- Expressions are formed from variables and constants.
Terms like monomials, binomials, trinomials and polynomials are discussed in detail with examples.
- Expression that contains only one term is called a monomial.
- Expression that contains two terms is called a binomial.
- An expression containing three terms is a trinomial.
- An expression containing, one or more terms with non-zero coefficient (with variables having non-negative exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one term.
- Like terms are formed from the same variables and the powers of these variables are the same, too. Coefficients of like terms need not be the same.
- There are a number of situations in which we need to multiply algebraic expressions: for example, in finding the area of a rectangle, the sides of which are given as expressions.
- Multiplying a Monomial by a Monomial
- Multiplying two monomials
- Multiplying three or more monomials
- Multiplying a Monomial by a Polynomial
- Multiplying a monomial by a binomial
- Multiplying a monomial by a trinomial
- Multiplying a Polynomial by a Polynomial
- Multiplying a binomial by a binomial
- Multiplying a binomial by a trinomial
- An identity is an equality, which is true for all values of the variables in equality. On the other hand, an equation is true only for certain values of its variables.
Unsolved exercises, questions in different patterns will make the chapter more comprehensible to students.
All important points of the chapter are cited in the end under the title- What Have We Discussed?
Page No 140:
Question 1:
Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 − 3zy
(ii) 1 + x + x2
(iii) 4x2y2 − 4x2y2z2 + z2
(iv) 3 − pq + qr − rp
(v)
(vi) 0.3a − 0.6ab + 0.5b
Answer:
The terms and the respective coefficients of the given expressions are as follows.
|
- |
Terms |
Coefficients |
|
(i) |
5xyz2 − 3zy |
5 − 3 |
|
(ii) |
1 x x2 |
1 1 1 |
|
(iii) |
4x2y2 − 4x2y2z2 z2 |
4 − 4 1 |
|
(iv) |
3 − pq qr − rp |
3 −1 1 −1 |
|
(v) |
− xy |
− 1 |
|
(vi) |
0.3a − 0.6ab 0.5b |
0.3 − 0.6 0.5 |
Page No 140:
Question 2:
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3 + x4, 7 + y + 5x, 2y − 3y2, 2y − 3y2 + 4y3, 5x − 4y + 3xy, 4z − 15z2, ab + bc + cd + da, pqr, p2q + pq2, 2p + 2q
Answer:
The given expressions are classified as
Monomials: 1000, pqr
Binomials: x + y, 2y − 3y2, 4z − 15z2, p2q + pq2, 2p + 2q
Trinomials: 7 + y + 5x, 2y − 3y2 + 4y3, 5x − 4y + 3xy
Polynomials that do not fit in any of these categories are
x + x2 + x3 + x4, ab + bc + cd + da
Page No 140:
Question 3:
Add the following.
(i) ab − bc, bc − ca, ca − ab
(ii) a − b + ab, b − c + bc, c − a + ac
(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Answer:
The given expressions written in separate rows, with like terms one below the other and then the addition of these expressions are as follows.
(i)
Thus, the sum of the given expressions is 0.
(ii)
Thus, the sum of the given expressions is ab + bc + ac.
(iii)
Thus, the sum of the given expressions is −p2q2 + 4pq + 9.
(iv)
Thus, the sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl).
Page No 140:
Question 4:
(a) Subtract 4a − 7ab + 3b + 12 from 12a − 9ab + 5b − 3
(b) Subtract 3xy + 5yz − 7zx from 5xy − 2yz − 2zx + 10xyz
(c) Subtract 4p2q − 3pq + 5pq2 − 8p + 7q − 10 from 18 − 3p − 11q + 5pq − 2pq2 + 5p2q
Answer:
The given expressions in separate rows, with like terms one below the other and then the subtraction of these expressions is as follows.
(a)
(b)
(c)
Page No 143:
Question 1:
Find the product of the following pairs of monomials.
(i) 4, 7p (ii) − 4p, 7p (iii) − 4p, 7pq
(iv) 4p3, − 3p (v) 4p, 0
Answer:
The product will be as follows.
