NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1.

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1

Question 1.

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) 4x^{2} – 3x + 7

(ii) y^{2} + √2

(iii) 3 √t + t√2

(iv) y+

(v) x^{10}+ y^{3}+t^{50}

Solution:

(i) We have 4x^{2} – 3x + 7 = 4x^{2} – 3x + 7x^{0}

It is a polynomial in one variable i.e., x

because each exponent of x is a whole number.

(ii) We have y^{2} + √2 = y^{2} + √2y^{0}

It is a polynomial in one variable i.e., y

because each exponent of y is a whole number.

(iii) We have 3 √t + t√2 = 3 √t^{1/2} + √2.t

It is not a polynomial, because one of the exponents of t is ,

which is not a whole number.

(iv) We have y + = y + 2.y^{-1}

It is not a polynomial, because one of the exponents of y is -1,

which is not a whole number.

(v) We have x^{10}+ y^{3 }+ t^{50}

Here, exponent of every variable is a whole number, but x^{10} + y^{3} + t^{50} is a polynomial in x, y and t, i.e., in three variables.

So, it is not a polynomial in one variable.

Question 2.

Write the coefficients of x^{2} in each of the following

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

(ii) 2 – x^{2} + x^{3}

(iii) x^{2} + x

(iv) √2 x – 1

Solution:

(i) The given polynomial is 2 + x^{2} + x.

The coefficient of x^{2} is 1.

(ii) The given polynomial is 2 – x^{2} + x^{3}.

The coefficient of x^{2} is -1.

(iii) The given polynomial is + x.

The coefficient of x^{2} is .

(iv) The given polynomial is √2 x – 1.

The coefficient of x^{2} is 0.

Question 3.

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution:

(i) Abmomial of degree 35 can be 3x^{35} -4.

(ii) A monomial of degree 100 can be √2y^{100}.

Question 4.

Write the degree of each of the following polynomials.

(i) 5x^{3}+4x^{2} + 7x

(ii) 4 – y^{2 }

(iii) 5t – √7

(iv) 3

Solution:

(i) The given polynomial is 5x^{3} + 4x^{2} + 7x.

The highest power of the variable x is 3.

So, the degree of the polynomial is 3.

(ii) The given polynomial is 4- y^{2}. The highest

power of the variable y is 2.

So, the degree of the polynomial is 2.

(iii) The given polynomial is 5t – √7 . The highest power of variable t is 1. So, the degree of the polynomial is 1.

(iv) Since, 3 = 3x° [∵ x°=1]

So, the degree of the polynomial is 0.

Question 5.

Classify the following as linear, quadratic and cubic polynomials.

(i) x^{2}+ x

(ii) x – x^{3}

(iii) y + y^{2}+4

(iv) 1 + x

(v) 3t

(vi) r^{2}

(vii) 7x^{3 }Solution:

(i) The degree of x^{2} + x is 2. So, it is a quadratic polynomial.

(ii) The degree of x – x^{3} is 3. So, it is a cubic polynomial.

(iii) The degree of y + y^{2} + 4 is 2. So, it is a quadratic polynomial.

(iv) The degree of 1 + x is 1. So, it is a linear polynomial.

(v) The degree of 3t is 1. So, it is a linear polynomial.

(vi) The degree of r^{2} is 2. So, it is a quadratic polynomial.

(vii) The degree of 7x^{3} is 3. So, it is a cubic polynomial.

### NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.

Find the value of the polynomial 5x – 4x^{2} + 3 at

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution:

1et p(x) = 5x – 4x^{2} + 3

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

Thus, the value of 5x – 4x^{2} + 3 at x = 0 is 3.

(ii) p(-1) = 5(-1) – 4(-1)^{2} + 3

= – 5x – 4x^{2} + 3 = -9 + 3 = -6

Thus, the value of 5x – 4x^{2} + 3 at x = -1 is -6.

(iii) p(2) = 5(2) – 4(2)^{2} + 3 = 10 – 4(4) + 3

= 10 – 16 + 3 = -3

Thus, the value of 5x – 4x^{2} + 3 at x = 2 is – 3.

Question 2.

Find p (0), p (1) and p (2) for each of the following polynomials.

(i) p(y) = y^{2} – y +1

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

(iii) P (x) = x^{3}

(iv) p (x) = (x-1) (x+1)

Solution:

(i) Given that p(y) = y^{2} – y + 1.

