3) = 5 - 4(3) + 2(3)²

= 5 - 4(3) + 2(9)

= 5 – 12 + 18

= 11

Hence,

P (3) = 11

(iii) We have,

Now,

Put x = -2

= 5 - 4(-2) + 2(4)

= 5 + 8 + 8

= 21

Hence,

P (-2) = 21

(i) We have,

Now,

Put y = 0

= 4 - 3(0) - (0) + 5(0)

= 4 – 0 – 0 + 0

= 4

Hence,

P (0) = 4

(ii) We have,

Now,

Put y = 2

²³

=

=

=

Hence,

P (2) = 46

(iii) We have,

Now,

Put y = -1

²³

=

=

=

Hence,

p(-1) = -5

(i) We have,

Now,

Put t = 0

f(0) = 4(0)² - 3(0) + 6

= 4(0) - 3(0) + 6

= 0 – 0 + 6

= 6

Hence,

f(0) = 6

(ii) We have,

Now,

Put t = 4

f(4) = 4(4)² -3(4) + 6

= 4(16) - 3(4) + 6

= 64 – 12 + 6

= 58

Hence,

f(4) = 58

(iii) We have,

Now,

Put t = -5

f(-5)=4(-5)² - 3(-5) + 6

= 4(25) - 3(-5) + 6

= 100 + 15 + 6

= 121

Hence,

f(0) = 6

(i) At first,

In order to find the zero of the polynomial we will,

Put p(x) = 0

Now,

We have,

P(x)=x-5

0 = x – 5

x = 5

Hence, 5 is the zero of the given polynomial.

(ii) At first,

In order to find the zero of the polynomial we will,

Put q(x) = 0

Now,

We have,

q(x)=x+4

0 = x + 4

x = - 4

Hence, -4 is the zero of the given polynomial.

(iii) At first,

In order to find the zero of the polynomial we will,

Put p(t) = 0

Now,

We have,

P(t)= 2t - 3

0 = 2t – 3

2t = 3

t = 3/2

Hence, 3/2 is the zero of the given polynomial.

(iv) At first,

In order to find the zero of the polynomial we will,

Put f(x) = 0

Now,

We have,

f(x)= 3x + 1

0 = 3x + 1

3x = -1

x =

Hence, -1/3 is the zero of the given polynomial.

(v) At first,

In order to find the zero of the polynomial we will,

Put g(x) = 0

Now,

We have,

g(x)= 5 – 4x

0 = 5 – 4x

4x = 5

x = 5/4

Hence, 5/4 is the zero of the given polynomial.

(vi) At first,

In order to find the zero of the polynomial we will,

Put h(x) = 6x - 1

Now,

We have,

h(x)= 6x- 1

0 = 6x – 1

6x = 1

x = 1/6

Hence, 1/6 is the zero of the given polynomial.

(vii) At first,

In order to find the zero of the polynomial we will,

Put p(x) = 0

Now,

We have,

p(x)= ax + b

0 = ax + b

ax = -b

x = (-b)/a

Hence, (-b)/a is the zero of the given polynomial.

(viii) At first,

In order to find the zero of the polynomial we will,

Put q(x) = 0

Now,

We have,

q(x)= 4x

0 = 4x

x = 0

Hence, 0 is the zero of the given polynomial.

(ix) At first,

In order to find the zero of the polynomial we will,

Put p(x) = 0

Now,

We have,

p(x)= ax

0 = ax

x = 0

Hence, 0 is the zero of the given polynomial.

(i) We have, p(x) = x – 4

In order to verify the zero of the polynomial,
Put p(x) = 4
put x = 4 in the expression , we get,
p(4) = 4 – 4
p(4) = 0
Since p(4) = 0

Hence, 4 is a zero of the polynomial p(x).

(ii) We have, p(x) = x + 3

In order to verify the zero of the polynomial,
Put p(x) = -3
p(-3) = - 3 + 3
p(-3) = 0
Since p(-3) = 0

Hence, -3 is a zero of the polynomial p(x).