(i) 4 × 7p = 4 × 7 × p = 28p
(ii) − 4p × 7p = − 4 × p × 7 × p = (− 4 × 7) × (p × p) = − 28 p2
(iii) − 4p × 7pq = − 4 × p × 7 × p × q = (− 4 × 7) × (p × p × q) = − 28p2q
(iv) 4p3 × − 3p = 4 × (− 3) × p × p × p × p = − 12 p4
(v) 4p × 0 = 4 × p × 0 = 0
Page No 143:
Question 2:
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)
Answer:
We know that,
Area of rectangle = Length × Breadth
Area of 1st rectangle = p × q = pq
Area of 2nd rectangle = 10m × 5n = 10 × 5 × m × n = 50 mn
Area of 3rd rectangle = 20x2 × 5y2 = 20 × 5 × x2 × y2 = 100 x2y2
Area of 4th rectangle = 4x × 3x2 = 4 × 3 × x × x2 = 12x3
Area of 5th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn2p
Page No 144:
Question 3:
Complete the table of products.
|
|
2x |
− 5y |
3x2 |
− 4xy |
7x2y |
− 9x2y2 |
|
2x |
4x2 |
… |
… |
… |
… |
… |
|
− 5y |
… |
… |
− 15x2y |
… |
… |
… |
|
3x2 |
… |
… |
… |
… |
… |
… |
|
− 4xy |
… |
… |
… |
… |
… |
… |
|
7x2y |
… |
… |
… |
… |
… |
… |
|
− 9x2y2 |
… |
… |
… |
… |
… |
… |
Answer:
The table can be completed as follows.
|
|
2x |
− 5y |
3x2 |
− 4xy |
7x2y |
− 9x2y2 |
|
2x |
4x2 |
− 10xy |
6x3 |
− 8x2y |
14x3y |
− 18x3y2 |
|
− 5y |
− 10xy |
25 y2 |
− 15x2y |
20xy2 |
− 35x2y2 |
45x2y3 |
|
3x2 |
6x3 |
− 15x2y |
9x4 |
− 12x3y |
21x4y |
− 27x4y2 |
|
− 4xy |
− 8x2y |
20xy2 |
− 12x3y |
16x2y2 |
− 28x3y2 |
36x3y3 |
|
7x2y |
14x3y |
− 35x2y2 |
21x4y |
− 28x3y2 |
49x4y2 |
− 63x4y3 |
|
− 9x2y2 |
− 18x3y2 |
45 x2y3 |
− 27x4y2 |
36x3y3 |
− 63x4y3 |
81x4y4 |
Page No 144:
Question 4:
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a2, 7a4 (ii) 2p, 4q, 8r (iii) xy, 2x2y, 2xy2
(iv) a, 2b, 3c
Answer:
We know that,
Volume = Length × Breadth × Height
(i) Volume = 5a × 3a2 × 7a4 = 5 × 3 × 7 × a × a2 × a4 = 105 a7
(ii) Volume = 2p × 4q × 8r = 2 × 4 × 8 × p × q × r = 64pqr
(iii) Volume = xy × 2x2y × 2xy2 = 2 × 2 × xy ×x2y × xy2 = 4x4y4
(iv) Volume = a × 2b × 3c = 2 × 3 × a × b × c = 6abc
Video Solution for algebraic expressions and identities (Page: 144 , Q.No.: 4)
NCERT Solution for Class 8 math - algebraic expressions and identities 144 , Question 4
Page No 144:
Question 5:
Obtain the product of
(i) xy, yz, zx (ii) a, − a2, a3 (iii) 2, 4y, 8y2, 16y3
(iv) a, 2b, 3c, 6abc (v) m, − mn, mnp
Answer:
(i) xy × yz × zx = x2y2z2
(ii) a × (− a2) × a3 = − a6
(iii) 2 × 4y × 8y2 × 16y3 = 2 × 4 × 8 × 16 × y × y2 × y3 = 1024 y6
(iv) a × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a2b2c2
(v) m × (− mn) × mnp = − m3n2p
Page No 146:
Question 1:
Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r (ii) ab, a − b (iii) a + b, 7a2b2
(iv) a2 − 9, 4a (v) pq + qr + rp, 0
Answer:
(i) (4p) × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr
(ii) (ab) × (a − b) = (ab × a) + [ab × (− b)] = a2b − ab2
(iii) (a + b) × (7a2 b2) = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3
(iv) (a2 − 9) × (4a) = (a2 × 4a) + (− 9) × (4a) = 4a3 − 36a
(v) (pq + qr + rp) × 0 = (pq × 0) + (qr × 0) + (rp × 0) = 0
Page No 146:
Question 2:
Complete the table
|
--- |
First expression |
Second Expression |
Product |
|
(i) |
a |
b + c + d |
- |
|
(ii) |
x + y − 5 |
5 xy |
- |
|
(iii) |
p |
6p2 − 7p + 5 |
- |
|
(iv) |
4p2q2 |
p2 − q2 |
- |
|
(v) |
a + b + c |
abc |
- |
Answer:
The table can be completed as follows.