∴ P(0) = (0)^{2} – 0 + 1 = 0 – 0 + 1 = 1

p(1) = (1)^{2} – 1 + 1 = 1 – 1 + 1 = 1

p(2) = (2)^{2} – 2 + 1 = 4 – 2 + 1 = 3

(ii) Given that p(t) = 2 + t + 2t^{2 } – t^{3 }

∴p(0) = 2 + 0 + 2(0)^{2 } – (0)^{3 }

= 2 + 0 + 0 – 0=2

P(1) = 2 + 1 + 2(1)^{2 } – (1)^{3}

= 2 + 1 + 2 – 1 = 4

p( 2) = 2 + 2 + 2(2)^{2 } – (2)^{3}

= 2 + 2 + 8 – 8 = 4

(iii) Given that p(x) = x^{3}

∴ p(0) = (0)^{3} = 0, p(1) = (1)^{3} = 1

p(2) = (2)^{3} = 8

(iv) Given that p(x) = (x – 1)(x + 1)

∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1

p(1) = (1 – 1)(1 +1) = (0)(2) = 0

P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3.

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1,x = –

(ii) p (x) = 5x – π, x =

(iii) p (x) = x^{2} – 1, x = x – 1

(iv) p (x) = (x + 1) (x – 2), x = – 1,2

(v) p (x) = x^{2}, x = 0

(vi) p (x) = 1x + m, x = –

(vii) P (x) = 3x^{2} – 1, x = – ,

(viii) p (x) = 2x + 1, x =

Solution:

(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π

∴

(iii) We have, p(x) = x^{2} – 1

∴ p(1) = (1)^{2} – 1 = 1 – 1=0

Since, p(1) = 0, so x = 1 is a zero of x^{2} -1.

Also, p(-1) = (-1)^{2} -1 = 1 – 1 = 0

Since p(-1) = 0, so, x = -1, is also a zero of x^{2} – 1.

(iv) We have, p(x) = (x + 1)(x – 2)

∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0

Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).

Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0

Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x^{2}

∴ p(o) = (0)^{2} = 0

Since, p(0) = 0, so, x = 0 is a zero of x^{2}.

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3x^{2} – 1

(viii) We have, p(x) = 2x + 1

∴

Since, ≠ 0, so, x = is not a zero of 2x + 1.

Question 4.

Find the zero of the polynomial in each of the following cases

(i) p(x)=x+5

(ii) p (x) = x – 5

(iii) p (x) = 2x + 5

(iv) p (x) = 3x – 2

(v) p (x) = 3x

(vi) p (x)= ax, a≠0

(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.

Solution:

(i) We have, p(x) = x + 5. Since, p(x) = 0

⇒ x + 5 = 0

⇒ x = -5.

Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.

Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5

Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0

⇒ 2x + 5 =0

⇒ 2x = -5

⇒ x =

Thus, zero of 2x + 5 is .

(iv) We have, p(x) = 3x – 2. Since, p(x) = 0

⇒ 3x – 2 = 0

⇒ 3x = 2

⇒ x =

Thus, zero of 3x – 2 is

(v) We have, p(x) = 3x. Since, p(x) = 0

⇒ 3x = 0 ⇒ x = 0

Thus, zero of 3x is 0.

(vi) We have, p(x) = ax, a ≠ 0.

Since, p(x) = 0 => ax = 0 => x-0

Thus, zero of ax is 0.

(vii) We have, p(x) = cx + d. Since, p(x) = 0

⇒ cx + d = 0 ⇒ cx = -d ⇒

Thus, zero of cx + d is

### NCERT So1utions for C1ass 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.

Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

(i) x + 1

(ii) x –

(iii) x

(iv) x + π

(v) 5 + 2x

Solution:

Let p(x) = x^{3} + 3x^{2} + 3x +1

(i) The zero of x + 1 is -1.

∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1

= -1 + 3- 3 + 1 = 0

Thus, the required remainder = 0

(ii) The zero of is

Thus, the required remainder =

(iii) The zero of x is 0.

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

= 0 + 0 + 0 + 1 = 1

Thus, the required remainder = 1.

(iv) The zero of x + π is -π.

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

= -π^{3} + 3π^{2} + (-3π) + 1

= – π^{3} + 3π^{2} – 3π +1

Thus, the required remainder is -π^{3} + 3π^{2} – 3π+1.

(v) The zero of 5 + 2x is .

Thus, the required remainder is .

Question 2.

