Class 9 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 2: Polynomials

Correct option: (d)

P(x) = 5x - 4x² + 3

⇒ p(-1) = 5(-1) - 4(-1)² + 3 = -5 - 4 + 3 = -6

Correct option: (d)

Let f(x) = x⁵¹ + 51

By the remainder theorem, when f(x) is divided by (x + 1), the remainder is f(-1).

Now, f(-1) = [(-1)ⁿ + 51] = -1 + 51 = 50

Correct option: (c)

Let p(x) = 2x² + kx

Since (x + 1) is a factor of p(x),

P(–1) = 0

⇒ 2(–1)² + k(–1) = 0

⇒ 2 – k = 0

⇒ k = 2

Correct option: (b)

p(x) = 2x + 5

Now, p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Correct option: (c)

p(x) = 2x² + 7x - 4

Now, p(x) = 0

⇒ 2x² + 7x - 4 = 0

⇒ 2x² + 8x - x - 4 = 0

⇒ 2x(x + 4) - 1(x + 4) = 0

⇒ (x + 4)(2x - 1) = 0

⇒ x + 4 = 0 and 2x - 1 = 0

⇒ x = -4 and x =  

Correct option: (d)

p(x) = 3x² - 1

Now, p(x) = 0

⇒ 3x² - 1 = 0

⇒ 3x² = 1

It is a polynomial, Degree = 2.

It is not a polynomial.

It is a polynomial, Degree = 0.

It is a polynomial, Degree = 0.

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 5. So, it is a polynomial of degree 5. 

The given expression is an expression having only non-negative integral powers of y. So, it is a polynomial.

The highest power of y is 3. So, it is a polynomial of degree 3. 

The given expression is an expression having only non-negative integral powers of t. So, it is a polynomial.

The highest power of t is 2. So, it is a polynomial of degree 2. 

X¹⁰⁰ - 1 

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 100. So, it is a polynomial of degree 100. 

The given expression can be written as  

It contains a term having negative integral power of x. So, it is not a polynomial.

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of degree 2. 

The given expression can be written as 2x^-2.

It contains a term having negative integral power of x. So, it is not a polynomial. 

The given expression contains a term containing x^1/2, where ½ is not a non-negative integer.

So, it is not a polynomial. 

The given expression is an expression having only non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of degree 2. 

x⁴ - x^3/2 + x - 3 

The given expression contains a term containing x^3/2, where 3/2 is not a non-negative integer.

So, it is not a polynomial. 

The given expression can be written as 2x³ + 3x² + x^1/2 - 1. 

The given expression contains a term containing x^1/2, where ½ is not a non-negative integer.

So, it is not a polynomial. 

-7 + x

The degree of a given polynomial is 1.

Hence, it is a linear polynomial.

6y

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. 

-z³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. 

1 - y - y³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. 

x - x³ + x⁴

The degree of a given polynomial is 4.

Hence, it is a quartic polynomial. 

1 + x + x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. 

-6x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. 

-13

The given polynomial contains only one term namely constant.

Hence, it is a constant polynomial. 

-p

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. 

The coefficient of x³ in x + 3x² - 5x³ + x⁴ is -5. 

The coefficient of x in . 

The given polynomial can be written as x³ + 0x² + 2x - 3.

Hence, the coefficient of x² in 2x - 3 + x³ is 0. 

The coefficient of x in . 

The constant term in . 

Hence, the degree of a given polynomial is 2. 

y²(y - y³)

= y³ - y⁵

Hence, the degree of a given polynomial is 5. 

(3x - 2)(2x³ + 3x²)

= 6x⁴ + 9x³ - 4x³ - 6x²

= 6x⁴ + 5x³ - 6x²

Hence, the degree of a given polynomial is 4. 

The degree of a given polynomial is 1. 

-8

This is a constant polynomial.

The degree of a non-zero constant polynomial is zero. 

x^-2(x⁴ + x²)

= x^-2.x²(x² + 1)

= x⁰ (x² + 1)

= x² + 1

Hence, the degree of a given polynomial is 2. 