|
- |
First expression |
Second Expression |
Product |
|
(i) |
a |
b + c + d |
ab + ac + ad |
|
(ii) |
x + y − 5 |
5 xy |
5x2y + 5xy2 − 25xy |
|
(iii) |
p |
6p2 − 7p + 5 |
6p3 − 7p2 + 5p |
|
(iv) |
4p2q2 |
p2 − q2 |
4p4q2 − 4p2q4 |
|
(v) |
a + b + c |
abc |
a2bc + ab2c + abc2 |
Page No 146:
Question 3:
Find the product.
(i) (a2) × (2a22) × (4a26)
(ii)
(iii)
(iv) x × x2 × x3 × x4
Answer:
(i) (a2) × (2a22) × (4a26) = 2 × 4 ×a2 × a22 × a26 = 8a50
(ii)
(iii)
(iv) x × x2 × x3 × x4 = x10
Page No 146:
Question 4:
(a) Simplify 3x
(4x −5) + 3 and find its values for (i) x = 3,
(ii)
.
(b) a (a2 + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 1, (iii) a = − 1.
Answer:
(a) 3x (4x − 5) + 3 = 12x2 − 15x + 3
(i) For x = 3, 12x2 − 15x + 3 = 12 (3)2 − 15(3) + 3
= 108 − 45 + 3
= 66
(ii) For
(b)a (a2 + a + 1) + 5 = a3 + a2 + a + 5
(i) For a = 0, a3 + a2 + a + 5 = 0 + 0 + 0 + 5 = 5
(ii) For a = 1, a3 + a2 + a + 5 = (1)3 + (1)2 + 1 + 5
= 1 + 1 + 1 + 5 = 8
(iii) For a = −1, a3 + a2 + a + 5 = (−1)3 + (−1)2 + (−1) + 5
= − 1 + 1 − 1 + 5 = 4
Page No 148:
Question 1:
Multiply the binomials.
(i) (2x + 5) and (4x − 3) (ii) (y − 8) and (3y − 4)
(iii) (2.5l − 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq − 2q2)
(vi)
Answer:
(i) (2x + 5) × (4x − 3) = 2x × (4x − 3) + 5 × (4x − 3)
= 8x2 − 6x + 20x − 15
= 8x2 + 14x −15 (By adding like terms)
(ii) (y − 8) × (3y − 4) = y × (3y − 4) − 8 × (3y − 4)
= 3y2 − 4y − 24y + 32
= 3y2 − 28y + 32 (By adding like terms)
(iii) (2.5l − 0.5m) × (2.5l + 0.5m) = 2.5l × (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)
= 6.25l2 + 1.25lm − 1.25lm − 0.25m2
= 6.25l2 − 0.25m2
(iv) (a + 3b) × (x + 5) = a × (x + 5) + 3b × (x + 5)
= ax + 5a + 3bx + 15b
(v) (2pq + 3q2) × (3pq − 2q2) = 2pq × (3pq − 2q2) + 3q2 × (3pq − 2q2)
= 6p2q2 − 4pq3 + 9pq3 − 6q4
= 6p2q2 + 5pq3 − 6q4
(vi)
Page No 148:
Question 2:
Find the product.
