# R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials

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## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise MCQ

Correct option: (d)

P(x)
= 5x - 4x^{2} + 3

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

Correct option: (d)

Let
f(x) = x^{51} + 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) = 2x^{2} + 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)
= 2x^{2} + 7x - 4

Now, p(x) = 0

⇒ 2x^{2} + 7x - 4 = 0

⇒ 2x^{2} + 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)
= 3x^{2} - 1

Now, p(x) = 0

⇒ 3x^{2 }-
1 = 0

⇒ 3x^{2}
= 1

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2A

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.

X^{100} - 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 highest power of x is 2. So, it is a polynomial of degree 2.

x^{4} - 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 2x^{3}
+ 3x^{2} + 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.

-z^{3}

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial.

1 - y - y^{3}

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial.

x - x^{3} + x^{4}

The degree of a given polynomial is 4.

Hence, it is a quartic polynomial.

1 + x + x^{2}

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial.

-6x^{2}

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 x^{3}
in x + 3x^{2} - 5x^{3} + x^{4} is -5.

The coefficient of x in .

The given polynomial can be
written as x^{3} + 0x^{2} + 2x - 3.

Hence, the coefficient of x^{2}
in 2x - 3 + x^{3} is 0.

The coefficient of x in .

The constant term in .

Hence, the degree of a given polynomial is 2.

y^{2}(y - y^{3})

= y^{3} - y^{5}

Hence, the degree of a given polynomial is 5.

(3x - 2)(2x^{3} + 3x^{2})

= 6x^{4} + 9x^{3}
- 4x^{3} - 6x^{2}

= 6x^{4} + 5x^{3}
- 6x^{2}

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}(x^{4} + x^{2})

= x^{-2}.x^{2}(x^{2}
+ 1)

= x^{0} (x^{2} +
1)

= x^{2} + 1

Hence, the degree of a given polynomial is 2.

Example of a monomial of degree 5:

3x^{5}^{ }

Example of a binomial of degree 8:

x - 6x^{8}^{ }

Example of a trinomial of degree 4:

7 + 2y + y^{4}^{ }

Example of a monomial of degree 0:

7

x - 2x^{2} + 8 + 5x^{3}
in standard form:

5x^{3} - 2x^{2} +
x + 8

^{ }

^{ }

6x^{3} + 2x - x^{5}
- 3x^{2} in standard form:

-x^{5} + 6x^{3} -
3x^{2} + 2x

2 + t - 3t^{3} + t^{4}
- t^{2} in standard form:

t^{4} - 3t^{3} - t^{2}
+ t + 2

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2B

p(x) = 5 - 4x + 2x^{2}

(i) p(0) = 5 - 4 _{} 0 + 2 _{} 0^{2} = 5

(ii) p(3) = 5 - 4 _{} 3 + 2 _{} 3^{2}

= 5 - 12 + 18

= 23 - 12 = 11

(iii) p(-2) = 5 - 4(-2) + 2 (-2)^{2}

= 5 + 8 + 8 = 21

p(y) = 4 + 3y - y^{2} + 5y^{3}

(i) p(0) = 4 + 3 _{} 0 - 0^{2} + 5 _{} 0^{3}

= 4 + 0 - 0 + 0 = 4

(ii) p(2) = 4 + 3 _{} 2 - 2^{2} + 5 _{} 2^{3}

= 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) = 4t^{2} - 3t + 6

(i) f(0) = 4 _{} 0^{2} - 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) = x^{3} - 3x^{2}
+ 2x

Thus, we have

p(0) = 0^{3} - 3(0)^{2}
+ 2(0) = 0

p(1) = 1^{3} - 3(1)^{2}
+ 2(1) = 1 - 3 + 2 = 0

p(2) = 2^{3} - 3(2)^{2}
+ 2(2) = 8 - 12 + 4 = 0

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

p(x) = x^{3} + x^{2}
- 9x - 9

Thus, we have

p(0) = 0^{3} + 0^{2}
- 9(0) - 9 = -9

p(3) = 3^{3} + 3^{2}
- 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) = x^{2} + x - 6

Then, p(2) = 2^{2} + 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) = x^{2} - 3x.

Then,p(0) = 0^{2} - 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) = 2x^{3} - 5x^{2}
+ 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.

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2C

Quotient = x^{3}
+ x^{2} + x + 1

Remainder = 2

Verification:

f(x) = x^{4} + 1

Then, f(1) = 1^{4}
+ 1 = 1 + 1 = 2 = Remainder

f(x) = x^{3} - 6x^{2} + 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) = 1^{3} - 6 _{} 1^{2} + 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) = 2x^{3} - 7x^{2} + 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) = 3x^{4} - 6x^{2} - 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) = 2x^{3} - 9x^{2} + 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) = x^{3} - 2x^{2} - 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) = 2x^{3} + x^{2} - 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) = 6x^{3} + 13x^{2} + 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) = x^{3} - 6x^{2} + 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) = 2x^{3} + 3x^{2} - 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) = x^{3} - ax^{2} + 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
= a^{3}- a^{3}+ 6a - a = 5a

Hence, the required remainder is 5a.

Let p(x) = 2x^{3} + x^{2}
- ax + 2 and q(x) = 2x^{3} - 3x^{2}
- 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) = (x^{4} - 2x^{3} + 3x^{2} - ax + b)

_{From the given information,}

f(1) = 1^{4} - 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) = x^{4} - 2x^{3} + 3x^{2} - ax + b

= x^{4} - 2x^{3} + 3x^{2} - 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).