Example of a monomial of degree 5:

3x⁵ 

Example of a binomial of degree 8:

x - 6x⁸ 

Example of a trinomial of degree 4:

7 + 2y + y⁴ 

Example of a monomial of degree 0:

7 

x - 2x² + 8 + 5x³ in standard form:

5x³ - 2x² + x + 8 

6x³ + 2x - x⁵ - 3x² in standard form:

-x⁵ + 6x³ - 3x² + 2x 

2 + t - 3t³ + t⁴ - t² in standard form:

t⁴ - 3t³ - t² + t + 2 

p(x) = 5 - 4x + 2x²

(i) p(0) = 5 - 4 0 + 2 0² = 5

(ii) p(3) = 5 - 4 3 + 2 3²

= 5 - 12 + 18

= 23 - 12 = 11

(iii) p(-2) = 5 - 4(-2) + 2 (-2)²

= 5 + 8 + 8 = 21

p(y) = 4 + 3y - y² + 5y³

(i) p(0) = 4 + 3 0 - 0² + 5 0³

= 4 + 0 - 0 + 0 = 4

(ii) p(2) = 4 + 3 2 - 2² + 5 2³

= 4 + 6 - 4 + 40

= 10 - 4 + 40 = 46

(iii) p(-1) = 4 + 3(-1) - (-1)² + 5(-1)³

= 4 - 3 - 1 - 5 = -5

f(t) = 4t² - 3t + 6

(i) f(0) = 4 0² - 3 0 + 6

= 0 - 0 + 6 = 6

(ii) f(4) = 4(4)² - 3 4 + 6

= 64 - 12 + 6 = 58

(iii) f(-5) = 4(-5)² - 3(-5) + 6

= 100 + 15 + 6 = 121

p(x) = x³ - 3x² + 2x

Thus, we have

p(0) = 0³ - 3(0)² + 2(0) = 0

p(1) = 1³ - 3(1)² + 2(1) = 1 - 3 + 2 = 0

p(2) = 2³ - 3(2)² + 2(2) = 8 - 12 + 4 = 0

Hence, 0, 1 and 2 are the zeros of the polynomial p(x) = x³ - 3x² + 2x.  

p(x) = x³ + x² - 9x - 9

Thus, we have 

p(0) = 0³ + 0² - 9(0) - 9 = -9 

p(3) = 3³ + 3² - 9(3) - 9 = 27 + 9 - 27 - 9 = 0

p(-3) = (-3)³ + (-3)² - 9(-3) - 9 = -27 + 9 + 27 - 9 = 0

p(-1) = (-1)³ + (-1)² - 9(-1) - 9 = -1 + 1 + 9 - 9 = 0

Hence, 0, 3 and -3 are the zeros of p(x).

Now, 0 is not a zero of p(x) since p(0) ≠ 0. 

p(x) = x - 4

Then, p(4) = 4 - 4 = 0

 4 is a zero of the polynomial p(x).

p(x) = x - 3

Then,p(-3) = -3 - 3 = -6

 -3 is not a zero of the polynomial p(x).

p(y) = 2y + 1

Then,

is a zero of the polynomial p(y).

p(x) = 2 - 5x

Then, 

 is a zero of the polynomial p(x).

p(x) = (x - 1) (x - 2)

Then,p(1)= (1 - 1) (1 - 2) = 0 -1 = 0

 1 is a zero of the polynomial p(x).

Also, p(2) = (2 - 1)(2 - 2) = 1 0 = 0

 2 is a zero of the polynomial p(x).

 Hence,1 and 2 are the zeroes of the polynomial p(x).

p(x) = x² + x - 6

Then, p(2) = 2² + 2 - 6

= 4 + 2 - 6

= 6 - 6 = 0

 2 is a zero of the polynomial p(x).

