In Class 9, Mathematics students are introduced to the fundamental concept of the polynomials in Chapter 2 of the NCERT textbook. This chapter explores the nature, properties, and applications of polynomials. Exercise 2.5 in Set 1 focuses on solving the problems related to polynomial expressions which helps in understanding their behavior and simplification techniques. Mastering this exercise is crucial as it builds the foundation for the more advanced algebraic concepts and problem-solving skills.
Polynomials
The Polynomials are algebraic expressions consisting of variables and coefficients connected by addition, subtraction, and multiplication operations but not division by a variable. In Class 9, polynomials are studied to understand their degree, roots, and various types. This foundational knowledge is essential as polynomials are widely used in algebra, calculus, and various real-life applications. Exercise 2.5 helps reinforce these concepts through practice problems enhancing students' problem-solving abilities and their grasp of the polynomial operations.
Question 1. Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
Solution:
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
[So, a = 4 and b = 10]
(x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)
= x2 + 14x + 40
(ii) (x + 8) (x - 10)
Solution:
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
[So, a = 8 and b = −10]
(x + 8) (x - 10) = x2 + (8 + (-10) )x + (8 × (-10))
= x2 + (8 - 10) x - 80
= x2 − 2x - 80
(iii) (3x + 4) (3x - 5)
Solution:
Using formula, (y + a) (y + b) = y2 + (a + b)y + ab
[So, y = 3x, a = 4 and b = −5]
(3x + 4) (3x − 5) = (3x)2 + [4 + (-5)]3x + 4 × (-5)
= 9x2 + 3x (4 - 5) - 20
= 9x2 – 3x – 20
(iv) (y2 +
Solution:
Using formula, (a + b) (a – b) = a2 – b2
[So, a = y2 and b =
\frac{3}{2} ](y 2 +
\frac{3}{2} ) (y2 –\frac{3}{2} ) = (y2)2 – (\frac{3}{2} )^2= y 4 –
\frac{9}{4}
Question 2. Evaluate the following products without multiplying directly:
(i) 103 × 107
Solution:
103 × 107 = (100 + 3) × (100 + 7)
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
Then,
x = 100
a = 3
b = 7
So, 103 × 107 = (100 + 3) × (100 + 7)
= (100)2 + (3 + 7)100 + (3 × 7)
= 10000 + 1000 + 21
= 11021
(ii) 95 × 96
Solution:
95 × 96 = (100 - 5) × (100 - 4)
Using formula, (x - a) (x - b) = x2 - (a + b)x + ab
Then, According to the identity
x = 100
a = 5
b = 4
So, 95 × 96 = (100 - 5) × (100 - 4)
= (100)2 - 100 (5+4) + (5 × 4)
= 10000 - 900 + 20
= 9120
(iii) 104 × 96
Solution:
104 × 96 = (100 + 4) × (100 – 4)
Using formula, (a + b) (a - b) = a2 - b2
Then,
a = 100
b = 4
So, 104 × 96 = (100 + 4) × (100 – 4)
= (100)2 – (4)2
= 10000 – 16
= 9984
Question 3. Factorize the following using appropriate identities:
(i) 9x2 + 6xy + y2
Solution:
9x2 + 6xy + y2 = (3x)2 + (2 × 3x × y) + y2
Using formula, a2 + 2ab + b2 = (a + b)2
Then,
a = 3x
b = y
9x2 + 6xy + y2 = (3x)2 + (2 × 3x × y) + y2
= (3x + y)2
= (3x + y) (3x + y)
(ii) 4y2 − 4y + 1
Solution:
4y2 − 4y + 1 = (2y)2 – (2 × 2y × 1) + 1
Using formula, a2 - 2ab + b2 = (a - b)2
Then,
a = 2y
b = 1
= (2y – 1)2
= (2y – 1) (2y – 1)
(iii) x2 -
Solution:
x2 –
\frac{y^2}{100} = x2 –(\frac{y}{10})^2 Using formula, a2 - b2 = (a - b) (a + b)
Then,
a = x
b =
\frac{y}{10} = (x –
\frac{y}{10} ) (x +\frac{y}{10} )
Question 4. Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = x
y = 2y
z = 4z
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + (2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
(ii) (2x − y + z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
