Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  NCERT Solutions: Polynomials (Exercise 2.4)

NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

Q1: Use suitable identities to find the following products: 
(i) (x + 4)(x + 10)

Ans: Using the identity (x + a)(x + b) = x2 + (a + b)x + ab,
[Here, a = 4 and b = 10]
We get,
(x + 4)(x + 10) = x2 + (4 + 10)x + (4 x 10)
= x2 + 14x + 40

(ii) (x + 8)(x – 10)

Ans: Using the identity, (x+a)(x+b) = x2+(a+b)x+ab
[Here, a = 8 and b = (–10)]
We get: (x + 8)(x – 10) = x2 + [8 + (–10)]x + [8 x (–10)]
= x+ [8-10]x + [–80]
= x2 – 2x – 80

(iii) (3x + 4)(3x – 5)

Ans: Using the identity (x + a)(x + b) = x2 + (a + b)x + ab,
[Here, x = 3x, a = 4 and b = −5]
we get
(3x + 4)(3x – 5) = (3x)2 + [4 + (–5)]3x + [4 x (–5)]
= 9x2 + 3x(4–5)–20
= 9x2 – 3x – 20

(iv) (y+ 3/2) (y- 3/2)

Ans: Using the identity (x + y)(x – y) = x2 – y2,
[Here, x = yand y = 3/2]
we get:
(y2+3/2)(y2–3/2) = (y2)2–(3/2)2
= y4–9/4

(v) (3 – 2x) (3 + 2x)

Ans: Using (a + b) (a - b) = a2 - b2,
putting a = 3 , b = 2x
= (3)- (2x)2
= 9 - 4x2

Q2: Evaluate the following products without multiplying directly: 
(i) 103 × 107

Ans: (100+3) × (100+7)
Using identity, [(x+a)(x+b) = x2+(a+b)x+ab
Here, x = 100
a = 3
b = 7
We get, 103×107 = (100+3)×(100+7)
= (100)2+(3+7)100+(3×7))
= 10000+1000+21
= 11021

(ii) 95×96

Ans: (100-5)×(100-4)
Using identity, [(x-a)(x-b) = x2-(a+b)x+ab
Here, x = 100
a = -5
b = -4
We get, 95×96 = (100-5)×(100-4)
= (100)2+100(-5+(-4))+(-5×-4)
= 10000-900+20
= 9120

(iii) 104 × 96

Ans: (100+4)×(100–4)
Using identity, [(a+b)(a-b)= a2-b2]
Here, a = 100
b = 4
We get, 104×96 = (100+4)×(100–4)
= (100)2–(4)2
= 10000–16
= 9984

Q3: Factorise the following using appropriate identities: 
(i) 9x+ 6xy + y2 

Ans: (3x)2+(2×3x×y)+y2
Using identity, x2+2xy+y2 = (x+y)2
Here, x = 3x
y = y
9x2+6xy+y2 = (3x)2+(2×3x×y)+y2
= (3x+y)2
= (3x+y)(3x+y)

(ii) 4y2 – 4y + 1

Ans: 4y2−4y+1 = (2y)2–(2×2y×1)+1
Using identity, x2 – 2xy + y2 = (x – y)2
Here, x = 2y
y = 1
4y2−4y+1 = (2y)2–(2×2y×1)+12
= (2y–1)2
= (2y–1)(2y–1)

(iii) x2– y2/100

Ans: x2–y2/100 = x2–(y/10)2
Using identity, x2-y2 = (x-y)(x+y)
Here, x = x
y = y/10
x2–y2/100 = x2–(y/10)2
= (x–y/10)(x+y/10)

 Q4: Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)2

Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = x
y = 2y
z = 4z
(x+2y+4z)2 = x22+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x2+4y2+16z2+4xy+16yz+8xz

(ii) (2x – y + z)2

Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = 2x
y = −y
z = z
(2x−y+z)2 = (2x)2+(−y)2+z2+(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x2+y2+z2–4xy–2yz+4xz

(iii) (–2x + 3y + 2z)2

Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = −2x
y = 3y
z = 2z
(−2x+3y+2z)2 = (−2x)2+(3y)2+(2z)2+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)
= 4x2+9y2+4z2–12xy+12yz–8xz