 Also, p(-3) = (-3)² - 3 - 6

= 9 - 3 - 6 = 0

 -3 is a zero of the polynomial p(x).

 Hence, 2 and -3 are the zeroes of the polynomial p(x).

p(x) = x² - 3x.

Then,p(0) = 0² - 3 0 = 0

p(3) = (3)²- 3 3 = 9 - 9 = 0

0 and 3 are the zeroes of the polynomial p(x).

p(x) = 0

x - 5 = 0

x = 5

_{5 is the zero of the polynomial p(x).}

q(x) = 0

 x + 4 = 0

x= -4

 _{-4 is the zero of the polynomial q(x).} 

f(x) = 0

 3x + 1= 0

3x=-1

 x =

  x =is the zero of the polynomial f(x).

g(x) = 0

 5 - 4x = 0

 -4x = -5

 x =

 x =  is the zero of the polynomial g(x).

p(x) = 0

 ax = 0

x = 0

0 is the zero of the polynomial p(x).

q(x) = 0

4x = 0

 x = 0

0 is the zero of the polynomial q(x).

r(x) = 2x + 5

Now, r(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

h(x) = 6x - 2

Now, h(x) = 0

⇒ 6x - 2 = 0

⇒ 6x = 2

f(x) = 2x³ - 5x² + ax + b

Now, 2 is a zero of f(x).

⇒ f(2) = 0

⇒ 2(2)³ - 5(2)² + a(2) + b = 0

⇒ 16 - 20 + 2a + b = 0

⇒ 2a + b - 4 = 0 ….(i)

Also, 0 is a zero of f(x).

⇒ f(0) = 0

⇒ 2(0)³ - 5(0)² + a(0) + b = 0

⇒ 0 - 0 + 0 + b = 0

⇒ b = 0

Substituting b = 0 in (i), we get

2a + 0 - 4 = 0

⇒ 2a = 4

⇒ a = 2

Thus, a = 2 and b = 0. 

f(x) = (x³ - 8)

By the Factor Theorem, (x - 2) will be a factor of f(x) if f(2) = 0.

Here, f(2) = (2)³ - 8

= 8 - 8 = 0

(x - 2) is a factor of (x³ - 8).

f(x) = (2x³ + 7x² - 24x - 45)

By the Factor Theorem, (x - 3) will be a factor of f(x) if f(3) = 0.

Here, f(3) = 2 3³ + 7 3² - 24 3 - 45

= 54 + 63 - 72 - 45

= 117 - 117 = 0

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

f(x) = (2x⁴ + 9x³ + 6x² - 11x - 6)

By the Factor Theorem, (x - 1) will be a factor of f(x) if f(1) = 0.

Here, f(1) = 2 1⁴ + 9 1³ + 6 1² - 11 1 - 6

= 2 + 9 + 6 - 11 - 6

= 17 - 17 = 0

(x - 1) is factor of (2x⁴ + 9x³ + 6x² - 11x - 6).

f(x) = (x⁴ - x² - 12)

By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.

Here, f(-2) = (-2)⁴ - (-2)² - 12

= 16 - 4 - 12

= 16 - 16 = 0

(x + 2) is a factor of (x⁴ - x² - 12).

By the factor theorem, g(x) = x + 3 will be a factor of p(x) if p(-3) = 0.

Now, p(x) = 69 + 11x - x² + x³

⇒ p(-3) = 69 + 11(-3) - (-3)² + (-3)³

= 69 - 33 - 9 - 27

= 0

Hence, g(x) = x + 3 is a factor of the given polynomial p(x). 

f(x) = 2x³ + 9x² - 11x - 30

By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.

Here, f(-5) = 2(-5)³ + 9(-5)² - 11(-5) - 30

= -250 + 225 + 55 - 30

= -280 + 280 = 0

(x + 5) is a factor of (2x³ + 9x² - 11x - 30).

f(x) = (2x⁴ + x³ - 8x² - x + 6)

By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0.