(iii) We have, p(y) = 2y + 1

In order to verify the zero of the polynomial,
Put p(y) = 1/2

p(1/2) = 2(1/2) + 1

p(1/2) = 1 + 1
p(1/2) = 2

Since p(1/2) ≠ 0

Hence, 1/2 is not a zero of the polynomial p(y)

(iv) We have, p(x) = 2 - 5x

In order to verify the zero of the polynomial,
Put p(x) = 2/5
p(2/5) = 2 - 5 (2/5)
p(2/5)= 2 - 2
p(2/5)= 0

Since p(2/5) = 0

Hence, 2/5 is a zero of the polynomial p(x)

(v) We have, p(x) = (x-1)(x-2)

In order to verify the zero of the polynomial,

Case 1: Put x = 1, we get,

p(1) = (1-1)(1-2)
p(1) = 0(-1)
p(1) = 0
Hence, 1 is a zero of the polynomial p(x)

Case 2: Put x = 2, we get,

p(3) = (2-1)(2-2)
p(3) = (1)0
p(3) = 0
since p(3) = 0
Hence, 2 is a zero of the polynomial p(x)

(vi) We have, p(x) = x² -3x
In order to verify the zero of the polynomial,

Case 1: Put x = 0

p(0) = (0)² – 3(0)
p(0) = 0 - 0
p(0) = 0

Since p(0) = 0
Hence, 0 is a zero of the polynomial p(x)

Case 2: Put x = 3
p(3) = (3)² – 3(3)
p(3) = 9 - 9
p(3) = 0

Since p(3) = 0
Hence, 3 is a zero of the polynomial p(x)

(vii) We have, p(x) =x² + x - 6

In order to verify the zero of the polynomial,

Case 1: Put x = 2
p(2) = (2)² + 2 - 6
p(2) = 4 + 2 - 6
p(2) = 0

Since p(2) = 0

Hence, 2 is a zero of the polynomial p(x)

Case 2: Put x = -3

p(3) = (-3)² + (–3) – 6
p(3) = 9 - 3 - 6
p(3) = 0

Since p(-3) = 0

Hence, -3 is a zero of the polynomial p(x)

Let, f(x) = x³ – 6x² + 9x + 3

Now,

As per the question,

x – 1 = 0

x = 1

Using Remainder theorem,

We know that when f(x)is divided by (x – 1), the remainder so obtained will be f(1).

Hence,

f(1) = (1)³ – 6(1)² + 9(1) + 3

= 1 – 6 + 9 + 3

= 13 – 6

= 7

Therefore,

The required remainder is 7

Let, f(x) = 2x³ – 5x² + 9x - 8

Now,

As per the question,

x – 3 = 0

x = 3

Using Remainder theorem,

We know that when f(x)is divided by (x – 3), the remainder so obtained will be f(3).

Hence,

f(3) = 2(3)³ – 5(3)² + 9(3) - 8

= 2(27) – 5(9) + 27 - 8

= 54 – 45 + 19

= 28

Therefore,

The required remainder is 28

Let, f(x) = 3x⁴ – 6x² - 8x + 2

Now,

As per the question,

x – 2 = 0

x = 2

Using Remainder theorem,

We know that when f(x)is divided by (x – 2), the remainder so obtained will be f(2).

Hence,

f(2) = 3(2)⁴ – 6(2)² - 8(2) + 2

= 3(16) – 6(4) -16 + 2

= 48 -24 – 14

= 10

Therefore,

The required remainder is 10.

Let, f(x) = x³ – 7x² + 6x + 4

Now,

As per the question,

x – 6 = 0

x = 6

Using Remainder theorem,

We know that when f(x) is divided by (x – 6), the remainder so obtained will be f(6).

Hence,

f(6) = (6)³ – 7(6)² + 6(6) + 4

= (216) – 7(36) + 36 + 4

= 256 -252

= 4

Therefore,

The required remainder is 4.

