The National Council of Educational Research and Training (NCERT) is an autonomous body of the Indian government that formulates the curricula for schools in India that are governed by the Central Board of Secondary Education (CBSE) and certain state boards. Therefore, students who will be taking the Class 10 tests administered by various boards should consult this NCERT Syllabus in order to prepare for those examinations, which in turn will assist those students get a passing score.

When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 9 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.

### NCERT Solutions for Class 9 Maths Exercise 2.1

**Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.**

(i)

(ii)

(iii)

(iv)

(v)

**Solution:**

(i)

We can observe that in the polynomial , we have x as the only variable and the powers of x in each term are a whole number.

Therefore, we conclude that is a polynomial in one variable.

(ii)

We can observe that in the polynomial , we have y as the only variable and the powers of y in each term are a whole number.

Therefore, we conclude that is a polynomial in one variable.

(iii)

We can observe that in the polynomial, we have t as the only variable and the powers of t in each term are not a whole number.

Therefore, we conclude thatis not a polynomial in one variable.

(iv)

We can observe that in the polynomial, we have y as the only variable and the powers of y in each term are not a whole number.

Therefore, we conclude thatis not a polynomial in one variable.

(v)

We can observe that in the polynomial, we have x, y and t as the variables and the powers of x, y and t in each term is a whole number.

Therefore, we conclude that is a polynomial but not a polynomial in one variable.

**Question 2. Write the coefficients of in each of the following :**

(i)

(ii)

(iii)

(iv)

**Solution:**

(i)

The coefficient ofin the polynomialis 1.

(ii)

The coefficient ofin the polynomialis.

(iii)

The coefficient ofin the polynomialis.

(iv)

The coefficient ofin the polynomial is 0.

**Question 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.****Solution:**

The binomial of degree 35 can be.

The binomial of degree 100 can be.

**Question 4. Write the degree of each of the following polynomials :**

(i)

(ii)

(iii)

(iv)3

**Solution:**

(i)

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomial, the highest power of the variable x is 3.

Therefore, we conclude that the degree of the polynomialis 3.

(ii)

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomial, the highest power of the variable y is 2.

Therefore, we conclude that the degree of the polynomialis 2.

(iii)

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We observe that in the polynomial, the highest power of the variable t is 1.

Therefore, we conclude that the degree of the polynomialis 1.

(iv)3

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomial 3, the highest power of the assumed variable x is 0.

Therefore, we conclude that the degree of the polynomial 3 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 Exercise 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:**

Let 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} – t3

∴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 x2.

(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 Solutions for Class 9 Maths Exercise 2.3

**Question 1. Find the remainder whenis divided by**

(i)

(ii)

(iii) x

(iv)

(v)

**Solution:**

(i)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

=-1+3-3+1

=0

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as 0.

(ii)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

(iii)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

=0+0+0+1

=1

Therefore, we conclude that on dividing the polynomialby x, we will get the remainder as 1.

(iv)

We need to find the zero of the polynomial.

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

(v)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomial in the polynomial, to get

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

**Question 2. Find the remainder whenis divided by.**

**Solution:**

We need to find the zero of the polynomial.

= 5a

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

**Question 3. Check whether is a factor of.**

**Solution:**

We know that if the polynomialis a factor of, then on dividing the polynomialby, we must get the remainder as 0.

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialinthe polynomial, to get

We conclude that on dividing the polynomialby, we will get the remainder as, which is not 0.

Therefore, we conclude thatis not a factor of.

### NCERT Solutions for Class 9 Maths Exercise 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, 6×2 + 5x – 6 = 6×2 + 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} + y2 – 2y – 1

= (y – 1)(y + 1)(2y +1)

### NCERT Solutions for Class 9 Maths Exercise 2.5

**Question 1. Use suitable identities to find the following products :**

(i)

(ii)

(iii)

(iv)

(v)

**Solution:**

(i)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

(ii)

We know that.

We need to apply the above identity to find the product

=

Therefore, we conclude that the productis.

(iii)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

(iv)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

(v)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

**Question 2. Evaluate the following products without multiplying directly :**

(i)

(ii)

(iii)

**Solution:**

(i)

We can observe that, we can apply the identity

=10000+1000+21

=11021

Therefore, we conclude that the value of the productis.

(ii)

We can observe that, we can apply the identity

=10000-900+20

=9120

Therefore, we conclude that the value of the productis .

(iii)

We can observe that, we can apply the identitywith respect to the expression, to get

=10000-16

=9984

Therefore, we conclude that the value of the productis.

**Question 3. Factorize the following using appropriate identities :**

(i)

(ii)

(iii)

**Solution:**

(i)

We can observe that, we can apply the identity

(ii)

We can observe that, we can apply the identity

(iii)

We can observe that, we can apply the identity

**Question 4. Expand each of the following, using suitable identities :**

(i)

(ii)

(iii)

(iv)

(v)

(vi)

**Solution:**

(i)

We know that.

We need to apply the above identity to expand the expression.

(ii)

We know that.

We need to apply the above identity to expand the expression.

(iii)

We know that.

We need to apply the above identity to expand the expression.

(iv)

We know that.

We need to apply the above identity to expand the expression.

(v)

We know that.

We need to apply the above identity to expand the expression.

(vi)

We know that.

**Question 5. Factorize :**

(i)

(ii)

**Solution:**

(i)

The expression can also be written as

We can observe that, we can apply the identitywith respect to the expression, to get

Therefore, we conclude that after factorizing the expression, we get.

(ii)

We need to factorize the expression.

The expression can also be written as

We can observe that, we can apply the identitywith respect to the expression, to get

Therefore, we conclude that after factorizing the expression, we get.

**Question 6. Write the following cubes in expanded form :**

(i)

(ii)

(iii)

(iv)

**Solution :**

(i)

We know that.

Therefore, the expansion of the expressionis .

(ii)

We know that.

Therefore, the expansion of the expressionis .

(iii)

We know that.

Therefore, the expansion of the expressionis.

(iv)

We know that.

Therefore, the expansion of the expressionis.

**Question 7. Evaluate the following using suitable identities**

(i) (99)^{3}

(ii) (102)^{3}

(iii) (998)^{3}**Solution:**

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

∴ 993 = (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

∴ 1023 = (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 }+ 12a2b+6ab^{2}

(ii) 8a^{3} -b^{3}-12a2b+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} – 12a^{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) = x3 – 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

x3 + y^{3} = (x + y)(x2 – xy + y^{2})

We have, 27y^{3} + 125z3 = (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 + y2)

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

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

= (4m – 7n)(16m^{2} + 28mn + 49n2)

**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}+ y2 + 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 – 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).

## Conclusions for NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 2 POLYNOMIALS

SWC academic staff has developed NCERT answers for this chapter of the ninth grade mathematics curriculum. We have solutions prepared for all of the exercises of this chapter. The answers, broken down into steps, to all of the questions included in the NCERT textbook’s chapter are provided here. Read this chapter on theory. Be certain that you have read the theory section of this chapter of the NCERT textbook and that you have learnt the formulas for the chapter that you are studying.