Here, 2x - 3 = 0 x =

is a factor of .

By the factor theorem, g(x) = 3x - 2 will be a factor of p(x) if  = 0.

Now, p(x) = 3x³ + x² - 20x + 12

Hence, g(x) = 3x - 2 is a factor of the given polynomial p(x). 

f(x) =

By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0. 

Here, 

= 14 - 8 - 6

= 14 - 14 = 0

f(x) =

By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0.

Here, 

Let q(p) = (p¹⁰ - 1) and f(p) = (p¹¹ - 1)

By the factor theorem, (p - 1) will be a factor of q(p) and f(p) if q(1) and f(1) = 0.

Now, q(p) = p¹⁰ - 1

⇒ q(1) = 1¹⁰ - 1 = 1 - 1 = 0

Hence, (p - 1) is a factor of p¹⁰ - 1.

And, f(p) = p¹¹ - 1

⇒ f(1) = 1¹¹ - 1 = 1 - 1 = 0

Hence, (p - 1) is also a factor of p¹¹ - 1. 

f(x) = (2x³ + 9x² + x + k)

x - 1 = 0 x = 1

f(1) = 2 1³ + 9 1² + 1 + k

= 2 + 9 + 1 + k

= 12 + k

Given that (x - 1) is a factor of f(x).

By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0 and therefore f(1) = 0.

f(1) = 12 + k = 0

k = -12.

f(x) = (2x³ - 3x² - 18x + a)

x - 4 = 0 x = 4

f(4) = 2(4)³ - 3(4)² - 18 4 + a

= 128 - 48 - 72 + a

= 128 - 120 + a

= 8 + a

Given that (x - 4) is a factor of f(x).

By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0 and therefore f(4) = 0.

 f(4) = 8 + a = 0

 a = -8

Let p(x) = ax³ + x² - 2x + 4a - 9

It is given that (x + 1) is a factor of p(x).

⇒ p(-1) = 0

⇒ a(-1)³ + (-1)² - 2(-1) + 4a - 9 = 0

⇒ -a + 1 + 2 + 4a - 9 = 0

⇒ 3a - 6 = 0

⇒ 3a = 6

⇒ a = 2 

Let p(x) = x⁵ - 4a²x³ + 2x + 2a + 3

It is given that (x + 2a) is a factor of p(x).

⇒ p(-2a) = 0

⇒ (-2a)⁵ - 4a²(-2a)³ + 2(-2a) + 2a + 3 = 0

⇒ -32a⁵ - 4a²(-8a³) - 4a + 2a + 3 = 0

⇒ -32a⁵ + 32a⁵ -2a + 3 = 0

⇒ 2a = 3

Let p(x) = 8x⁴ + 4x³ - 16x² + 10x + m

It is given that (2x - 1) is a factor of p(x).

Let p(x) = x⁴ - x³ - 11x² - x + a

It is given that p(x) is divisible by (x + 3).

⇒ (x + 3) is a factor of p(x).

⇒ p(-3) = 0

⇒ (-3)⁴ - (-3)³ - 11(-3)² - (-3) + a = 0

⇒ 81 + 27 - 99 + 3 + a = 0

⇒ 12 + a = 0

⇒ a = -12 

Let f(x) = x³ - 3x² - 13x + 15

Now, x² + 2x - 3 = x² + 3x - x - 3

= x (x + 3) - 1 (x + 3)

= (x + 3) (x - 1)

Thus, f(x) will be exactly divisible by x² + 2x - 3 = (x + 3) (x - 1) if (x + 3) and (x - 1) are both factors of f(x), so by factor theorem, we should have f(-3) = 0 and f(1) = 0.

Now, f(-3) = (-3)³ - 3 (-3)² - 13 (-3) + 15

= -27 - 3 9 + 39 + 15

= -27 - 27 + 39 + 15

= -54 + 54 = 0

And, f(1) = 1³ - 3 1² - 13 1 + 15

= 1 - 3 - 13 + 15

= 16 - 16 = 0

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

So, x² + 2x - 3 divides f(x) exactly.