Let, f(x) = x³ – 6x² + 13x + 60

Now,

As per the question,

x + 2 = 0

x = -2

Using Remainder theorem,

We know that when f(x)is divided by (x + 2), the remainder so obtained will be f(-2).

Hence,

f(-2) = (-2)³ – 6(-2)² + 13(-2) + 60

= - 8 – 6(4) - 26 + 60

= 60 - 58

= 2

Therefore,

The required remainder is 2.

Let, f(x) = 2x⁴ + 6x³ + 2x² + x - 8

Now,

As per the question,

x - 3 = 0

x = 3

Using Remainder theorem,

We know that when f(x)is divided by (x – 3), the remainder so obtained will be f(3).

Hence,

f(3) = 2(3)⁴ + 6(3)³ + 2(3)² + 3 - 8

= 2(81) + 6(27) +18 – 5

= 18 - 11

= 7

Therefore,

The required remainder is 7.

Let, f(x) = 4x³ – 12x² + 11x -5

Now,

As per the question,

2x – 1 = 0

2x = 1

x =

Using Remainder theorem,

We know that when f(x)is divided by (2x – 1), the remainder so obtained will be .

Hence,

= -4/2

= -2

Therefore,

The required remainder is -2.

Let, f(x) = 81x⁴ + 54x³ – 9x² - 3x + 2

Now,

As per the question,

3x + 2 = 0

3x = -2

x =

Using Remainder theorem,

We know that when f(x)is divided by (2x – 1), the remainder so obtained will be .

Hence,

= 16 – 16 + 4 - 4

= 0

Therefore,

The required remainder is 0.

Let, f(x) = x³ - ax² + 2x - a

Now,

As per the question,

x - a = 0

x = a

Using Remainder theorem,

We know that when f(x)is divided by (x – a), the remainder so obtained will be f(a).

Hence,

f(a) = a³ – a(a)² + 2(a) – a

= a³ – a³ + 2a – a

= 2a - a

= a

Therefore,

The required remainder is a.

Let f(x) = ax³ + 3x² – 3

And,

g(x) = 2x³ - 5x + a

Now,

f(4) = a(4)³ + 3(4)² – 3

= 64a + 48 - 3

= 64a + 45

And,

g(4) = 2(4)³ – 5(4) + a

= 128 – 20 + a

= 108 + a

According to the question,

f(4) = g(4)

64a + 45 = 108 + a

64a – a = 108 – 45

63a = 63

a = 1

Hence, the value of ‘a’ is 1.

Let f(x) = x⁴ – 2x³ + 3x² – ax + b

Now,

f(1) = 1⁴ – 2(1)³ + 3(1)² – a(1) + b

5 = 1 – 2 + 3 – a + b

3 = - a + b (i)

And,

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

19 = 1 + 2 + 3 + a + b

13 = a + b (ii)

Now,

Adding (i) and (ii),

8 + 2b = 24

2b = 16

b = 8

Now,

Using the value of b in (i)

3 = - a + 8

a = 5

Hence,

a = 5 and b = 8

Hence,

f(x) = x⁴ – 2(x)³ + 3(x)² – a(x) + b

= x⁴ – 2x³ + 3x² – 5x + 8

f(2) = (2)⁴ – 2(2)³ + 3(2)² – 5(2) + 8

= 16 – 16 + 12 – 10 + 8

= 20 – 10

= 10

Therefore, remainder is 10

From the factor theorem, we have

(x – 2) is the factor of f(x) if f(2) = 0

Here, we have

f(2) = (2)³ – 8

= 8 – 8

= 0

Therefore,

(x – 2) is a factor of (x³ – 8)

From the factor theorem, we have

(x – 3) is the factor of f(x) if f(3) = 0

Here, we have

f(3) = 2 × 3³ + 7 × 3² – 24 × 3 - 45

= 54 + 63 – 72 – 45

= 117 – 117

= 0

Therefore,

(x – 3) is a factor of (2x³ + 7x² – 24x – 45)