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

Solution:

We have, p(x) = x^{3} – ax^{2} + 6x – a and zero of x – a is a.

∴ p(a) = (a)^{3} – a(a)^{2} + 6(a) – a

= a^{3} – a^{3} + 6a – a = 5a

Thus, the required remainder is 5a.

Question 3.

Check whether 7 + 3x is a factor of 3x^{3}+7x.

Solution:

We have, p(x) = 3x^{3}+7x. and zero of 7 + 3x is .

Since,( ) ≠ 0

i.e. the remainder is not 0.

∴ 3x^{3} + 7x is not divisib1e by 7 + 3x.

Thus, 7 + 3x is not a factor of 3x^{3} + 7x.

### NCERT So1utions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

Question 1.

Determine which of the following polynomials has (x +1) a factor.

(i) x^{3}+x^{2}+x +1

(ii) x^{4} + x^{3} + x^{2} + x + 1

(iii) x^{4} + 3x^{3} + 3x^{2} + x + 1

(iv) x^{3} – x^{2} – (2 +√2 )x + √2

Solution:

The zero of x + 1 is -1.

(i) Let p (x) = x^{3} + x^{2} + x + 1

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

= -1 + 1 – 1 + 1

⇒ p (- 1) = 0

So, (x+ 1) is a factor of x^{3} + x^{2} + x + 1.

(ii) Let p (x) = x^{4} + x^{3} + x^{2} + x + 1

∴ P(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1)+1

= 1 – 1 + 1 – 1 + 1

⇒ P (-1) ≠ 1

So, (x + 1) is not a factor of x^{4} + x^{3} + x^{2} + x+ 1.

(iii) Let p (x) = x^{4} + 3x^{3} + 3x^{2} + x + 1 .

∴ p (-1)= (-1)^{4} + 3 (-1)^{3} + 3 (-1)^{2} + (- 1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x^{4} + 3x^{3} + 3x^{2} + x+ 1.

(iv) Let p (x) = x^{3} – x^{2} – (2 + √2) x + √2

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

= -1 – 1 + 2 + √2 + √2

= 2√2

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x^{3} – x^{2} – (2 + √2) x + √2.

Question 2.

Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases

(i) p (x)= 2x^{3} + x^{2} – 2x – 1, g (x) = x + 1

(ii) p(x)= x^{3} + 3x^{2} + 3x + 1, g (x) = x + 2

(iii) p (x) = x^{3} – 4x^{2} + x + 6, g (x) = x – 3

Solution:

(i) We have, p (x)= 2x^{3} + x^{2} – 2x – 1 and g (x) = x + 1

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

= 2(-1) + 1 + 2 – 1

= -2 + 1 + 2 -1 = 0

⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x^{3} + 3x^{2} + 3x + 1 and g(x) = x + 2

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

= -8 + 12 – 6 + 1

= -14 + 13

= -1

⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

(iii) We have, = x^{3} – 4x^{2} + x + 6 and g (x) = x – 3

∴ p(3) = (3)^{3} – 4(3)^{2} + 3 + 6

= 27 – 4(9) + 3 + 6

= 27 – 36 + 3 + 6 = 0

⇒ p(3) = 0, so g(x) is a factor of p(x).

Question 3.

Find the value of k, if x – 1 is a factor of p (x) in each of the following cases

(i) p (x) = x^{2} + x + k

(ii) p (x) = 2x^{2} + kx + √2

(iii) p (x) = kx^{2} – √2 x + 1

(iv) p (x) = kx^{2} – 3x + k

Solution:

For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x^{2} + x + k

Since, p(1) = (1)^{2} +1 + k

⇒ p(1) = k + 2 = 0

⇒ k = -2.

(ii) Here, p (x) = 2x^{2} + kx + √2

Since, p(1) = 2(1)^{2} + k(1) + √2

= 2 + k + √2 =0

k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx^{2} – √2 x + 1

Since, p(1) = k(1)^{2} – (1) + 1

= k – √2 + 1 = 0

⇒ k = √2 -1

(iv) Here, p(x) = kx^{2} – 3x + k

p(1) = k(1)^{2} – 3(1) + k

= k – 3 + k

= 2k – 3 = 0

⇒ k =

Question 4.

