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# NCERT Solution for Class 9 Mathematics Chapter 2 - Polynomials

NCERT Textbook Solutions are considered extremely helpful when preparing for your CBSE Class 9 Mathematics exams. TopperLearning study resources infuse profound knowledge, and our Textbook Solutions compiled by our subject experts are no different. Here you will find all the answers to the NCERT textbook questions of Chapter 2 - Polynomials.

All our solutions for Chapter 2 - Polynomials are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

Exercise/Page

## NCERT Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2.1

Solution 1

(i)

Yes, this expression is a polynomial in one variable x.

(ii)

Yes, this expression is a polynomial in one variable y.

(iii)

No, here the exponent of variable t in term  which is not a whole number.

So this expression is not a polynomial.

(iv)

No, here the exponent of variable t in term is -1, which is not a whole number. So this expression is not a polynomial.

(v)

No, this expression is a polynomial in 3 variables x, y and t.

Concept Insight: In such problems to check whether the given algebraic expressions is a polynomial or not, check the exponents of variable to be a whole number. The second step is to look for the number of variables the expression has. Any alphabet used in the expression is the variable unless specified as constant.

Solution 2

(i)              In the above expression coefficient of x2 is 1   (ii)               In the above expression coefficient of x2 is - 1.   (iii)                  In the above expression coefficient of    (iv)                     In the above expression coefficient of x2  is 0.   Concept Insight: The constant/ variable multiplied with the variable is the coefficient of the variable. Also consider the sign (+ve or the -ve) of the term while writing the coefficient.

Solution 3

Degree of a polynomial is the highest power of variable in the polynomial.Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .   Monomial has only one term in it. So monomial of degree 100 can be written as 7x100.   Concept Insight: Mono, bi and tri means one, two and three respectively. So, monomial is a polynomial having one term similarly for binomials and trinomials. Degree is the highest exponent of variable.  The answer is not unique in such problems . Remember that the terms are always separated by +ve or -ve sign and not with  .

Solution 4

Degree of a polynomial is the highest power of variable in the polynomial.             This is a polynomial in variable x and highest power of variable x is 3. So, degree of           this polynomial is 3              This is a polynomial in variable y and the highest power of variable y is 2. So degree            of this polynomial is 2.            This is a polynomial in variable t and the highest power of variable t is 1. So degree            of this polynomial is 1. (iv)      3            This is a constant polynomial. Degree of a constant polynomial is always 0.   Concept Insight: Degree is the highest  exponent of the variable. While finding the degree of a polynomial express the polynomial in standard form  Combine the like terms and Remember the result that xo = 1.

Solution 5

(i)       2 + x2 + x  is a quadratic polynomial as its degree is 2.

(ii)      x - x3 is a cubic polynomial as  its degree is 3.

(iii)     y + y2 + 4  is a quadratic polynomial as its degree is 2.

(iv)     1 + x is a linear polynomial as its degree is 1.

(v)      3t is a linear polynomial as its degree is 1.

(vi)     r2 is a quadratic polynomial as its degree is 2.

(vii)     7x3 is a cubic polynomial as its degree is 3.     Concept Insight: Linear polynomial, quadratic polynomial and cubic polynomial has its degrees as 1, 2, and 3 respectively.

## NCERT Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2.2

Solution 1

Concept Insight: Given Polynomial is p(x) to find the  value of given polynomial at any particular value of x replace the variable x with its corresponding value. Remember for odd power of negative number the negative sign remains while for even power of negative numbers the negative sign vanishes.  Also check for  calculation errors; there are chances of making calculation mistake while computing square, cubes and higher powers of numbers.