From the factor theorem, we have

(x – 1) is the factor of f(x) if f(1) = 0

Here, we have

f(1) = 2 × 1⁴ + 9 × 1³ + 6 × 1² – 11 × 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17

= 0

Therefore,

(x – 1) is the factor of (2x⁴ + 9x³ + 6x² – 11x – 6)

From the factor theorem, we have

(x + 2) is the factor of f(x) if (f-2) = 0

Here, we have

f(-2) = (-2)⁴ – (-2)² – 12

= 16 – 4 – 12

= 16 – 16

= 0

Therefore,

(x + 2) is a factor of (x⁴ – x² – 12)

From the factor theorem, we have

(x + 5) is the factor of f(x) if f(-5) = 0

Here, we have

f(-5) = 2(-5)³ + 9(-5)² – 11(-5) – 30

= -250 + 225 + 55 – 30

= -280 + 280

= 0

Therefore,

(x + 5) is the factor of (2x³ + 9x² – 11x – 30)

From the factor theorem, we have

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

Here, we have

2x – 3 = 0

x =

= 0

Therefore,

(2x – 3) is a factor of (2x⁴ + x³ – 8x² – x + 6)

From the factor theorem, we have

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

Here, we have

= 14 – 8 – 6

= 14 – 14

= 0

Therefore,

(x - ) is a factor of (7x² - 4√2x – 6)

From the factor theorem, we have

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

Here, we have

= 0

Therefore,

(x + √2) is the factor of (2√2x² + 5x + √2)

Let, f(x) = 2x³ + 9x² + x + k

Now, we have

x – 1 = 0

x = 1

Hence,

f(1) = 2 × 1³ + 9 × 1² + 1 + k

= 2 + 9 + 1 + k

= 12 + k

As per the question,

(x – 1) is the factor of f(x)

Now, by using factor theorem we get

(x – a) will be a factor of f(x) only if f(a) = 0 and hence f(1) = 0

So,

f(1) = 0

0 = 12 + k

k = -12

Let, f(x) = 2x³ – 3x² – 18x + a

Now, we have

x - 4 = 0

x = 4

Hence,

f(4) = 2(4)³ – 3(4)² – 18 × 4 + a

= 128 – 48 – 72 + a

= 128 – 120 + 8

= 8 + a

As per question,

(x - 4) is the factor of f(x)

Now, by using factor theorem we get

(x – a) will be the factor of f(x) if f(a) = 0 and hence f(4) = 0

So,

f(4) = 8 + a = 0

a = -8

Let, f(x) = x⁴ – x³ – 11x² - x + a

Now, we have

x + 3 = 0

x = -3

Hence,

f(-3) = (-3)⁴ – (-3)³ – 11(-3)² – (-3) + a

= 81 + 27 – 11 × 9 + 3 + a

= 81 + 27 – 99 + 3 + a

= 111 – 99 + a

= 12 + a

As per question,

(x + 3) is the factor of f(x)

Now, by using factor theorem we get

(x – a) will be the factor of f(x) if f(a) = 0 and hence f(-3) = 0

So,

f(-3) = 12 + a = 0

a = -12

Let, f(x) = 2x³ + ax² + 11x + a + 3

Now, we have

2x – 1 = 0

x = 1/2

As per the question,

f(x) is exactly divisible by 2x – 1 which means that 2x – 1 is a factor of f(x)

Hence,

Using factor theorem,

(x – a) will be a factor of f(x) if f(a) = 0 and hence,

f() 0

Hence,

= 5a = -35

a = -7

Thus, the value of a is -7

Let f(x) = x³ – 10x² + ax + b

Now,

By using factor theorem,

(x – 1) and (x - 2) will be the factors of f(x) if f(1) = 0 and f(2) = 0

Hence,

f(1) = 1³ – 10 (1)² + a × 1 + b
0 = 1 – 10 + a + b

a + b = 9 (i)