Factorise

(i) 12x^{2} – 7x +1

(ii) 2x^{2} + 7x + 3

(iii) 6x^{2} + 5x – 6

(iv) 3x^{2} – x – 4

Solution:

(i) We have,

12x^{2} – 7x + 1 = 12x^{2} – 4x- 3x + 1

= 4x (3x – 1 ) -1 (3x – 1)

= (3x -1) (4x -1)

Thus, 12x^{2} -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x^{2} + 7x + 3 = 2x^{2} + x + 6x + 3

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

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

Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6x^{2} + 5x – 6 = 6x^{2} + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3)(3x – 2)

Thus, 6x^{2} + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3x^{2} – x – 4 = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)

Thus, 3x^{2} – x – 4 = (3x – 4)(x + 1)

Question 5.

Factorise

(i) x^{3} – 2x^{2} – x + 2

(ii) x^{3} – 3x^{2} – 9x – 5

(iii) x^{3} + 13x^{2} + 32x + 20

(iv) 2y^{3} + y^{2} – 2y – 1

Solution:

(i) We have, x^{3} – 2x^{2} – x + 2

Rearranging the terms, we have x^{3} – x – 2x^{2} + 2

= x(x^{2} – 1) – 2(x^{2} -1) = (x^{2} – 1)(x – 2)

= [(x)^{2} – (1)^{2}](x – 2)

= (x – 1)(x + 1)(x – 2)

[∵ (a^{2} – b^{2}) = (a + b)(a-b)]

Thus, x^{3} – 2x^{2} – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x^{3} – 3x^{2} – 9x – 5

= x^{3} + x^{2} – 4x^{2} – 4x – 5x – 5 ,

= x^{2} (x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x^{2} – 4x – 5)

= (x + 1)(x^{2} – 5x + x – 5)

= (x + 1)[x(x – 5) + 1(x – 5)]

= (x + 1)(x – 5)(x + 1)

Thus, x^{3} – 3x^{2} – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x^{3} + 13x^{2} + 32x + 20

= x^{3} + x^{2} + 12x^{2} + 12x + 20x + 20

= x^{2}(x + 1) + 12x(x +1) + 20(x + 1)

= (x + 1)(x^{2} + 12x + 20)

= (x + 1)(x^{2} + 2x + 10x + 20)

= (x + 1)[x(x + 2) + 10(x + 2)]

= (x + 1)(x + 2)(x + 10)

Thus, x^{3} + 13x^{2} + 32x + 20

= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y^{3} + y^{2} – 2y – 1

= 2y^{3} – 2y^{2} + 3y^{2} – 3y + y – 1

= 2y^{2}(y – 1) + 3y(y – 1) + 1(y – 1)

= (y – 1)(2y^{2} + 3y + 1)

= (y – 1)(2y^{2} + 2y + y + 1)

= (y – 1)[2y(y + 1) + 1(y + 1)]

= (y – 1)(y + 1)(2y + 1)

Thus, 2y^{3} + y^{2} – 2y – 1

= (y – 1)(y + 1)(2y +1)

### NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 1.

Use suitable identities to find the following products

(i) (x + 4)(x + 10)

(ii) (x+8) (x -10)

(iii) (3x + 4) (3x – 5)

(iv) (y^{2}+ ) (y^{2}– )

(v) (3 – 2x) (3 + 2x)

Solution:

(i) We have, (x+ 4) (x + 10)

Using identity,

(x+ a) (x+ b) = x^{2} + (a + b) x+ ab.

We have, (x + 4) (x + 10) = x^{2}+(4 + 10) x + (4 x 10)

= x^{2} + 14x+40

(ii) We have, (x+ 8) (x -10)

Using identity,

(x + a) (x + b) = x^{2} + (a + b) x + ab

We have, (x + 8) (x – 10) = x^{2} + [8 + (-10)] x + (8) (- 10)

= x^{2} – 2x – 80

(iii) We have, (3x + 4) (3x – 5)

Using identity,

(x + a) (x + b) = x^{2} + (a + b) x + ab

We have, (3x + 4) (3x – 5) = (3x)^{2} + (4 – 5) x + (4) (- 5)

= 9x^{2} – x – 20

Question 2.