Letf(x) = (x³ + ax² + bx + 6)

Now, by remainder theorem, f(x) when divided by (x - 3) will leave a remainder as f(3).

So, f(3) = 3³ + a 3² + b 3 + 6 = 3

27 + 9a + 3b + 6 = 3

9 a + 3b + 33 = 3

9a + 3b = 3 - 33

9a + 3b = -30

3a + b = -10(i)

Given that (x - 2) is a factor of f(x).

By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0.

 f(2) =  2³ + a 2² + b 2 + 6 = 0

                       8 + 4a+ 2b + 6 = 0

                               4a + 2b = -14

                                   2a + b = -7(ii)

Subtracting (ii) from (i), we get,

a = -3

Substituting the value of a = -3 in (i), we get,

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

a = -3 and b = -1.

Let f(x) = (x³ - 10x² + ax + b), then by factor theorem

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

f(1) = 1³ - 10 1² + a 1 + b = 0

1 - 10 + a + b = 0

a + b = 9(i)

Andf(2) = 2³ - 10 2² + a 2 + b = 0

8 - 40 + 2a + b = 0

2a + b = 32(ii)

Subtracting (i) from (ii), we get

a = 23

Substituting the value of a = 23 in (i), we get

23 + b = 9

b = 9 - 23

b = -14

a = 23 and b = -14.

Letf(x)= (x⁴ + ax³ - 7x² - 8x + b)

Now, x + 2 = 0 x = -2 and x + 3 = 0 x = -3

By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if f(-2) = 0 and f(-3) = 0

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

16 - 8a - 28 + 16 + b = 0

-8a + b = -4

8a - b = 4(i)

And, f(-3) = (-3)⁴ + a (-3)³ - 7 (-3)² - 8 (-3) + b = 0

81 - 27a - 63 + 24 + b = 0

-27a + b = -42

27a - b = 42(ii)

Subtracting (i) from (ii), we get,

19a = 38

So, a = 2

Substituting the value of a = 2 in (i), we get

8 2 - b = 4

16 - b = 4

-b = -16 + 4

-b = -12

b = 12

a = 2 and b = 12.

Let f(x) = px² + 5x + r

Now, (x - 2) is a factor of f(x).

⇒ f(2) = 0

⇒ p(2)² + 5(2) + r = 0

⇒ 4p + 10 + r = 0

⇒ 4p + r = -10

Also,  is a factor of f(x).

From (i) and (ii), we have

4p + r = p + 4r

⇒ 4p - p = 4r - r

⇒ 3p = 3r

⇒ p = r

Let f(x) = 2x⁴ - 5x³ + 2x² - x + 2

and g(x) = x² - 3x + 2

= x² - 2x - x + 2

= x(x - 2) - 1(x - 2)

= (x - 2)(x - 1)

Clearly, (x - 2) and (x - 1) are factors of g(x).

In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that f(x) is exactly divisible by (x - 2) and (x - 1).

Thus, we will show that (x - 2) and (x - 1) are factors of f(x).

Now,

f(2) = 2(2)⁴ - 5(2)³ + 2(2)² - 2 + 2 = 32 - 40 + 8 = 0 and

f(1) = 2(1)⁴ - 5(1)³ + 2(1)² - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0

Therefore, (x - 2) and (x - 1) are factors of f(x).

⇒ g(x) = (x - 2)(x - 1) is a factor of f(x).

Hence, f(x) is exactly divisible by g(x). 

Let the required number to be added be k.

Then, p(x) = 2x⁴ - 5x³ + 2x² - x - 3 + k and g(x) = x - 2

Thus, we have

p(2) = 0

⇒ 2(2)⁴ - 5(2)³ + 2(2)² - 2 - 3 + k = 0

⇒ 32 - 40 + 8 - 5 + k = 0

⇒ k - 5 = 0

⇒ k = 5

Hence, the required number to be added is 5. 