(iv) (3a – 7b – c)2

Ans: Using identity (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = 3a
y = – 7b
z = – c
(3a –7b– c)2 = (3a)2+(– 7b)2+(– c)2+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
= 9a2 + 49b2 + c2– 42ab+14bc–6ca

(v) (–2x + 5y – 3z)2 

Ans: Using identity, (x+y+z)2= x2+y2+z2+2xy+2yz+2zx
Here, x = –2x
y = 5y
z = – 3z
(–2x+5y–3z)2 = (–2x)2+(5y)2+(–3z)2+(2×–2x × 5y)+(2× 5y×– 3z)+(2×–3z ×–2x)
= 4x2+25y2 +9z2– 20xy–30yz+12zx

(vi)  ((1/4)a-(1/2)b+1)2

Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = (1/4)a
y = (-1/2)b
z = 1
NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

Q5: Factorise: 

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2
4x2+9y2+16z2+12xy–24yz–16xz = (2x)2+(3y)2+(−4z)2+(2×2x×3y)+(2×3y×−4z) +(2×−4z×2x)
= (2x+3y–4z)2
= (2x+3y–4z)(2x+3y–4z)

(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2 yz – 8xz

Ans: Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2
2x2+y2+8z2–2√2xy+4√2yz–8xz
= (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z)2
= (−√2x+y+2√2z)(−√2x+y+2√2z)

Q6: Write the following cubes in expanded form: 
(i) (2x + 1)

Ans: Using identity,(x+y)3 = x3+y3+3xy(x+y)
(2x+1)3= (2x)3+13+(3×2x×1)(2x+1)
= 8x3+1+6x(2x+1)
= 8x3+12x2+6x+1

(ii) (2a – 3b)3 

Ans: Using identity,(x–y)3 = x3–y3–3xy(x–y)
(2a−3b)3 = (2a)3−(3b)3–(3×2a×3b)(2a–3b)
= 8a3–27b3–18ab(2a–3b)
= 8a3–27b3–36a2b+54ab2

(iii)  ((3/2)x+1)3

Ans: Using identity,(x+y)3 = x3+y3+3xy(x+y)
((3/2)x+1)3=((3/2)x)3+13+(3×(3/2)x×1)((3/2)x +1)
NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

(iv) (x−(2/3)y)3

Ans: Using identity, (x –y)3 = x3–y3–3xy(x–y)
NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

Q7: Evaluate the following using suitable identities: 
(i) (99)3

Ans: We can write 99 as 100–1 
Using identity, (x –y)3 = x3–y3–3xy(x–y) 
(99)= (100–1)3 
= (100)3–13–(3×100×1)(100–1) 
= 1000000 –1–300(100 – 1) 
= 1000000–1–30000+300 
= 970299 
= 970299

(ii) (102)3

Ans: We can write 102 as 100+2 
Using identity,(x+y)3 = x3+y3+3xy(x+y) 
(100+2)3 =(100)3+23+(3×100×2)(100+2)
= 1000000 + 8 + 600[100 + 2]
= 1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)3

Ans: We can write 99 as 1000–2 
Using identity,(x–y)3 = x3–y3–3xy(x–y) 
(998)=(1000–2)3 
=(1000)3–23–(3×1000×2)(1000–2) 
= 1000000000–8–6000(1000– 2) 
= 1000000000–8- 6000000+12000 
= 994011992 

Q8: Factorise each of the following:  
(i) 8a3 + b3 + 12a2b + 6ab2 

Ans: The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2
8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2
= (2a+b)3
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used.

(ii) 8a– b3 – 12a2b + 6ab2

Ans: The expression, 8a3–b3−12a2b+6ab2 can be written as (2a)3–b3–3(2a)2b+3(2a)(b)2
8a3–b3−12a2b+6ab2 = (2a)3–b3–3(2a)2b+3(2a)(b)2
= (2a–b)3
= (2a–b)(2a–b)(2a–b)
Here, the identity,(x–y)3 = x3–y3–3xy(x–y) is used.