Evaluate the following products without multiplying directly

(i) 103 x 107

(ii) 95 x 96

(iii) 104 x 96

Solution:

(i)We have, 103 x 107 = (100 + 3) (100 + 7)

= ( 100)^{2} + (3 + 7) (100)+ (3 x 7)

[Using (x + a)(x + b) = x^{2} + (a + b)x + ab]

= 10000 + (10) x 100 + 21

= 10000 + 1000 + 21=11021

(ii) We have, 95 x 96 = (100 – 5) (100 – 4)

= ( 100)^{2} + [(- 5) + (- 4)] 100 + (- 5 x – 4)

[Using (x + a)(x + b) = x^{2} + (a + b)x + ab]

= 10000 + (-9) + 20 = 9120

= 10000 + (-900) + 20 = 9120

(iii) We have 104 x 96 = (100 + 4) (100 – 4)

= (100)^{2}-4^{2 }

[Using (a + b)(a -b) = a^{2}– b^{2}]

= 10000 – 16 = 9984

Question 3.

Factorise the following using appropriate identities

(i) 9x^{2 }+ 6xy + y^{2}

(ii) 4y^{2}-4y + 1

(iii) x^{2} –

Solution:

(i) We have, 9x^{2} + 6xy + y^{2}

= (3x)^{2} + 2(3x)(y) + (y)^{2}

= (3x + y)^{2}

[Using a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (3x + y)(3x + y)

(ii) We have, 4y^{2} – 4y + 1^{2}

= (2y)^{2} + 2(2y)(1) + (1)^{2}

= (2y -1)^{2}

[Using a^{2} – 2ab + b^{2} = (a- b)^{2}]

= (2y – 1)(2y – 1 )

Question 4.

Expand each of the following, using suitable identity

(i) (x+2y+ 4z)^{2}

(ii) (2x – y + z)^{2 }

(iii) (- 2x + 3y + 2z)^{2}

(iv) (3a -7b – c)^{z }

(v) (- 2x + 5y – 3z)^{2}

(vi) [ a –b + 1] ^{2 }

Solution:

We know that

(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx

(i) (x + 2y + 4z)^{2}

= x^{2} + (2y)^{2} + (4z)^{2} + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)

= x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8 zx

(ii) (2x – y + z)^{2} = (2x)^{2} + (- y)^{2} + z^{2} + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)

= 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)^{2} = (- 2x)^{2} + (3y)^{2} + (2z)^{2} + 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x)

= 4x^{2} + 9y^{2} + 4z^{2} – 12xy + 12yz – 8zx

(iv) (3a -7b- c)^{2 }= (3a)^{2} + (- 7b)^{2} + (- c)^{2} + 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a)

= 9a^{2} + 49b^{2} + c^{2} – 42ab + 14bc – 6ac

(v)(- 2x + 5y- 3z)^{2} = (- 2x)^{2} + (5y)^{2} + (- 3z)^{2} + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)

= 4x^{2} + 25y^{2} + 9z^{2} – 20xy – 30yz + 12zx

Question 5.

Factorise

(i) 4 x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

(ii) 2x^{2} + y^{2} + 8z^{2} – 2√2xy + 4√2yz – 8xz

Solution:

(i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

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

= (2x + 3y – 4z)^{2} = (2x + 3y + 4z) (2x + 3y – 4z)

(ii) 2x^{2} + y^{2} + 8z^{2} – 2√2xy + 4√2yz – 8xz

= (- √2x)^{2} + (y)^{2} + (2 √2z)^{2}y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x)

= (- √2x + y + 2 √2z)^{2}

= (- √2x + y + 2 √2z) (- √2x + y + 2 √2z)

Question 6.

Write the following cubes in expanded form

Solution:

We have, (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y) …(1)

and (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y) …(2)

(i) (2x + 1)^{3} = (2x)^{3} + (1)^{3} + 3(2x)(1)(2x + 1) [By (1)]

= 8x^{3} + 1 + 6x(2x + 1)

= 8x^{3} + 12x^{2} + 6x + 1

(ii) (2a – 3b)^{3} = (2a)^{3} – (3b)^{3} – 3(2a)(3b)(2a – 3b) [By (2)]

= 8a^{3} – 27b^{3} – 18ab(2a – 3b)

= 8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}

Question 7.

Evaluate the following using suitable identities

(i) (99)^{3}

(ii) (102)^{3}

(iii) (998)^{3}

Solution:

(i) We have, 99 = (100 -1)

∴ 99^{3} = (100 – 1)^{3}

= (100)^{3} – 1^{3} – 3(100)(1)(100 -1)

[Using (a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)]

= 1000000 – 1 – 300(100 – 1)

= 1000000 -1 – 30000 + 300

= 1000300 – 30001 = 970299

(ii) We have, 102 =100 + 2

∴ 102^{3} = (100 + 2)^{3}

= (100)^{3} + (2)^{3} + 3(100)(2)(100 + 2)

[Using (a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)]

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200 = 1061208

(iii) We have, 998 = 1000 – 2

∴ (998)^{3} = (1000-2)^{3}

= (1000)^{3}– (2)^{3} – 3(1000)(2)(1000 – 2)

[Using (a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)]

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8 – 6000000 +12000

= 994011992

Question 8.