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
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
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
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
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
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
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) (
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x =
\frac{1}{4} ay =
\frac{-1}{2} bz = 1
(
\frac{1}{4} a - (\frac{-1}{2} )b + 1)2 = [\frac{1}{4} a]2 + [\frac{-1}{2} b]2 + 12 + [2 x\frac{1}{4} }a x\frac{-1}{2} b] + [2 x\frac{-1}{2} b x 1] + [2 x 1 x\frac{1}{4} a]=
\frac{1}{16} a2 +\frac{1}{4} b2 + 1 -\frac{1}{4} ab - b +\frac{1}{2} a
Question 5. Factorize:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then, 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√2yz – 8xz
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then, 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)
Question 6. Write the following cubes in expanded form:
(i) (2x + 1)3
Solution:
Using formula,(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
Solution:
Using formula, (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) (
Solution:
Using formula, (x + y)3 = x3 + y3 + 3xy(x + y)
(
\frac{3}{2} x+ 1)3 = (\frac{3}{2} x)3 + 13 + (3 ×\frac{3}{2} x × 1) (\frac{3}{2} x + 1)=
\frac{27}{8} x3 + 1 +\frac{27}{4} x2 +\frac{9}{2} x=
\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x +1
(iv) (x −
Solution:
Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)
(x −
\frac{2}{3} y)3 = x3 − [\frac{2}{3} y]3 - 3(x)\frac{2}{3} y[x −\frac{2}{3} y]= x3 -
\frac{8}{27} y3 - 2x2y +\frac{4}{3} xy2
Question 7. Evaluate the following using suitable identities:
(i) (99)3
Solution:
99 = 100 – 1
Using formula, (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
(ii) (102)3
Solution:
102 = 100 + 2
Using formula, (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
Solution:
998 = 1000 – 2
Using formula, (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
Question 8. Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
Solution:
8a3 + b3 +12a2b + 6ab2 can also 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
Formula used, (x + y)3 = x3 + y3 + 3xy(x + y)
= (2a + b)3
= (2a + b) (2a + b) (2a + b)
(ii) 8a3 – b3 – 12a2b + 6ab2
Solution:
8a3 – b3 − 12a2b + 6ab2 can also 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
formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (2a – b)3
= (2a – b) (2a – b) (2a – b)
(iii) 27 – 125a3 – 135a + 225a2
Solution:
27 – 125a3 – 135a +225a2 can be also 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
Formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (3 – 5a)3
= (3 – 5a) (3 – 5a) (3 – 5a)
(iv) 64a3 – 27b3 – 144a2b + 108ab2
Solution:
64a3 – 27b3 – 144a2b + 108ab2 can also 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
Formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (4a – 3b)3
= (4a – 3b) (4a – 3b) (4a – 3b)
(v) 7p3 –
Solution:
27p3 –
\frac{1}{216} − (\frac{9}{2} ) p2 + (\frac{1}{4} )p can also be written as (3p)3 –(\frac{1}{6})^3 – 3(3p)2(\frac{1}{6} ) + 3(3p)(\frac{1}{6} )227p3 – (
\frac{1}{216} ) − (\frac{9}{2} ) p2 + (\frac{1}{4} )p = (3p)3 – (\frac{1}{6} )3 – 3(3p)2(\frac{1}{6} ) + 3(3p)(\frac{1}{6} )2Formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (3p –
\frac{1}{6} )3= (3p –
\frac{1}{6} ) (3p –\frac{1}{6} ) (3p –\frac{1}{6} )
Chapter 2 Polynomials - Exercise 2.5 | Set 2
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