# 2-POLYNOMIALS

Here we are going through the ncert solutions for class 9 maths chapter 2 – polynomials in the zoo so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

## (ncert solutions for class 9 maths chapter 2)

### EXERCISE 2.1 (ncert solutions for class 9 maths chapter 2)

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)

(i)

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

(ii)

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

(iii)

No. It can be observed that the exponent of variable t in term   is  , which is not a whole number. Therefore, this expression is not a polynomial.

(iv)

No. It can be observed that the exponent of variable y in term is −1, which is not a whole number. Therefore, this expression is not a polynomial.

(v)

No. It can be observed that this expression is a polynomial in 3 variables xy, and t. Therefore, it is not a polynomial in one variable.

Question 2:

Write the coefficients of  in each of the following:

(i)   (ii)

(iii)   (iv)

(i)

In the above expression, the coefficient of  is 1.

(ii)

In the above expression, the coefficient of  is −1.

(iii)

In the above expression, the coefficient of   is .

(iv)

In the above expression, the coefficient of   is 0.

Question 3:                                        (ncert solutions for class 9 maths chapter 2)

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

Degree of a polynomial is the highest power of the variable in the polynomial.

Binomial has two terms in it. Therefore, binomial of degree 35 can be written as  .

Monomial has only one term in it. Therefore, monomial of degree 100 can be written as x100.

Question 4:

Write the degree of each of the following polynomials:

(i)   (ii)

(iii)   (iv) 3

Degree of a polynomial is the highest power of the variable in the polynomial.

(i)

This is a polynomial in variable x and the highest power of variable x is 3. Therefore, the degree of this polynomial is 3.

(ii)

This is a polynomial in variable y and the highest power of variable y is 2. Therefore, the degree of this polynomial is 2.

(iii)

This is a polynomial in variable t and the highest power of variable t is 1. Therefore, the degree of this polynomial is 1.

(iv) 3

This is a constant polynomial. Degree of a constant polynomial is always 0.

Question 5:

Classify the following as linear, quadratic and cubic polynomial:

(i)  (ii)   (iii)   (iv)   (v)

(vi)   (vii)

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively.

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

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

(iii)   is a quadratic polynomial as its degree is 2.

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

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

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

(vii)  is a cubic polynomial as its degree is 3.

### EXERCISE 2.2 (ncert solutions for class 9 maths chapter 2)

Question 1:

Find the value of the polynomial   at

(i) x = 0 (ii) x = −1 (iii) x = 2

(i)

(ii)

(iii)

Question 2:                                          (ncert solutions for class 9 maths chapter 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 + + 2t2 − t3

(iii) p(x) = x3 (iv) p(x) = (x − 1) (x + 1)

(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 + + 2t2 − t3

p(0) = 2 + 0 + 2 (0)2 − (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) 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

Question 3:

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

(i)   (ii)

(iii) p(x) = x2 − 1, x = 1, − 1 (iv) p(x) = (x + 1) (x − 2), x = − 1, 2

(v) p(x) = xx = 0 (vi) p(x) = lx m

(vii)   (viii)

(i) If  is a zero of given polynomial p(x) = 3x + 1, then   should be 0.

Therefore,  is a zero of the given polynomial.

(ii) If  is a zero of polynomial p(x) = 5x − π , then should be 0.

Therefore,  is not a zero of the 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.

Here, p(1) = (1)2 − 1 = 0, and

p(− 1) = (− 1)2 − 1 = 0

Hence, x = 1 and −1 are zeroes of the given 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.

Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and

p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0

Therefore, x = −1 and = 2 are zeroes of the given polynomial.

(v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.

Here, p(0) = (0)= 0

Hence, x = 0 is a zero of the given polynomial.

(vi) If  is a zero of polynomial p(x) = lx + m, then  should be 0.

Here,

Therefore,  is a zero of the given polynomial.

(vii) If  and  are zeroes of polynomial p(x) = 3x2 − 1, then

Hence,  is a zero of the given polynomial. However,  is not a zero of the given polynomial.

(viii) If  is a zero of polynomial p(x) = 2x + 1, then  should be 0.

Therefore,  is not a zero of the given polynomial.

Question 4:                                            (ncert solutions for class 9 maths chapter 2)

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≠ 0

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

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) p(x) = x + 5

p(x) = 0

x + 5 = 0

x = − 5

Therefore, for x = −5, the value of the polynomial is 0 and hence, x = −5 is a zero of the given polynomial.

(ii) p(x) = x − 5

p(x) = 0

x − 5 = 0

x = 5

Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.

(iii) p(x) = 2x + 5

p(x) = 0

2x + 5 = 0

2x = − 5

Therefore, for , the value of the polynomial is 0 and hence,   is a zero of the given polynomial.

(iv) p(x) = 3x − 2

p(x) = 0

3x − 2 = 0

Therefore, for , the value of the polynomial is 0 and hence,   is a zero of the given polynomial.