Factorise each of the following

(i) 8a^{3} +b^{3} + 12a^{2}b+6ab^{2}

(ii) 8a^{3} -b^{3}-12a^{2}b+6ab^{2 }(iii) 27-125a^{3} -135a+225a^{2 } (iv) 64a^{3} -27b^{3} -144a^{2}b + 108ab^{2}

Solution:

(i) 8a^{3} +b^{3} +12a^{2}b+6ab^{2}

= (2a)^{3} + (b)^{3} + 6ab(2a + b)

= (2a)^{3} + (b)^{3} + 3(2a)(b)(2a + b)

= (2 a + b)^{3}

[Using a^{3} + b^{3} + 3 ab(a + b) = (a + b)^{3}]

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

(ii) 8a^{3} – b^{3} – 12o^{2}b + 6ab^{2}

= (2a)^{3} – (b)^{3} – 3(2a)(b)(2a – b)

= (2a – b)^{3}

[Using a^{3} + b^{3} + 3 ab(a + b) = (a + b)^{3}]

= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a^{3} – 135a + 225a^{2}

= (3)^{3} – (5a)^{3} – 3(3)(5a)(3 – 5a)

= (3 – 5a)^{3}

[Using a^{3} + b^{3} + 3 ab(a + b) = (a + b)^{3}]

= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a^{3} -27b^{3} -144a^{2}b + 108ab^{2}

= (4a)^{3} – (3b)^{3} – 3(4a)(3b)(4a – 3b)

= (4a – 3b)^{3}

[Using a^{3} – b^{3} – 3 ab(a – b) = (a – b)^{3}]

= (4a – 3b)(4a – 3b)(4a – 3b)

Question 9.

Verify

(i) x^{3} + y^{3} = (x + y)-(x^{2} – xy + y^{2})

(ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})

Solution:

(i) ∵ (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)

⇒ (x + y)^{3} – 3(x + y)(xy) = x^{3} + y^{3}

⇒ (x + y)[(x + y)2-3xy] = x^{3} + y^{3}

⇒ (x + y)(x^{2} + y^{2} – xy) = x^{3} + y^{3}

Hence, verified.

(ii) ∵ (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)

⇒ (x – y)^{3} + 3xy(x – y) = x^{3} – y^{3}

⇒ (x – y)[(x – y)^{2} + 3xy)] = x^{3} – y^{3}

⇒ (x – y)(x^{2} + y^{2} + xy) = x^{3} – y^{3}

Hence, verified.

Question 10.

Factorise each of the following

(i) 27y^{3} + 125z^{3}

(ii) 64m^{3} – 343n^{3}

[Hint See question 9]

Solution:

(i) We know that

x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2})

We have, 27y^{3} + 125z^{3} = (3y)^{3} + (5z)^{3}

= (3y + 5z)[(3y)^{2} – (3y)(5z) + (5z)^{2}]

= (3y + 5z)(9y^{2} – 15yz + 25z^{2})

(ii) We know that

x^{3} – y^{3} = (x – y)(x^{2} + xy + y^{2})

We have, 64m^{3} – 343n^{3} = (4m)^{3} – (7n)^{3}

= (4m – 7n)[(4m)^{2} + (4m)(7n) + (7n)^{2}]

= (4m – 7n)(16m^{2} + 28mn + 49n^{2})

Question 11.

Factorise 27x^{3} +y^{3} +z^{3} -9xyz.

Solution:

We have,

27x^{3} + y^{3} + z^{3} – 9xyz = (3x)^{3} + (y)^{3} + (z)^{3} – 3(3x)(y)(z)

Using the identity,

x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)

We have, (3x)^{3} + (y)^{3} + (z)^{3} – 3(3x)(y)(z)

= (3x + y + z)[(3x)^{3} + y^{3} + z^{3} – (3x × y) – (y × 2) – (z × 3x)]

= (3x + y + z)(9x^{2} + y^{2} + z^{2} – 3xy – yz – 3zx)

Question 12.