Let p(x) = x⁴ + 2x³ - 2x² + 4x + 6 and q(x) = x² + 2x - 3.

When p(x) is divided by q(x), the remainder is a linear expression in x.

So, let r(x) = ax + b be subtracted from p(x) so that p(x) - r(x) is divided by q(x).

Let f(x) = p(x) - r(x) = p(x) - (ax + b)

= (x⁴ + 2x³ - 2x² + 4x + 6) - (ax + b)

= x⁴ + 2x³ - 2x² + (4 - a)x + 6 - b

We have,

q(x) = x² + 2x - 3

= x² + 3x - x - 3

= x(x + 3) - 1(x + 3)

= (x + 3)(x - 1)

Clearly, (x + 3) and (x - 1) are factors of q(x).

Therefore, f(x) will be divisible by q(x) if (x + 3) and (x - 1) are factors of f(x).

i.e., f(-3) = 0 and f(1) = 0

Consider, f(-3) = 0

⇒ (-3)⁴ + 2(-3)³ - 2(-3)² + (4 - a)(-3) + 6 - b = 0

⇒ 81 - 54 - 18 - 12 + 3a + 6 - b = 0

⇒ 3 + 3a - b = 0

⇒ 3a - b = -3 ….(i)

And, f(1) = 0

⇒ (1)⁴ + 2(1)³ - 2(1)² + (4 - a)(1) + 6 - b = 0

⇒ 1 + 2 - 2 + 4 - a + 6 - b = 0

⇒ 11 - a - b = 0

⇒ -a - b = -11 ….(ii)

Subtracting (ii) from (i), we get

4a = 8

⇒ a = 2

Substituting a = 2 in (i), we get

3(2) - b = -3

⇒ 6 - b = -3

⇒ b = 9

Putting the values of a and b in r(x) = ax + b, we get

r(x) = 2x + 9

Hence, p(x) is divisible by q(x), if r(x) = 2x + 9 is subtracted from it. 

Let f(x) = xⁿ + aⁿ

In order to prove that (x + a) is a factor of f(x) for any odd positive integer n, it is sufficient to show that f(-a) = 0.

Now,

f(-a) = (-a)ⁿ + aⁿ

= (-1)ⁿ aⁿ + aⁿ

= [(-1)ⁿ + 1] aⁿ

= [-1 + 1] aⁿ …[n is odd ⇒ (-1)ⁿ = -1]

= 0 × aⁿ

= 0

Hence, (x + a) is a factor of xⁿ + aⁿ for any odd positive integer n. 

Quotient = x³ + x² + x + 1

Remainder = 2

Verification:

f(x) = x⁴ + 1

Then, f(1) = 1⁴ + 1 = 1 + 1 = 2 = Remainder 

f(x) = x³ - 6x² + 9x + 3

Now, x - 1 = 0 x = 1

By the remainder theorem, we know that when f(x) is divided by (x - 1) the remainder is f(1).

Now, f(1) = 1³ - 6 1² + 9 1 + 3

= 1 - 6 + 9 + 3

= 13 - 6 = 7

The required remainder is 7.

x - 3 = 0

⇒ x = 3

By the remainder theorem, we know that when p(x) = 2x³ - 7x² + 9x - 13 is divided by g(x) = x - 3, the remainder is g(3).

Now,

g(3) = 2(3)³ - 7(3)² + 9(3) - 13 = 54 - 63 + 27 - 13 = 5

Hence, the required remainder is 5.

x - 2 = 0

⇒ x = 2

By the remainder theorem, we know that when p(x) = 3x⁴ - 6x² - 8x - 2 is divided by g(x) = x - 2, the remainder is g(2).

Now,

g(2) = 3(2)⁴ - 6(2)² - 8(2) - 2 = 48 - 24 - 16 - 2 = 6

Hence, the required remainder is 6. 

