NCERT Solutions: Polynomials (Exercise 2.4)

Polynomials (Exercise 2.4) NCERT Solutions - Mathematics (Maths) Class 9

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

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)

Q.6. 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)

(iv) (x−(2/3)y)3
Ans: Using identity, (x –y)3 = x3–y3–3xy(x–y)

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)

Q.9. 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 Polynomials (Exercise 2.4) NCERT Solutions | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

FAQs on Polynomials (Exercise 2.4) NCERT Solutions - Mathematics (Maths) Class 9

 1. What are polynomials?
Ans. Polynomials are algebraic expressions consisting of variables and coefficients, combined using addition, subtraction, multiplication, and exponentiation operations. They are commonly represented in the form of terms, where each term has a variable raised to a non-negative integer power, multiplied by a coefficient.
 2. How do you classify polynomials based on the number of terms?
Ans. Polynomials can be classified based on the number of terms they have. A polynomial with one term is called a monomial, with two terms is called a binomial, with three terms is called a trinomial, and a polynomial with more than three terms is called a polynomial with more than three terms.
 3. What is the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable in the polynomial. It helps in determining the behavior and characteristics of the polynomial. For example, a polynomial with a degree of 2 is called a quadratic polynomial, and a polynomial with a degree of 3 is called a cubic polynomial.
 4. How do you add or subtract polynomials?
Ans. To add or subtract polynomials, we combine like terms, which have the same variable raised to the same power. We add or subtract the coefficients of these like terms while keeping the variable and its power unchanged. It is important to align the terms properly before performing the addition or subtraction operation.
 5. How do you multiply polynomials?
Ans. To multiply polynomials, we use the distributive property and multiply each term of one polynomial with each term of the other polynomial. We then collect like terms, if any, and simplify the resulting expression by combining the like terms. The resulting expression is the product of the two polynomials.

Mathematics (Maths) Class 9

62 videos|426 docs|102 tests

Mathematics (Maths) Class 9

62 videos|426 docs|102 tests

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