Verify that

x^{3} +y^{3} +z^{3} – 3xyz = (x + y+z)[(x-y)^{2} + (y – z)^{2} +(z – x)^{2}]

Solution:

R.H.S

= (x + y + z)[(x – y)^{2}+(y – z)^{2}+(z – x)^{2}]

= (x + y + 2)[(x^{2} + y^{2} – 2xy) + (y^{2} + z^{2} – 2yz) + (z^{2} + x^{2} – 2zx)]

= (x + y + 2)(x^{2} + y^{2} + y^{2} + z^{2} + z^{2} + x^{2} – 2xy – 2yz – 2zx)

= (x + y + z)[2(x^{2} + y^{2} + z^{2} – xy – yz – zx)]

= 2 x x (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)

= (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)

= x^{3} + y^{3} + z^{3} – 3xyz = L.H.S.

Hence, verified.

Question 13.

If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3 xyz.

Solution:

Since, x + y + z = 0

⇒ x + y = -z (x + y)^{3} = (-z)^{3}

⇒ x^{3} + y^{3} + 3xy(x + y) = -z^{3}

⇒ x^{3} + y^{3} + 3xy(-z) = -z^{3} [∵ x + y = -z]

⇒ x^{3} + y^{3} – 3xyz = -z^{3}

⇒ x^{3} + y^{3} + z^{3} = 3xyz

Hence, if x + y + z = 0, then

x^{3} + y^{3} + z^{3} = 3xyz

Question 14.

Without actually calculating the cubes, find the value of each of the following

(i) (- 12)^{3} + (7)^{3} + (5)^{3 } (ii) (28)^{3} + (- 15)^{3} + (- 13)^{3 }

Solution:

(i) We have, (-12)^{3} + (7)^{3} + (5)^{3}

Let x = -12, y = 7 and z = 5.

Then, x + y + z = -12 + 7 + 5 = 0

We know that if x + y + z = 0, then, x^{3} + y^{3} + z^{3} = 3xyz

∴ (-12)^{3} + (7)^{3} + (5)^{3} = 3[(-12)(7)(5)]

= 3[-420] = -1260

(ii) We have, (28)^{3} + (-15)^{3} + (-13)^{3}

Let x = 28, y = -15 and z = -13.

Then, x + y + z = 28 – 15 – 13 = 0

We know that if x + y + z = 0, then x^{3} + y^{3} + z^{3} = 3xyz

∴ (28)^{3} + (-15)^{3} + (-13)^{3} = 3(28)(-15)(-13)

= 3(5460) = 16380

Question 15.

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given

(i) Area 25a^{2} – 35a + 12

(ii) Area 35y^{2} + 13y – 12

Solution:

Area of a rectangle = (Length) x (Breadth)

(i) 25a^{2} – 35a + 12 = 25a^{2} – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)

Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y^{2}+ 13y -12 = 35y^{2} + 28y – 15y -12

= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)

Thus, the possible length and breadth are (7y – 3) and (5y + 4).

Question 16.

What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume 3x^{2} – 12x

(ii) Volume 12ky^{2} + 8ky – 20k

Solution:

Volume of a cuboid = (Length) x (Breadth) x (Height)

(i) We have, 3x^{2} – 12x = 3(x^{2} – 4x)

= 3 x x x (x – 4)

∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky^{2} + 8ky – 20k

= 4[3ky^{2} + 2ky – 5k] = 4[k(3y^{2} + 2y – 5)]

= 4 x k x (3y^{2} + 2y – 5)

= 4k[3y^{2} – 3y + 5y – 5]

= 4k[3y(y – 1) + 5(y – 1)]

= 4k[(3y + 5) x (y – 1)]

= 4k x (3y + 5) x (y – 1)

Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

### NCERT Solutions for Class 9 Maths

- Chapter 1 Number systems
- Chapter 2 Polynomials
- Chapter 3 Coordinate Geometry
- Chapter 4 Linear Equations in Two Variables
- Chapter 5 Introduction to Euclid Geometry
- Chapter 6 Lines and Angles
- Chapter 7 Triangles
- Chapter 8 Quadrilaterals
- Chapter 9 Areas of Parallelograms and Triangles
- Chapter 10 Circles
- Chapter 11 Constructions
- Chapter 12 Heron’s Formula
- Chapter 13 Surface Areas and Volumes
- Chapter 14 Statistics
- Chapter 15 Probability
- Class 9 Maths (Download PDF)

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