2x - 3 = 0

⇒ x =  

By the remainder theorem, we know that when p(x) = 2x³ - 9x² + x + 15 is divided by g(x) = 2x - 3, the remainder is g.

Now,

Hence, the required remainder is 3. 

x + 1 = 0

⇒ x = -1

By the remainder theorem, we know that when p(x) = x³ - 2x² - 8x - 1 is divided by g(x) = x + 1, the remainder is g(-1).

Now,

g(-1) = (-1)³ - 2(-1)² - 8(-1) - 1 = -1 - 2 + 8 - 1 = 4

Hence, the required remainder is 4. 

x + 2 = 0

⇒ x = -2

By the remainder theorem, we know that when p(x) = 2x³ + x² - 15x - 12 is divided by g(x) = x + 2, the remainder is g(-2).

Now,

g(-2) = 2(-2)³ + (-2)² - 15(-2) - 12 = -16 + 4 + 30 - 12 = 6

Hence, the required remainder is 6. 

3x + 2 = 0

⇒ x =  

By the remainder theorem, we know that when p(x) = 6x³ + 13x² + 3 is divided by g(x) = 3x + 2, the remainder is g.

Now,

Hence, the required remainder is 7. 

By the remainder theorem, we know that when p(x) = x³ - 6x² + 2x - 4 is divided by g(x) = , the remainder is g.

Now,

Hence, the required remainder is . 

By the remainder theorem, we know that when p(x) = 2x³ + 3x² - 11x - 3 is divided by g(x) = , the remainder is g.

Now,

Hence, the required remainder is 3. 

x - a = 0

⇒ x = a

By the remainder theorem, we know that when p(x) = x³ - ax² + 6x - a is divided by g(x) = x - a, the remainder is g(a).

Now,

g(a) = (a)³ - a(a)² + 6(a) - a = a³- a³+ 6a - a = 5a

Hence, the required remainder is 5a. 

Let p(x) = 2x³ + x² - ax + 2 and q(x) = 2x³ - 3x² - 3x + a be the given polynomials.

The remainders when p(x) and q(x) are divided by (x - 2) are p(2) and q(2) respectively.

By the given condition, we have

p(2) = q(2)

⇒ 2(2)³ + (2)² - a(2) + 2 = 2(2)³ - 3(2)² - 3(2) + a

⇒ 16 + 4 - 2a + 2 = 16 - 12 - 6 + a

⇒ 22 - 2a = -2 + a

⇒ a + 2a = 22 + 2

⇒ 3a = 24

⇒ a = 8 

Letf(x) = (x⁴ - 2x³ + 3x² - ax + b)

_{From the given information,}

f(1) = 1⁴ - 2(1)³ + 3(1)² - a 1 + b = 5

1 - 2 + 3 - a + b = 5

2 - a + b = 5(i)

And,

f(-1) = (-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + b = 19

1 + 2 + 3 + a + b = 19

6 + a + b = 19(ii)

Adding (i) and (ii), we get

8 + 2b = 24

2b= 24 - 8 = 16

b =

Substituting the value of b = 8 in (i), we get

2 - a + 8 = 5

-a + 10 = 5

-a = -10 + 5

-a = -5

a = 5

a = 5 and b = 8

f(x) = x⁴ - 2x³ + 3x² - ax + 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

The required remainder is 10.

The polynomial p(x) will be a multiple of g(x) if g(x) divides p(x) completely.

i.e. when p(x) is divided by g(x), it does not leave any remainder.

Now, x - 2 = 0 ⇒ x = 2

Also,

p(2) = (2)³ - 5(2)² + 4(2) - 3 = 8 - 20 + 8 - 3 = -7 ≠ 0

Thus, p(x) is not a multiple of g(x).

The polynomial g(x) will be a factor of p(x) if g(x) divides p(x) completely.

i.e. when p(x) is divided by g(x), it does not leave any remainder.

Now, 2x + 1 = 0 ⇒ x =

Also,

Thus, g(x) is not a factor of p(x).