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

NCERT Solutions: 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)]
= x2 + [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) (y2 + 3/2) (y2 - 3/2)

Ans: Using the identity (x + y)(x - y) = x2 - y2,
[Here, x = y2 and 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)2 - (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) 9x2 + 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: 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)3 

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: Polynomials (Exercise 2.4)

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

Ans: Using identity, (x -y)3 = x3-y3-3xy(x-y)
NCERT Solutions: 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)3 = (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)3 =(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) 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.

(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 - 3(4a)2(3b)+3(4a)(3b)2
64a3-27b3 - 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+y3 = (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-y3 = (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 27x3 + y3 + 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)3 + (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)3 

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: Polynomials (Exercise 2.4) is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on NCERT Solutions: Polynomials (Exercise 2.4)

1. How do I find the remainder when a polynomial is divided by a linear expression without doing the full division?
Ans. The Remainder Theorem states that when a polynomial p(x) is divided by (x - a), the remainder equals p(a). Instead of performing long division, substitute the value directly into the polynomial to get your answer instantly. This method saves time and reduces calculation errors during exams.
2. What's the difference between a factor and a root of a polynomial, and why does it matter?
Ans. A root is a value that makes the polynomial equal to zero; a factor is the expression (x - root) that divides the polynomial evenly. If r is a root, then (x - r) is a factor. Understanding this relationship is crucial for factorisation problems and solving polynomial equations in CBSE Class 9 Mathematics.
3. Can I use the Factor Theorem to check if an expression divides a polynomial without actually dividing?
Ans. Yes. The Factor Theorem states that (x - a) is a factor of p(x) if and only if p(a) = 0. Simply substitute the value into the polynomial; if the result is zero, it's a factor. This verification method is faster than synthetic or long division and essential for Exercise 2.4 problems.
4. Why do some polynomials have no real roots even though they look factorisable?
Ans. Not all polynomials with real coefficients factor into real linear expressions. Quadratic and higher-degree polynomials may have complex or irrational roots that cannot be expressed as simple factors. Understanding this distinction prevents students from assuming every polynomial can be factorised over real numbers.
5. How should I approach factorisation of cubic polynomials when I don't immediately see a pattern?
Ans. Use the Rational Root Theorem to identify potential factors, then apply the Remainder Theorem to test values. Once you find one root using p(a) = 0, use polynomial division to extract that factor, reducing the cubic to a quadratic. This systematic approach makes Exercise 2.4 cubic polynomial questions manageable and logical.
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NCERT Solutions: Polynomials (Exercise 2.4) Notes

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NCERT Solutions: Polynomials (Exercise 2.4) Class 9 Questions

The "NCERT Solutions: Polynomials (Exercise 2.4) Class 9 Questions" guide is a valuable resource for all aspiring students preparing for the Class 9 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

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