(v) p(x) = 3x

p(x) = 0

3x = 0

x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vi) p(x) = ax

p(x) = 0

ax = 0

x = 0

Therefore, for = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vii) p(x) = cx + d

p(x) = 0

cx+ d = 0

Therefore, for , the value of the polynomial is 0 and hence,  is a zero of the given polynomial.

### EXERCISE 2.3 (ncert solutions for class 9 maths chapter 2)

Question 1:

Find the remainder when x3 + 3x2 + 3x + 1 is divided by

(i) x + 1 (ii)   (iii) x

(iv) x + π (v) 5 + 2x

(i) x + 1

By long division,

Therefore, the remainder is 0.

(ii)

By long division,

Therefore, the remainder is .

(iii) x

By long division,

Therefore, the remainder is 1.

(iv) x + π

By long division,

Therefore, the remainder is

(v) 5 + 2x

By long division,

Therefore, the remainder is

Question 2:                                              (ncert solutions for class 9 maths chapter 2)

Find the remainder when x3 − ax2 + 6x − is divided by x − a.

By long division,

Therefore, when x3 − ax2 + 6x − is divided by x − a, the remainder obtained is 5a.

Question 3:

Check whether 7 + 3x is a factor of 3x3 + 7x.

Let us divide (3x3 + 7x) by (7 + 3x). If the remainder obtained is 0, then 7 + 3will be a factor of 3x3 + 7x.

By long division,

As the remainder is not zero, therefore, 7 + 3x is not a factor of 3x3 + 7x.

### EXERCISE 2.4 (ncert solutions for class 9 maths chapter 2)

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)

(i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

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, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x4 + x3 + x2 + x + 1

p(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1

= 1 − 1 + 1 −1 + 1 = 1

As p(− 1) ≠ 0,

Therefore, 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, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

p(−1) = (−1)4 + 3(−1)3 + 3(−1)2 + (−1) + 1

= 1 − 3 + 3 − 1 + 1 = 1

As p(−1) ≠ 0,

Therefore, x + 1 is not a factor of this polynomial.

(iv) If(x + 1) is a factor of polynomial p(x) =  , then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

As p(−1) ≠ 0,

Therefore, (x + 1) is not a factor of this polynomial.

Question 2:                                              (ncert solutions for class 9 maths chapter 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 − 4 x2 + x + 6, g(x) = x − 3

(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then 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 the given polynomial.

(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then 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

As p(−2) ≠ 0,

Hence, g(x) = x + 2 is not a factor of the given polynomial.

(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then 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

Hence, g(x) = x − 3 is a factor of the given polynomial.

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 + (ii)

(iii)  (iv) p(x) = kx2 − 3x + k

If x − 1 is a factor of polynomial p(x), then p(1) must be 0.

(i) p(x) = x2 + x + k

p(1) = 0

⇒ (1)2 + 1 + k = 0

⇒ 2 + k = 0
⇒ k = −2

Therefore, the value of k is −2.

(ii)

p(1) = 0

(iii)

p(1) = 0

(iv) p(x) = kx2 − 3x + k

⇒ p(1) = 0

⇒ k(1)2 − 3(1) + k = 0

⇒ k − 3 + k = 0

⇒ 2− 3 = 0

Therefore, the value of k is .

Question 4:                                                  (ncert solutions for class 9 maths chapter 2)

Factorise:

(i) 12x2 − 7x + 1 (ii) 2x2 + 7x + 3

(iii) 6x2 + 5x − 6 (iv) 3x2 − x − 4

(i) 12x2 − 7x + 1

We can find two numbers such that pq = 12 × 1 = 12 and q = −7. They are p = −4 and = −3.

Here, 12x2 − 7x + 1 = 12x2 − 4− 3x + 1

= 4(3− 1) − 1 (3− 1)

= (3− 1) (4− 1)

(ii) 2x2 + 7x + 3

We can find two numbers such that pq = 2 × 3 = 6 and q = 7.

They are p = 6 and = 1.

Here, 2x2 + 7x + 3 = 2x2 + 6x + x + 3

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

= (x + 3) (2x+ 1)

(iii) 6x2 + 5x − 6

We can find two numbers such that pq = −36 and q = 5.

They are p = 9 and = −4.

Here,

6x2 + 5x − 6 = 6x2 + 9x − 4x − 6

= 3(2+ 3) − 2 (2+ 3)

= (2x + 3) (3− 2)

(iv) 3x2 − x − 4

We can find two numbers such that pq = 3 × (− 4) = −12

and q = −1.

They are p = −4 and = 3.

Here,

3x2 − x − 4 = 3x2 − 4x + 3x − 4

(3− 4) + 1 (3− 4)

= (3x − 4) (+ 1)

Question 5:

Factorise:

(i) x3 − 2x2 − x + 2 (ii) x3 + 3x2 −9− 5

(iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 − 2y − 1

(i) Let p(x) = x3 − 2x2 − x + 2

All the factors of 2 have to be considered. These are ± 1, ± 2.

By trial method,

p(−1) = (−1)3 − 2(−1)2 − (−1) + 2

= −1 − 2 + 1 + 2 = 0

Therefore, (x +1 ) is factor of polynomial p(x).