(iii) 27 – 125a3 – 135a + 225a2 

Ans: The expression, 27–125a3–135a +225a2 can be written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2
27–125a3–135a+225a2 =
33–(5a)3–3(3)2(5a)+3(3)(5a)2
= (3–5a)3
= (3–5a)(3–5a)(3–5a)
Here, the identity, (x–y)3 = x3–y3-3xy(x–y) is used.

(iv) 64a3 – 27b3 – 144 a2b + 108 ab2 

Ans: The expression, 64a3–27b3–144a2b+108ab2 can be written as
(4a)3–(3b)- 3(4a)2(3b)+3(4a)(3b)2
64a3–27b– 144a2b+108ab2
= (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2
=(4a–3b)3
=(4a–3b)(4a–3b)(4a–3b)
Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.

(v) 27p3– (1/216)−(9/2) p2+(1/4)p

Ans: The expression, 27p3–(1/216)−(9/2) p2+(1/4)p can be written as 
(3p)3–(1/6)3−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6) 
Using (x – y)3 = x3 – y3 – 3xy (x – y) 
27p3–(1/216)−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6) 
Taking x = 3p and y = 1/6 
= (3p–1/6)3 
= (3p–1/6)(3p–1/6)(3p–1/6) 

Q9: Verify: 
(i) x3+y= (x+y)(x2–xy+y2) 

Ans: We know that, (x+y)3 = x3+y3+3xy(x+y)
⇒ x3+y3 = (x+y)3–3xy(x+y)
⇒ x3+y3 = (x+y)[(x+y)2–3xy]
Taking (x+y) common ⇒ x3+y3 = (x+y)[(x2+y2+2xy)–3xy]
x3+y3 = (x+y)(x2+y2–xy)


(ii) x3–y= (x–y)(x2+xy+y2)  

Ans: We know that,(x–y)3 = x3–y3–3xy(x–y)
⇒ x3−y3 = (x–y)3+3xy(x–y)
⇒ x3−y3 = (x–y)[(x–y)2+3xy]
Taking (x+y) common ⇒ x3−y3 = (x–y)[(x2+y2–2xy)+3xy]
x3+y3 = (x–y)(x2+y2+xy)

 Q10: Factorise each of the following: 
(i) 27y3+125z3

Ans: The expression, 27y3+125z3 can be written as (3y)3+(5z)3
27y3+125z3 = (3y)3+(5z)3
We know that, x3+y3 = (x+y)(x2–xy+y2)
27y3+125z3 = (3y)3+(5z)3
= (3y+5z)[(3y)2–(3y)(5z)+(5z)2]
= (3y+5z)(9y2–15yz+25z2)

(ii) 64m3–343n3

Ans: The expression, 64m3–343n3can be written as (4m)3–(7n)3
64m3–343n3 = (4m)3–(7n)3
We know that, x3–y3 = (x–y)(x2+xy+y2)
64m3–343n3 = (4m)3–(7n)3
= (4m-7n)[(4m)2+(4m)(7n)+(7n)2]
= (4m-7n)(16m2+28mn+49n2)

Q11: Factorise 27x+ y+ z3 – 9xyz.

Solution: The expression27x3+y3+z3–9xyz can be written as (3x)3+y3+z3–3(3x)(y)(z)
27x3+y3+z3–9xyz  = (3x)3+y3+z3–3(3x)(y)(z)
We know that, x3+y3+z3–3xyz = (x+y+z)(x2+y2+z2–xy –yz–zx)
27x3+y3+z3–9xyz  = (3x)3+y3+z3–3(3x)(y)(z)
= (3x+y+z)[(3x)2+y2+z2–3xy–yz–3xz]
= (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)

Q12: Verify that x3 + y3 + z3 – 3xyz = (1/2) (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]

Ans: We know that,
x3+y3+z3−3xyz = (x+y+z)(x2+y2+z2–xy–yz–xz)
⇒ x3+y3+z3–3xyz = (1/2)(x+y+z)[2(x2+y2+z2–xy–yz–xz)]
= (1/2)(x+y+z)(2x2+2y2+2z2–2xy–2yz–2xz)
= (1/2)(x+y+z)[(x2+y2−2xy)+(y2+z2–2yz)+(x2+z2–2xz)]
= (1/2)(x+y+z)[(x–y)2+(y–z)2+(z–x)2]