Solution 2

(i)       p(y) = y2 - y + 1
p(0) = (0)2 - (0) + 1 = 1
p(1) = (1)2 - (1) + 1 = 1
p(2) = (2)2 - (2) + 1 = 3   (ii)      p(t) = 2 + t + 2t2 - t3
p(0) = 2 + 0 + 2 (0)2 - (0)3  = 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)     p(x) = x3
p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8   (iv)     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   Concept Insight: Replace the variable with 0, 1 or  2 in the given polynomials to obtain the required value.  Be careful about the calculations, there are chances of making calculation mistake while computing square, cubes and higher powers of numbers. Carefully apply the properties of addition, subtraction and multiplication of numbers. While multiplying two binomials multiply each term of the binomial to each term of the other binomial.

Solution 3

(i)    If  is a zero of given polynomial p(x) = 3x + 1, then                   So,  is a zero of given polynomial   (ii)    If  is a zero of polynomial p(x) = 5x -  ,                                                   then  should be 0                      So,  is not a zero of given polynomial   (iii)    If x = 1 and x = - 1 are zeroes of polynomial p(x) = x2 - 1 then p(1) and p(- 1)           should be 0
Now, p(1) = (1)2 - 1 = 0
p(- 1) = (- 1)2 - 1 = 0
Hence x = 1 and - 1 are zeroes of polynomial. (iv)    If x = - 1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x - 2), then p(- 1) and          p(2)should be 0.
Now, p(- 1) = (- 1 + 1) (- 1 - 2) = 0 (-3) = 0
p(2) = (2 + 1) (2 -  2 ) = 3 (0) = 0
So, x = - 1 and x = 2 are zeroes of given polynomial.   (v)     If x = 0 is a zero of polynomial p(x) = x2 then p(0) should be zero.
Now, p(0) = (0)2 = 0
Hence x = 0 is a zero of given polynomial   (vi)    If  is a zero of polynomial p(x) = lx + m, thenis 0.            (vii)    If  and  are zeroes of polynomial p(x) = 3x2 - 1, then                               Hence,  is a zero of given polynomial but  is not a zero of given                 polynomial.   (viii)      If    is a zero of polynomial p(x) = 2x + 1 then  should be 0.                             So,  is not a zero of polynomial.     Concept Insight: Key idea here is  Zero of the polynomial is not the real number zero but it is that value of the variable which makes the  value of the polynomial equal to zero. A polynomial can have more than one zeroes.

Solution 4

Zero of a polynomial is that value of variable at which value of polynomial comes to 0.
(i)     p(x) = x + 5
p(x) = 0
x + 5 = 0
x = - 5
So, for x = - 5, value of polynomial is 0 and hence x = - 5 is a zero of polynomial. (ii)    p(x) = x - 5
p(x) = 0
x - 5 = 0
x = 5
So, for x = 5 value of polynomial is 0 and hence x = 5 is a zero of polynomial. (iii)   p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = - 5

So, for   value of polynomial is 0 and hence   is a zero of polynomial.   (iv)   p(x) = 3x - 2
p(x) = 0
3x - 2 = 0

So, for  , value of polynomial is 0 and hence   is a zero of polynomial.   (v)    p(x) = 3x
p(x) = 0
3x = 0
x = 0
So, for x = 0 value of polynomial is 0 and hence x = 0 is a zero of polynomial.   (vi)   p(x) = ax
p(x) = 0
ax = 0
x = 0
So, for x = 0, value of polynomial is 0. Hence x = 0 is a zero of polynomial.   (vii)   p(x) = cx + d
p(x) = 0
cx+ d = 0
So, for , value of polynomial is 0. Hence  is a zero of polynomial.   Concept Insight: Equate the polynomial to zero and solve the corresponding linear equation to get the value of variable. Be careful while transposing the terms to the other side. For verification substitute the value of the variable obtained in the polynomial.

## NCERT Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2.3

Solution 1

Let p(x) = x3 + 3x2 + 3x + 1. (i)     x + 1
Zero of x +1 is-1.
i.e. p(-1) = (- 1)3 + 3 (- 1)2 + 3 (-1) + 1 = 0
So, the remainder is 0.   (ii)               Zero of   is           (iii)     x
Zero of x is 0.
p(0) = (0)3 + 3(0)2 + 3(0) + 1 = 1
So, the remainder is 1.   (iv)     x +
Zero of x + is:
x + = 0 x = -
p (- ) = (- )3 + 3(- )2 + 3(- ) + 1 = - 3 + 3 2 - 3 + 1
So, the remainder is - 3 + 3 2 - 3 + 1   (v)      5 + 2x
Zero of 5+2x  is:
5 + 2x = 0 2x = - 5
i.e. x = -                                              OR   (i)     x + 1         By long division                      So, remainder is 0. (ii).
By long division

So, remainder is  . (iii)     x           By long division                       So, remainder is 1.   (iv)     x +
By long division                        So, the remainder is (v)     5 + 2x
By long division                       So the remainder is -.   Concept Insight: The remainder of any polynomial p(x) when divided by another polynomial (ax+b) where a and b are real numbers  is p(-b/a).
Note that here -b/a  is the zero of polynomial ax+b.
This problem can also be solved using long division. For long division  first write the divisor and dividend in the standard form, i.e. arrange the terms in the descending order of their powers. The process of division is continued till the remainder is constant or the degree of new dividend is less than the degree of divisor. Do not forget to change the sign of terms while subtraction. For cross verification division algorithm
Dividend = Quotient  Divisor + Remainder can be used.

Solution 2

According to the remainder theorem, if p(x) is any polynomial of degree  1 and a is any real number, then when p(x) is divided by the linear polynomial x - a, then the remainder is p(a).

Here p(x) = x3 - ax2 + 6x - a
p(a) = (a)3 - a(a)2 + 6a - a
= 5a
So when x3 - ax2 + 6x - a is divided by x - a, remainder comes to 5a. OR
By long division       So when x3 - ax2 + 6x - a is divided by x - a, remainder comes to 5a.   Concept Insight:  The remainder of any polynomial p(x) when divided by another polynomial (ax+b) where a and b are real numbers  is
p(-b/a).
Note that here -b/a  is the zero of polynomial ax+ b.
This question can also be solved using long division method however it is long and time consuming. Chances of making computational error are high in that method.

Solution 3

Zero of 7 + 3x is:
7 + 3x = 0
Therefore, 7+3x can be a factor of p(x) = 3x3 + 7x only if
Here p(x) = 3x3 + 7x   7 + 3x is not a factor of 3x3 + 7x.   OR Let us divide (3x3 + 7x) by (7 + 3x). If remainder comes out to be 0 then  7 + 3x will be a factor of 3x3 + 7x.

By long division       As remainder is not zero so 7 + 3x is not a factor of 3x3 + 7x.

Concept Insight: Any linear polynomial 'ax+b' where a and b are real numbers is a factor of the polynomial p(x) iff p(-b/a) = 0 i.e  -b/a is a zero of p(x) or both the polynomials has a common zero -b/a. This question can also be solved using long division method. Do not forget to change the sign of terms while subtraction in the long division.

## NCERT Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2.4

Solution 1

(i)    If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, p (- 1) must be zero.

Here, p(x) = x3 + x2 + x + 1
p(-1) = (- 1)3 + (- 1)2 + (- 1) + 1
= - 1 + 1 - 1 + 1 = 0
Hence, x + 1 is a factor of this polynomial (ii)    If (x + 1) is a factor of p(x) = x4 + x3 + x2 + x + 1, p (- 1) must be zero.

Here, p(x) = x4 + x3 + x2 + x + 1
p( -1) = (- 1)4 + (- 1)3 + (- 1)2 + (- 1) + 1
= 1 - 1 + 1 -1 + 1 = 1
As,           So, x + 1 is not a factor of this polynomial   (iii)    If (x + 1) is a factor of polynomial p(x) = x4 + 3x3 + 3x2 + x + 1, p(- 1) must be 0.

p(- 1) = (- 1)4 + 3(- 1)3 + 3(- 1)2 + (- 1) + 1
= 1 - 3 + 3 - 1 + 1 = 1
As,            So, x + 1 is not a factor of this polynomial   (iv)    If (x + 1) is a factor of polynomial
p(x) =  ,  p(- 1) must be 0.

As,
So, (x + 1) is not a factor of this polynomial.   Concept Insight: A linear polynomial 'x-a' is a factor of the polynomial p(x) iff p(a) = 0. Note that 'a' is a zero of polynomials x-a and  p(x) . Be careful while squaring and cubing the numbers.

Solution 2

(i)    If g(x) = x + 1 is a factor of given polynomial p(x), p(- 1)  must be zero.
p(x) = 2x3 + x2 - 2x - 1
p(- 1) = 2(- 1)3 + (- 1)2 - 2(- 1) - 1
= 2(- 1) + 1 + 2 - 1 = 0
Hence, g(x) = x + 1 is a factor of given polynomial.   (ii)    If g(x) = x + 2 is a factor of given polynomial p(x), p(- 2) must be 0.
p(x) = x3 +3x2 + 3x + 1
p(- 2) = (- 2)3 + 3(- 2)2 + 3(- 2) + 1
= - 8 + 12 - 6 + 1
= - 1

Hence g(x) = x + 2 is not a factor of given polynomial.   (iii)    If g(x) = x - 3 is a factor of given polynomial p(x), p(3) must be 0.
p(x) = x3 - 4 x2 + x + 6
p(3) = (3)3 - 4(3)2 + 3 + 6
= 27 - 36 + 9 = 0
So, g(x) = x - 3 is a factor of given polynomial.   Concept Insight: The problem is a direct application of Factor theorem. g(x) will be the factor of the polynomial p(x) iff the zero of the linear polynomial g(x) when put in place of the variable of polynomial results to zero. Be careful while squaring and cubing the numbers.

Solution 3

If x - 1 is a factor of polynomial p(x), then p(1) = 0   (i)    p(x) = x2 + x + k
p(1) = 0
(1)2 + 1 + k = 0
2 + k = 0
k = - 2
So, value of k is - 2.         Concept Insight: x-1 is a factor of the given polynomial p(x) iff p(1) = 0 thus equating p(1) to zero will give the required value of constant k. Be careful with arithmetic simplifications.

Solution 4

(i)    12x2 - 7x + 1
The two numbers such that pq = 12  1 = 12 and p + q = - 7. They are p = - 4 and        q = - 3
Now, 12x2 - 7x + 1 = 12x2 - 4x - 3x + 1
= 4x (3x - 1) - 1 (3x - 1)
= (3x - 1) (4x - 1)   (ii)    2x2 + 7x + 3
The two numbers such that pq = 2  3 = 6 and p + q = 7.
They are p = 6 and q = 1
Now, 2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x+ 1)   (iii)    6x2 + 5x - 6
The two numbers such that pq = - 36 and p + q = 5.
They are p = 9 and q = - 4
Now,
6x2 + 5x - 6 = 6x2 + 9x - 4x - 6
= 3x (2x + 3) - 2 (2x + 3)
= (2x + 3) (3x - 2)
(iv)    3x2 - x - 4
The two numbers such that pq = 3  (- 4) = - 12          and p + q = - 1.
They are p = - 4 and q = 3.
Now,
3x2 - x - 4 = 3x2 - 4x + 3x - 4
= x (3x - 4) + 1 (3x - 4)
= (3x - 4) (x + 1)

Concept Insight: To factorise the polynomial ax2+bx+c, by splitting the middle term,
b is expressed as the sum of two numbers whose product is ac.
Do not forget to consider the sign of the terms while splitting.
Remember

 ac>0 b>0 b =(p+q) where p>0,q>0 ac>0 b<0 b =(p+q) where p<0,q<0 ac<0 b>0 b =(p+q) where p > q then p>0 and q<0 ac<0 b<0 b =(p+q) where p > q then p<0 and q>0

Solution 5

(i)    Let p(x) = x3 - 2x2 - x + 2
Factors of 2 are 1, 2.
By hit and trial method
p(2) = (2)3 - 2(2)2 - 2 + 2
= 8 - 8 - 2 + 2 = 0
So, (x - 2) is factor of polynomial p(x)

By long division                 Now,     Dividend = Divisor  Quotient + Remainder
x3 - 2x2 - x + 2 = (x + 1) (x2 - 3x + 2) + 0
= (x + 1) [x2 - 2x - x + 2]
= (x + 1) [x (x - 2) - 1 (x - 2)]
= (x + 1) (x - 1) (x - 2)
= (x - 2) (x - 1) (x + 1)   (ii)    Let p(x) = x3 - 3x2 - 9x - 5
Factors of 5 are 1, 5.
By hit and trial method
p(- 1) = (- 1)3 - 3(- 1)2 - 9(- 1) - 5
= - 1 - 3 + 9 - 5 = 0
So x + 1 is a factor of this polynomial
Let us find the quotient while dividing x3 + 3x2 - 9x - 5 by x + 1
By long division                    Now, Dividend = Divisor  Quotient + Remainder
x3 - 3x2 - 9x - 5 = (x + 1) (x2 - 4 x - 5) + 0
= (x + 1) (x2 - 5 x + x - 5)
= (x + 1) [(x (x - 5) +1 (x - 5)]
= (x + 1) (x - 5) (x + 1)
= (x - 5) (x + 1) (x + 1)   (iii)    Let p(x) = x3 + 13x2 + 32x + 20
The factors of 20 are 1, 2, 4, 5 ... ...
By hit and trial method
p(- 1) = (- 1)3 + 13(- 1)2 + 32(- 1) + 20
= - 1 + 13 - 32 + 20
= 33 - 33 = 0
As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).

Let us find the quotient while dividing x3 + 13x2 + 32x + 20 by (x + 1)
By long division                       We know that
Dividend = Divisor  Quotient + Remainder
x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0
= (x + 1) (x2 + 10x + 2x + 20)
= (x + 1) [x (x + 10) + 2 (x + 10)]
= (x + 1) (x + 10) (x + 2)
= (x + 1) (x + 2) (x + 10)   (iv)    Let p(y) = 2y3 + y2 - 2y - 1
By hit and trial method
p(1) = 2 ( 1)3 + (1)2 - 2( 1) - 1
= 2 + 1 - 2 - 1= 0
So, y - 1 is a factor of this polynomial
By long division method,                      p(y) = 2y3 + y2 - 2y - 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)

Concept Insight: To factorise p(x) when its degree is greater than or equal to 3 note down all the factors of constant term considering both negative and positive sign.
Check the obtained factors for the possible zeroes of the polynomial p(x)  Using Factor theorem one zero can be obtained continue the process till all the zeroes are obtained or use long division method. To obtain the other quadratic factor use long division to determine the other factors. The degree of the polynomial is  less than or  equal to the number of real factors the polynomial.

## NCERT Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2.5

Solution 1

(i).    By using identity           (ii).    By using identity              (iii).                   By using the identity               (iv).    By using identity
(v).    By using identity                Concept Insight: If the value of the two terms of the binomials are equal then use the algebraic identity (x+a) (x-a) = x2 - a2 else use (x+a) (x+b) = x2+(a+b)x+ab to obtain required product.

Solution 2

(i).    103  107 = (100 + 3) (100 + 7)
= (100)2 + (3 + 7) 100 + (3) (7)
[By using the identity, , where
x = 100, a = 3 and b = 7]
= 10000 + 1000 + 21
= 11021 (ii).   95  96 = (100 - 5) (100 - 4)
= (100)2 + (- 5 - 4) 100 + (- 5) (- 4)
[By using the identity, , where
x = 100, a = - 5 and b = - 4]
= 10000 - 900 + 20
= 9120 (iii).  104  96 = (100 + 4) (100 - 4)
= (100)2 - (4)2

= 10000 - 16
= 9984   Concept Insight: The key is to use the algebraic identity (x+a) (x+b) = x2+(a+b)x+ab or (x+a) (x-a) = x2 - a2  for such questions. Write each of the numeral as  100 k ,  or any other suitable number whose square can be easily computed.

Solution 3

Concept Insight: Use the appropriate square identity.  If the polynomial has only two terms, reduce each term to the perfect square and use the algebraic identity . When the polynomial has three terms and the term having   unit power of each variable has negative sign use the square identity  else use  .

Solution 4

Concept Insight: Use the algebraic identity     . Do consider the sign of terms while multiplying and squaring.

Solution 5

Concept Insight: Use the algebraic identity       in the reverse order. Write each term as per the terms of the standard identity. Do consider the sign of terms involved.

Solution 6

Concept Insight: Since the expressions involves cube so cubic identity will be used.  If the terms of the given polynomial are separated by positive sign use the identity   or if  negative signs are used  then use . Carefully apply the mathematical operations.

Solution 7

We know that     (i)    (99)3 = (100 - 1)3
= (100)3 - (1)3 - 3(100) (1) (100 - 1)
= 1000000 - 1 - 300(99)
= 1000000 - 1 - 29700
= 970299

(ii)    (102)3 = (100 + 2)3
= (100)3 + (2)3 + 3(100) (2) (100 + 2)
= 1000000 + 8 + 600 (102)
= 1000000 + 8 + 61200
= 1061208

(iii)    (998)3 = (1000 - 2)3
= (1000)3 - (2)3 - 3(1000) (2) (1000 - 2)
= 1000000000 - 8 - 6000(998)
= 1000000000 - 8 - 5988000
= 1000000000 - 5988008
= 994011992     Concept Insight: Use the cubic identity  and  . Write the numerical term as something added or   subtracted from 10,100, 1000 or higher powers of 10 as it's easy to compute higher powers of 10. Carefully apply the mathematical operations.

Solution 8

Concept Insight: Since all the polynomial given here have degree 3 so cubic identities would be used here. Now if all the terms of the given polynomial are positive  then use identity   while if any two terms has negative sign reduce each of   the term of the polynomial as per the standard cubic identity

Solution 9

Concept Insight: When the two terms of the polynomial are separated by positive sign use the identity   and when by negative sign use .   Carefully take the common term out.

Solution 10

Concept Insight: Reduce the terms of the polynomial to perfect cube and then if the two terms of the polynomial are separated by positive sign use the identity   and when by negative sign use  .

Solution 11

We Know that            Concept Insight: Reduce each terms of the polynomial as per the left hand side of the standard identity,  .

Solution 12

We know that                              Concept Insight: Since the left hand side of the identity resembles the left hand side of identity,  , so this identity   will be applicable here. Now the right hand side of the above identity can be written into many forms we need to look at what is required to proved, Accordingly apply  mathematical simplifications and square identities to get the desired result.

Solution 13

Concept Insight: Use the result that    for x + y + z = 0.

Solution 14

Concept Insight: Use the result   since x + y + z = 0. Also consider the   sign of the term. Carefully do the computation.

Solution 15

We know that,
Area = length  breadth   Concept Insight: For such questions factorise the expression, given for the area of rectangle by splitting the middle term. One of its factors will be its length and the other will be its breadth.

Solution 16

We know that,
Volume of cuboid = length  breadth  height (i).
So, the possible solutions is
Length = 3, breadth = x, height = x - 4             Concept Insight: For such questions factorise the expression, given for the volume of the cuboid by taking the common term out if it has two terms and by splitting the middle term if the polynomial has three terms. Three factors obtained will be its length  breadth and  height.

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