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 of
in 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) x2+ x
(ii) x – x3
(iii) y + y2+4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3
Solution :
(i) The degree of x2 + x is 2. So, it is a quadratic polynomial.
(ii) The degree of x – x3 is 3. So, it is a cubic polynomial.
(iii) The degree of y + y2 + 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 r2 is 2. So, it is a quadratic polynomial.
(vii) The degree of 7x3 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 – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let p(x) = 5x – 4x2 + 3
(i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.
(ii) p(-1) = 5(-1) – 4(-1)2 + 3
= – 5x – 4x2 + 3 = -9 + 3 = -6
Thus, the value of 5x – 4x2 + 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 – 4x2 + 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) = y2 – y +1
(ii) p (t) = 2 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)
Solution:
(i) Given that p(y) = y2 – 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 + 2t2 – 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) = x3
∴ 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) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = – 
(vii) P (x) = 3x2 – 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) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x2 -1.
Also, p(-1) = (-1)2 -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x2 – 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) = x2
∴ 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) = 3x2 – 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.
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
.
(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
.
While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial
, to get
= 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 polynomial
is 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) x3+x2+x +1
(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 – x2 – (2 +√2 )x + √2
Solution :
The zero of x + 1 is -1.
(i) Let p (x) = x3 + x2 + x + 1
∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x3 + x2 + x + 1.
(ii) Let p (x) = x4 + x3 + x2 + 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 x4 + x3 + x2 + x+ 1.
(iii) Let p (x) = x4 + 3x3 + 3x2 + 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 x4 + 3x3 + 3x2 + x+ 1.
(iv) Let p (x) = x3 – x2 – (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 x3 – x2 – (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)= 2x3 + x2 – 2x – 1, g (x) = x + 1
(ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2
(iii) p (x) = x3 – 4x2 + x + 6, g (x) = x – 3
Solution :
(i) We have, p (x)= 2x3 + x2 – 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) x3 + 3x2 + 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, = x3 – 4x2 + 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) = x2 + x + k
(ii) p (x) = 2x2 + kx + √2
(iii) p (x) = kx2 – √2 x + 1
(iv) p (x) = kx2 – 3x + k
Solution :
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.
(i) Here, p(x) = x2 + x + k
Since, p(1) = (1)2 +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.
(ii) Here, p (x) = 2x2 + kx + √2
Since, p(1) = 2(1)2 + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)
(iii) Here, p (x) = kx2 – √2 x + 1
Since, p(1) = k(1)2 – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1
(iv) Here, p(x) = kx2 – 3x + k
p(1) = k(1)2 – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k = 
Question 4. Factorise
(i) 12x2 – 7x +1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
Solution :
(i) We have,
12x2 – 7x + 1 = 12x2 – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x2 -7x + 3 = (2x – 1) (x + 3)
(ii) We have, 2x2 + 7x + 3 = 2x2 + 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, 6x2 + 5x – 6 = (2x + 3)(3x – 2)
(iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x2 – x – 4 = (3x – 4)(x + 1)
Question 5. Factorise
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
Solution:
(i) We have, x3 – 2x2 – x + 2
Rearranging the terms, we have x3 – x – 2x2 + 2
= x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2)
= [(x)2 – (1)2](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a2 – b2) = (a + b)(a-b)]
Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)
(ii) We have, x3 – 3x2 – 9x – 5
= x3 + x2 – 4x2 – 4x – 5x – 5 ,
= x2 (x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x2 – 4x – 5)
= (x + 1)(x2 – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1)
(iii) We have, x3 + 13x2 + 32x + 20
= x3 + x2 + 12x2 + 12x + 20x + 20
= x2(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x2 + 12x + 20)
= (x + 1)(x2 + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Thus, x3 + 13x2 + 32x + 20
= (x + 1)(x + 2)(x + 10)
(iv) We have, 2y3 + y2 – 2y – 1
= 2y3 – 2y2 + 3y2 – 3y + y – 1
= 2y2(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y2 + 3y + 1)
= (y – 1)(2y2 + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Thus, 2y3 + 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
.png "NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 2 POLYNOMIALS 175")
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 identity
with 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 identity
with 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 identity
with 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 expression
is
.
(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 – 13 – 3(100)(1)(100 -1)
[Using (a – b)3 = a3 – b3 – 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 = a3 + b3 + 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 = a3 – b3 – 3ab (a – b)]
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992
Question 8. Factorise each of the following
(i) 8a3 +b3 + 12a2b+6ab2
(ii) 8a3 -b3-12a2b+6ab2
(iii) 27-125a3 -135a+225a2
(iv) 64a3 -27b3 -144a2b + 108ab2
Solution:
(i) 8a3 +b3 +12a2b+6ab2
= (2a)3 + (b)3 + 6ab(2a + b)
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2 a + b)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (2a + b)(2a + b)(2a + b)
(ii) 8a3 – b3 – 12a2b + 6ab2
= (2a)3 – (b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (2a – b) (2a – b) (2a – b)
(iii) 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (3 – 5a) (3 – 5a) (3 – 5a)
(iv) 64a3 -27b3 -144a2b + 108ab2
= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
[Using a3 – b3 – 3 ab(a – b) = (a – b)3]
= (4a – 3b)(4a – 3b)(4a – 3b)
Question 9. Verify
(i) x3 + y3 = (x + y)-(x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
(i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y)
⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3
⇒ (x + y)[(x + y)2-3xy] = x3 + y3
⇒ (x + y)(x2 + y2 – xy) = x3 + y3
Hence, verified.
(ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y)
⇒ (x – y)3 + 3xy(x – y) = x3 – y3
⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3
⇒ (x – y)(x2 + y2 + xy) = x3 – y3
Hence, verified.
Question 10. Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
[Hint See question 9]
Solution:
(i) We know that
x3 + y3 = (x + y)(x2 – xy + y2)
We have, 27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2)
(ii) We know that
x3 – y3 = (x – y)(x2 + xy + y2)
We have, 64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2)
Question 11. Factorise 27x3 +y3 +z3 -9xyz.
Solution:
We have,
27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
Using the identity,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx)
Question 12. Verify that
x3 +y3 +z3 – 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)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)]
= (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx)
= (x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)]
= 2 x x (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = L.H.S.
Hence, verified.
Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.
Solution:
Since, x + y + z = 0
⇒ x + y = -z (x + y)3 = (-z)3
⇒ x3 + y3 + 3xy(x + y) = -z3
⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z]
⇒ x3 + y3 – 3xyz = -z3
⇒ x3 + y3 + z3 = 3xyz
Hence, if x + y + z = 0, then
x3 + y3 + z3 = 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, x3 + y3 + z3 = 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 x3 + y3 + z3 = 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 25a2 – 35a + 12
(ii) Area 35y2 + 13y – 12
Solution:
Area of a rectangle = (Length) x (Breadth)
(i) 25a2 – 35a + 12 = 25a2 – 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) 35y2+ 13y -12 = 35y2 + 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 3x2 – 12x
(ii) Volume 12ky2 + 8ky – 20k
Solution:
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) We have, 3x2 – 12x = 3(x2 – 4x)
= 3 x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).
(ii) We have, 12ky2 + 8ky – 20k
= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 x k x (3y2 + 2y – 5)
= 4k[3y2 – 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.