Q13: If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Ans: We know that,
x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2–xy–yz–xz)
Now, according to the question, let (x+y+z) = 0,
then, x3+y3+z3 -3xyz = (0)(x2+y2+z2–xy–yz–xz)
⇒ x3+y3+z3–3xyz = 0
⇒x3+y3+z3 = 3xyz
Hence Proved

Q14: Without actually calculating the cubes, find the value of each of the following: 
(i) (–12)3 + (7)+ (5)3

Ans: Let a = −12
b = 7
c = 5
We know that if x+y+z = 0, then x3+y3+z3=3xyz.
Here, −12+7+5=0
(−12)3+(7)3+(5)3 = 3xyz
= 3×-12×7×5
= -1260

(ii) (28)3 + (–15)3 + (–13)

Ans: Let a = 28
b = −15
c = −13
We know that if x+y+z = 0, then x3+y3+z3 = 3xyz.
Here, x+y+z = 28–15–13 = 0
(28)3+(−15)3+(−13)3 = 3xyz
= 0+3(28)(−15)(−13)
= 16380

Q15: 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

Ans: Using the splitting the middle term method,
We have to find a number whose sum = -35 and product =25×12=300
We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20=300]
25a2–35a+12 = 25a2–15a−20a+12
= 5a(5a–3)–4(5a–3)
= (5a–4)(5a–3)
Possible expression for length  = 5a–4
Possible expression for breadth  = 5a –3


(ii) Area : 35y2+13y–12

Ans: Using the splitting the middle term method,
We have to find a number whose sum = 13 and product = 35×-12 = 420
We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420]
35y2+13y–12 = 35y2–15y+28y–12
= 5y(7y–3)+4(7y–3)
= (5y+4)(7y–3)
Possible expression for length  = (5y+4)
Possible expression for breadth  = (7y–3)


Q16: What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: 3x2–12x

Ans: 3x2–12x can be written as 3x(x–4) by taking 3x out of both the terms.
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x–4)

(ii) Volume: 12ky2+8ky–20k

Ans: 12ky2+8ky–20k can be written as 4k(3y2+2y–5) by taking 4k out of both the terms.
12ky2+8ky–20k = 4k(3y2+2y–5)
[Here, 3y2+2y–5 can be written as 3y2+5y–3y–5 using splitting the middle term method.]
= 4k(3y2+5y–3y–5)
= 4k[y(3y+5)–1(3y+5)]
= 4k(3y+5)(y–1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1)

The document NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4) is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
44 videos|412 docs|55 tests

Top Courses for Class 9

FAQs on NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

1. What are polynomials?
Ans. Polynomials are algebraic expressions consisting of variables, coefficients, and exponents. They are formed by adding or subtracting terms, where each term is a variable raised to a non-negative integer power multiplied by a coefficient.
2. How do you classify polynomials?
Ans. Polynomials can be classified based on the number of terms they have. If a polynomial has only one term, it is called a monomial. A polynomial with two terms is a binomial, and a polynomial with three terms is called a trinomial. Polynomials with more than three terms are generally referred to as polynomials.
3. What is the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable present in the polynomial. For example, if a polynomial has the highest power as 3, it is said to be a polynomial of degree 3. The degree helps in determining the behavior and properties of the polynomial.
4. How can we add or subtract polynomials?
Ans. To add or subtract polynomials, we combine like terms. Like terms have the same variables raised to the same powers. We simply add or subtract the coefficients of these like terms while keeping the variables and their exponents unchanged.
5. How can we multiply polynomials?
Ans. To multiply polynomials, we use the distributive property of multiplication over addition. We multiply each term of one polynomial with each term of the other polynomial and then combine like terms if any. The resulting expression is the product of the two polynomials.
44 videos|412 docs|55 tests
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

video lectures

,

Important questions

,

practice quizzes

,

Exam

,

Sample Paper

,

Extra Questions

,

Viva Questions

,

past year papers

,

Semester Notes

,

study material

,

ppt

,

NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

,

Previous Year Questions with Solutions

,

shortcuts and tricks

,

pdf

,

NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

,

mock tests for examination

,

Summary

,

NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials (Exercise 2.4)

,

Objective type Questions

,

Free

,

MCQs

;