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2D

f(x) = (x^{3} - 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 (x^{3} - 8).

f(x) = (2x^{3} + 7x^{2} - 24x - 45)

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

Here, f(3) = 2 _{} 3^{3} + 7 _{} 3^{2} - 24 _{} 3 - 45

= 54 + 63 - 72 - 45

= 117 - 117 = 0

_{} (x - 3) is a factor of (2x^{3} + 7x^{2} - 24x - 45).

f(x) = (2x^{4} + 9x^{3} + 6x^{2} - 11x - 6)

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

Here, f(1) = 2 _{} 1^{4} + 9^{ }_{} 1^{3} + 6 _{} 1^{2} - 11 _{} 1 - 6

= 2 + 9 + 6 - 11 - 6

= 17 - 17 = 0

_{} (x - 1) is factor of (2x^{4} + 9x^{3} + 6x^{2} - 11x - 6).

f(x) = (x^{4} - x^{2} - 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 (x^{4} - x^{2} - 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 - x^{2}
+ x^{3}

⇒ 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) = 2x^{3} + 9x^{2} - 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 (2x^{3} + 9x^{2} - 11x - 30).

f(x) = (2x^{4} + x^{3} - 8x^{2} - 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) = 3x^{3} + x^{2}
- 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) = (p^{10} - 1)
and f(p) = (p^{11} - 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) = p^{10} - 1

⇒
q(1) = 1^{10} - 1 = 1 - 1 = 0

Hence, (p - 1) is a factor of p^{10}
- 1.

And, f(p) = p^{11} - 1

⇒
f(1) = 1^{11} - 1 = 1 - 1 = 0

Hence, (p - 1) is also a factor of
p^{11} - 1.

f(x) = (2x^{3} + 9x^{2} + x + k)

x - 1 = 0 _{} x = 1

_{} f(1) = 2 _{} 1^{3} + 9 _{} 1^{2} + 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) = (2x^{3} - 3x^{2} - 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) = ax^{3} + x^{2} -
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) = x^{5} - 4a^{2}x^{3}
+ 2x + 2a + 3

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

⇒ p(-2a) = 0

⇒
(-2a)^{5} - 4a^{2}(-2a)^{3} + 2(-2a) + 2a + 3 = 0

⇒ -32a^{5}
- 4a^{2}(-8a^{3}) - 4a + 2a + 3 = 0

⇒
-32a^{5} + 32a^{5} -2a + 3 = 0

⇒ 2a = 3

Let
p(x) = 8x^{4} + 4x^{3} -
16x^{2} + 10x + m

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

Let
p(x) = x^{4} - x^{3} -
11x^{2} - 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) = x^{3}
- 3x^{2} - 13x + 15

Now, x^{2} + 2x - 3 = x^{2}
+ 3x - x - 3

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

= (x + 3) (x - 1)

Thus,
f(x) will be exactly divisible by x^{2} + 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) = 1^{3} - 3 _{} 1^{2} -
13 _{} 1 + 15

= 1 - 3 - 13 + 15

= 16 - 16 = 0

_{} f(-3) = 0 and f(1) = 0

So,
x^{2} + 2x - 3 divides f(x) exactly.

Letf(x) = (x^{3} + ax^{2} + bx + 6)

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

So, f(3) = 3^{3} + a _{}3^{2} + 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) = 2^{3} + a _{}2^{2} + 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) = (x^{3} - 10x^{2} + 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) = 1^{3} - 10 _{}1^{2} + a _{}1 + b = 0

_{}1 - 10 + a + b = 0

_{}a + b = 9(i)

Andf(2) = 2^{3} - 10 _{}2^{2} + 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)= (x^{4} + ax^{3} - 7x^{2} - 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) = px^{2} + 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) = 2x^{4} - 5x^{3}
+ 2x^{2} - x + 2

and g(x) = x^{2} - 3x + 2

=
x^{2} - 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) = 2x^{4} - 5x^{3}
+ 2x^{2} - 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) = x^{4} + 2x^{3}
- 2x^{2} + 4x + 6 and q(x) = x^{2} + 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)

= (x^{4} + 2x^{3}
- 2x^{2} + 4x + 6) - (ax + b)

= x^{4} + 2x^{3} -
2x^{2} + (4 - a)x + 6 - b

We have,

q(x) = x^{2} + 2x - 3

=
x^{2} + 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) = x^{n}
+ a^{n}

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} + a^{n}

=
(-1)^{n} a^{n} + a^{n}

=
[(-1)^{n} + 1] a^{n}

=
[-1 + 1] a^{n} …[n is odd ⇒
(-1)^{n} = -1]

=
0 ×
a^{n}

= 0

Hence, (x + a) is a factor of x^{n} + a^{n} for any odd positive
integer n.

## R. S. Aggarwal and V. Aggarwal - Mathematics - IX Class 9 Chapter Solutions

- Chapter 1 - Number Systems
- Chapter 2 - Polynomials
- Chapter 3 - Factorisation of Polynomials
- Chapter 4 - Linear Equations in Two Variables
- Chapter 5 - Coordinate Geometry
- Chapter 6 - Introduction to Euclid's Geometry
- Chapter 7 - Lines And Angles
- Chapter 8 - Triangles
- Chapter 9 - Congruence of Triangles and Inequalities in a Triangle
- Chapter 10 - Quadrilaterals
- Chapter 11 - Areas of Parallelograms and Triangles
- Chapter 12 - Circles
- Chapter 13 - Geometrical Constructions
- Chapter 14 - Areas of Triangles and Quadrilaterals
- Chapter 15 - Volume and Surface Area of Solids
- Chapter 16 - Presentation of Data in Tabular Form
- Chapter 17 - Bar Graph, Histogram and Frequency Polygon
- Chapter 18 - Mean, Median and Mode of Ungrouped Data
- Chapter 19 - Probability

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