And,

f(2) = 2³ – 10 × 2² + a × 1 + b

0 = 8 – 40 + 2a + b

2a + b = 32 (ii)

Now, subtracting (i) from (ii)

a = 23

Using the value of a in (i), we get

23 + b = 9

b = 9 – 23

b = -14

Hence,

a = 23 and b = -14

Let f(x) = x⁴ + ax³ – 7x² - 8x + b

Now,

(x + 2) = 0

x = -2

And,

(x + 3) = 0

x = -3

Now,

By using factor theorem,

(x + 2) and (x + 3) will be the factors of f(x) if f(-2) = 0 and f(-3) = 0

Hence,

f(-2) = (-2)⁴ + a(-2)³- 7 (-2)² -8 (-2) + b
0 = 16 – 8a - 28 + 16 + b

8a - b = 4 (i)

And,

f(-3) = (-3)⁴ + a (-3)³ – 10 (-3)² – 8 (-3) + b

0 = 81 – 27a – 63 + 24 + b

27a – b = 42 (ii)

Now, subtracting (i) from (ii)

19a = 38

a = 2

Using the value of a in (i), we get

8 (2) – b = 4

16 – b = 4

b = 12

Therefore,

a = 2 and b = 12

Let f(x) = x³ – 3x² – 13x + 15

Now, we have

x² + 2x – 3 = 0

x² + 3x – x – 3 = 0

(x + 3)(x – 1)

Hence, f(x) will be exactly divisible by x² + 2x – 3 = (x + 3)(x – 1)

Now,

Using factor theorem,

If (x + 3) and (x – 1) are both the factors of f(x), then f(-3) = 0 and f(1) = 0

Now,

f(-3) = (-3)³ – 3(-3)² – 13(-3) + 15

= - 27 – 27 + 39 + 15

= - 54 + 54

= 0

And,

f(1) = (1)³ – 3(1)² – 13(1) + 15

= 1 – 3 - 13 + 15

= 16 - 16

= 0

Now,

Since, f(-3) = 0 and f(1) = 0

Therefore, x² + 2x -3 divides f(x) completely.

Let f(x) = (x³ + ax² + bx + 6)

By using remainder theorem,

If we divide f(x) by (x – 3) then it will leave a remainder as f(3)

So,

f(3) = 3² + a × 3² + b × 3 + 6 = 3

27 + 9a + 3b + 6 = 3

9a + 3b + 33 = 3

9a + 3b = 3 – 33

9a + 3b = -30

3a + b = -10 (i)

It is also given that,

(x – 2) is a factor of f(x)

Therefore,

By using factor theorem, we get

(x – a) is the factor of f(x) if f(a) = 0and also f(2) = 0

Now,

f(2) = 2³ + a × 2² + b × 2 + 6 = 0

8 + 4a + 2b + 6 = 0

4a + 2b = -14

2a + b = -7 (ii)

Now by subtracting (ii) from (i), we get

a = -3

Putting the value of a in (i), we get

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

Therefore,

b = -1 and a = -3

We have,

At first, we’ll take common from the expression

Hence,

The given expression can be factorized as:

3x(3x + 4y)

We have,

At first we’ll take common from the expression

Hence,

The given expression can be factorized as:

We have,

At first we’ll take common from the expression

Hence,

The given expression can be factorized as:

We have,

At first, we’ll take common from the expression

Hence,

The given expression can be factorized as:

We have,

At first, we’ll take common from the expression

=2[x (p² + q²) + 2y (p² + q²)]

=

Hence,

The given expression can be factorized as:

We have,

At first we’ll take common from the expression

= (a – 5) (x – y)

Hence,

The given expression can be factorized as:

We have,

At first, we’ll take (a+b) common from the expression.

= (a + b) [4 – 6 (a + b)]

= 2(a + b) (2 – 3a – 3b)

Hence,

The given expression can be factorized as: