# Class 9 NCERT Solutions Maths Chapter 2 - Polynomials

Find the best ways to practice NCERT Solutions for CBSE class 9 maths chapter 2 Polynomials. It is one of the crucial chapters for students of CBSE class 9 maths**. **Polynomial is the second chapter of CBSE Class 9 Maths which includes a detailed explanation of the Polynomial and different factors that include whole numbers, integers, and rational numbers.

The chapter begins with Polynomial in one variable- Linear, Quadratic and Cubic Polynomial, then it is followed by representing Real numbers on the number line, operation on Real numbers and concluded with laws of exponents for Real numbers.

You can go to the NCERT CBSE Class 9 Maths Textbook solutions of a specific chapter by selecting the chapter from our easy-to-use drop-down chapter list. Whether you want to revise an exercise in the chapter on Polynomials or check out Polynomials problems, we have got you covered. TopperLearning’s NCERT solutions allow you to improve your problem-solving techniques, revise formulae, and enable to score more marks.

Learn how to identify polynomials in one variable, find out the coefficient of Polynomial, Degree of the Polynomial and classifying Polynomials as linear, quadratic and cubic with our NCERT solutions for CBSE class 9 maths. The TopperLearning Expert Team provides the portal UNDOUBT where students can ask all their doubts. These doubts are answered by the academic experts making students understand every step clearly. Students use this portal to get a quick resolution to their questions.

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## Polynomials Exercise Ex. 2.1

### Solution 1

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

^{2}is 1

^{2}is - 1.

^{2}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

^{35}+ 7 . Monomial has only one term in it. So monomial of degree 100 can be written as 7x

^{100}. 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

**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 x

^{o}= 1.

### Solution 5

^{2}+ x is a quadratic polynomial as its degree is 2.

(ii) x - x

^{3}is a cubic polynomial as its degree is 3.

(iii) y + y

^{2}+ 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) r

^{2}is a quadratic polynomial as its degree is 2.

(vii) 7x

^{3}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.

## Polynomials Exercise Ex. 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

^{2}- y + 1

p(0) = (0)

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

p(1) = (1)

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

p(2) = (2)

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

^{2}- t

^{3}

p(0) = 2 + 0 + 2 (0)

^{2}- (0)

^{3}= 2

p(1) = 2 + (1) + 2(1)

^{2}- (1)

^{3}

= 2 + 1 + 2 - 1 = 4

^{2}- (2)

^{3}

= 2 + 2 + 8 - 8 = 4

^{3}

p(0) = (0)

^{3}= 0

p(1) = (1)

^{3}= 1

p(2) = (2)

^{3}= 8

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

^{2}- 1 then p(1) and p(- 1)

Now, p(1) = (1)

^{2}- 1 = 0

p(- 1) = (- 1)

^{2}- 1 = 0

Hence x = 1 and - 1 are zeroes of polynomial.

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.

^{2}then p(0) should be zero.

Now, p(0) = (0)

^{2}= 0

Hence x = 0 is a zero of given polynomial

^{2}- 1, then

**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

(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.

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.

p(x) = 0

2x + 5 = 0

2x = - 5

So, for value of polynomial is 0 and hence is a zero of polynomial.

p(x) = 0

3x - 2 = 0

So, for , value of polynomial is 0 and hence is a zero of polynomial.

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.

p(x) = 0

ax = 0

x = 0

So, for x = 0, value of polynomial is 0. Hence x = 0 is a zero of polynomial.

p(x) = 0

cx+ d = 0

**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.

## Polynomials Exercise Ex. 2.3

### Solution 1

^{3}+ 3x

^{2}+ 3x + 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.

Zero of x is 0.

p(0) = (0)

^{3}+ 3(0)

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

So, the remainder is 1.

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

Zero of 5+2x is:

5 + 2x = 0 2x = - 5

i.e. x = -

**OR**

By long division

By long division

By long division

**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

Here p(x) = x

^{3}- ax

^{2}+ 6x - a

p(a) = (a)

^{3}- a(a)

^{2}+ 6a - a

= 5a

So when x

^{3}- ax

^{2}+ 6x - a is divided by x - a, remainder comes to 5a.

**OR**

By long division

^{3}- ax

^{2}+ 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

7 + 3x = 0

Therefore,

^{3}+ 7x only if

Here p(x) = 3x

^{3 }+ 7x

^{3}+ 7x.

^{3}+ 7x) by (7 + 3x). If remainder comes out to be 0 then 7 + 3x will be a factor of

^{3}+ 7x.

By long division

^{3}+ 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.

## Polynomials Exercise Ex. 2.4

### Solution 1

^{3}+ x

^{2}+ x + 1, p (- 1) must be zero.

Here, p(x) = x

^{3}+ x

^{2}+ x + 1

p(-1) = (- 1)

^{3}+ (- 1)

^{2}+ (- 1) + 1

= - 1 + 1 - 1 + 1 = 0

Hence, x + 1 is a factor of this polynomial

^{4}+ x

^{3}+ x

^{2}+ x + 1, p (- 1) must be zero.

Here, 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 = 1

As,

^{4}+ 3x

^{3}+ 3x

^{2}+ 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,

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

p(x) = 2x

^{3}+ x

^{2}- 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.

p(x) = x

^{3}+3x

^{2}+ 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.

p(x) = x

^{3}- 4 x

^{2}+ 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

^{2}+ 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

^{2}- 7x + 1

The two numbers such that pq = 12 1 = 12 and p + q = - 7. They are p = - 4 and

Now, 12x

^{2}- 7x + 1 = 12x

^{2}- 4x - 3x + 1

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

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

^{2}+ 7x + 3

The two numbers such that pq = 2 3 = 6 and p + q = 7.

They are p = 6 and q = 1

Now, 2x

^{2}+ 7x + 3 = 2x

^{2}+ 6x + x + 3

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

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

^{2}+ 5x - 6

The two numbers such that pq = - 36 and p + q = 5.

They are p = 9 and q = - 4

Now,

6x

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

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

= 3x (2x + 3) - 2 (2x + 3)

= (2x + 3) (3x - 2)

(iv) 3x

^{2}- x - 4

The two numbers such that pq = 3 (- 4) = - 12

They are p = - 4 and q = 3.

Now,

3x

^{2}- x - 4 = 3x

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

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

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

**Concept Insight:**To factorise the polynomial ax

^{2}+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 |

ac<0 |
b<0 |
b =(p+q) where |

### Solution 5

^{3}- 2x

^{2}- 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

x

^{3}- 2x

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

^{2}- 3x + 2) + 0

= (x + 1) [x

^{2}- 2x - x + 2]

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

= (x + 1) (x - 1) (x - 2)

= (x - 2) (x - 1) (x + 1)

^{3}- 3x

^{2}- 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 x

^{3}+ 3x

^{2}- 9x - 5 by x + 1

By long division

x

^{3}- 3x

^{2}- 9x - 5 = (x + 1) (x

^{2}- 4 x - 5) + 0

= (x + 1) (x

^{2}- 5 x + x - 5)

= (x + 1) [(x (x - 5) +1 (x - 5)]

= (x + 1) (x - 5) (x + 1)

= (x - 5) (x + 1) (x + 1)

^{3}+ 13x

^{2 }+ 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 x

^{3}+ 13x

^{2}+ 32x + 20 by (x + 1)

By long division

Dividend = Divisor Quotient + Remainder

x

^{3}+ 13x

^{2}+ 32x + 20 = (x + 1) (x

^{2}+ 12x + 20) + 0

= (x + 1) (x

^{2}+ 10x + 2x + 20)

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

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

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

^{3}+ y

^{2}- 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,

^{3}+ y

^{2}- 2y - 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)

**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.

## Polynomials Exercise Ex. 2.5

### Solution 1

**Concept Insight:**If the value of the two terms of the binomials are equal then use the algebraic identity (x+a) (x-a) = x

^{2}- a

^{2}else use (x+a) (x+b) = x

^{2}+(a+b)x+ab to obtain required product.

### Solution 2

= (100)

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

[By using the identity, , where

= 10000 + 1000 + 21

= 11021

= (100)

^{2}+ (- 5 - 4) 100 + (- 5) (- 4)

[By using the identity, , where

x = 100, a = - 5 and b = - 4]

= 10000 - 900 + 20

= 9120

= (100)

^{2}- (4)

^{2}

= 10000 - 16

= 9984

**Concept Insight:**The key is to use the algebraic identity (x+a) (x+b) = x

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

^{2}- a

^{2}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

### Solution 4

**Concept Insight:**Use the algebraic identity

### Solution 5

**Concept Insight:**Use the algebraic identity

### 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

^{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

### 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

### 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 .

### 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

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

### Solution 12

**Concept Insight:**Since the left hand side of the identity resembles the left hand side of identity, , so this identity

### 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

### Solution 15

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

Volume of cuboid = length breadth height

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.