Let us find the quotient on dividing x3 − 2x2 − x + 2 by x + 1.

By long division,

It is known that,

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 − 9− 5

All the factors of 5 have to be considered. These are ±1, ± 5.

By trial method,

p(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5

= − 1 − 3 + 9 − 5 = 0

Therefore, x + 1 is a factor of this polynomial.

Let us find the quotient on dividing x3 + 3x2 − 9− 5 by x + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ x3 − 3x2 − 9− 5 = (+ 1) (x2 − 4x − 5) + 0

= (+ 1) (x2 − 5x + x − 5)

(x + 1) [((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

All the factors of 20 have to be considered. Some of them are ±1,

± 2, ± 4, ± 5 ……

By trial method,

p(−1) = (−1)3 + 13(−1)2 + 32(−1) + 20

= − 1 +13 − 32 + 20

= 33 − 33 = 0

As p(−1) is zero, therefore, + 1 is a factor of this polynomial p(x).

Let us find the quotient on dividing x3 + 13x2 + 32x + 20 by (x + 1).

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

x3 + 13x2 + 32x + 20 = (+ 1) (x2 + 12x + 20) + 0

= (+ 1) (x2 + 10x + 2+ 20)

= (x + 1) [x (+ 10) + 2 (+ 10)]

= (x + 1) (+ 10) (+ 2)

= (x + 1) (x + 2) (x + 10)

(iv) Let p(y) = 2y3 + y2 − 2y − 1

By trial method,

p(1) = 2 ( 1)3 + (1)2 − 2( 1) − 1

= 2 + 1 − 2 − 1= 0

Therefore, y − 1 is a factor of this polynomial.

Let us find the quotient on dividing 2y3 + y2 − 2y − 1 by y − 1.

p(y) = 2y3 + y2 − 2y − 1

= (− 1) (2y2 +3y + 1)

= (− 1) (2y2 +2y + y +1)

= (− 1) [2(+ 1) + 1 (+ 1)]

= (− 1) (+ 1) (2+ 1)

### MISCELLANEOUS EXERCISE

Question 1:

Use suitable identities to find the following products:

(i)   (ii)

(iii)   (iv)

(v)

(i) By using the identity  ,

(ii) By using the identity  ,

(iii)

By using the identity  ,

(iv) By using the identity  ,

(v) By using the identity  ,

Question 2:                                              (ncert solutions for class 9 maths chapter 2)

Evaluate the following products without multiplying directly:

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

(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

Question 3:

Factorise the following using appropriate identities:

(i) 9x2 + 6xy + y2

(ii)

(iii)

(i)

(ii)

(iii)

Question 4:                                                (ncert solutions for class 9 maths chapter 2)

Expand each of the following, using suitable identities:

(i)   (ii)

(iii)   (iv)

(v)   (vi)

It is known that,

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 5:

Factorise:

(i)

(ii)

It is known that,

(i)

(ii)

Question 6:                                                        (ncert solutions for class 9 maths chapter 2)

Write the following cubes in expanded form:

(i)   (ii)

(iii)   (iv)

It is known that,

(i)

(ii)

(iii)

(vi)

Question 7:

Evaluate the following using suitable identities:

(i) (99)3 (ii) (102)3 (iii) (998)3

It is known that,

(i) (99)= (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

Question 8:

Factorise each of the following:

(i)   (ii)

(iii)   (iv)

(v)

It is known that,

(i)

(ii)

(iii)

(iv)

(v)

Question 9:                                                (ncert solutions for class 9 maths chapter 2)

Verify:

(i)

(ii)

(i) It is known that,

(ii) It is known that,

Question 10:

Factorise each of the following:

(i)

(ii)

[Hint: See question 9.]

(i)

(ii)

Question 11:                                              (ncert solutions for class 9 maths chapter 2)

Factorise:

It is known that,

Question 12:

Verify that

It is known that,

Question 13:

If x + y + z = 0, show that x3 + y3 + z3 = 3xyzx3 + y3 + z3 = 3xyz.

It is known that,

Put x + y + z = 0,

Question 14:

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

(i)

(ii)

(i)

Let x = −12, y = 7, and z = 5

It can be observed that,

x + y + z = − 12 + 7 + 5 = 0

It is known that if x + y + z = 0, then

∴

= −1260

(ii)

Let x = 28, y = −15, and z = −13

It can be observed that,

x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0

It is known that if x + y + z = 0, then

Question 15:                                  (ncert solutions for class 9 maths chapter 2)

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

Area = Length × Breadth

The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.

(i)

Therefore, possible length = 5a − 3

And, possible breadth = 5a − 4

(ii)

Therefore, possible length = 5y + 4

And, possible breadth = 7y − 3

Question 16:

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

Volume of cuboid = Length × Breadth × Height

The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.

(i)

One of the possible solutions is as follows.

Length = 3, Breadth = x, Height = x − 4

(ii)

One of the possible solutions is as follows.

Length = 4k, Breadth = 3y + 5, Height = y − 1

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