(i) (5 − 2x) (3 + x) (ii) (x + 7y) (7x − y)
(iii) (a2 + b) (a + b2) (iv) (p2 − q2) (2p + q)
Answer:
(i) (5 − 2x) (3 + x) = 5 (3 + x) − 2x (3 + x)
= 15 + 5x − 6x − 2x2
= 15 − x − 2x2
(ii) (x + 7y) (7x − y) = x (7x − y) + 7y (7x − y)
= 7x2 − xy + 49xy − 7y2
= 7x2 + 48xy − 7y2
(iii) (a2 + b) (a + b2) = a2 (a + b2) + b (a + b2)
= a3 + a2b2 + ab + b3
(iv) (p2 − q2) (2p + q) = p2 (2p + q) − q2 (2p + q)
= 2p3 + p2q − 2pq2 − q3
Page No 148:
Question 3:
Simplify.
(i) (x2 − 5) (x + 5) + 25
(ii) (a2 + 5) (b3 + 3) + 5
(iii) (t + s2) (t2 − s)
(iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x − y)
(vi) (x + y) (x2 − xy + y2)
(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y
(viii) (a + b + c) (a + b − c)
Answer:
(i) (x2 − 5) (x + 5) + 25
= x2 (x + 5) − 5 (x + 5) + 25
= x3 + 5x2 − 5x − 25 + 25
= x3 + 5x2 − 5x
(ii) (a2 + 5) (b3 + 3) + 5
= a2 (b3 + 3) + 5 (b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20
(iii) (t + s2) (t2 − s)
= t (t2 − s) + s2 (t2 − s)
= t3 − st + s2t2 − s3
(iv) (a + b) (c − d) + (a − b) (c + d) + 2 (ac + bd)
= a (c − d) + b (c − d) + a (c + d) − b (c + d) + 2 (ac + bd)
= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd
= (ac + ac + 2ac) + (ad − ad) + (bc − bc) + (2bd − bd − bd)
= 4ac
(v) (x + y) (2x + y) + (x + 2y) (x − y)
= x (2x + y) + y (2x + y) + x (x − y) + 2y (x − y)
= 2x2 + xy + 2xy + y2 + x2 − xy + 2xy − 2y2
= (2x2 + x2) + (y2 − 2y2) + (xy + 2xy − xy + 2xy)
= 3x2 − y2 + 4xy
(vi) (x + y) (x2 − xy + y2)
= x (x2 − xy + y2) + y (x2 − xy + y2)
= x3 − x2y + xy2 + x2y − xy2 + y3
= x3 + y3 + (xy2 − xy2) + (x2y − x2y)
= x3 + y3
(vii) (1.5x − 4y) (1.5x + 4y + 3) − 4.5x + 12y
= 1.5x (1.5x + 4y + 3) − 4y (1.5x + 4y + 3) − 4.5x + 12y
= 2.25 x2 + 6xy + 4.5x − 6xy − 16y2 − 12y − 4.5x + 12y
= 2.25 x2 + (6xy − 6xy) + (4.5x − 4.5x) − 16y2 + (12y − 12y)
= 2.25x2 − 16y2
(viii) (a + b + c) (a + b − c)
= a (a + b − c) + b (a + b − c) + c (a + b − c)
= a2 + ab − ac + ab + b2 − bc + ca + bc − c2
= a2 + b2 − c2 + (ab + ab) + (bc − bc) + (ca − ca)
= a2 + b2 − c2 + 2ab
Page No 151:
Question 1:
Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5)
(iii) (2a −
7) (2a − 7) (iv)
(v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a2 + b2) (− a2 + b2)
(vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c)
(ix) (x) (7a
− 9b) (7a − 9b)
Answer:
The products will be as follows.
(i) (x + 3) (x + 3) = (x + 3)2
= (x)2 + 2(x) (3) + (3)2 [(a + b)2 = a2 + 2ab + b2]
= x2 + 6x + 9
(ii) (2y + 5) (2y + 5) = (2y + 5)2
= (2y)2 + 2(2y) (5) + (5)2 [(a + b)2 = a2 + 2ab + b2]
= 4y2 + 20y + 25
(iii) (2a − 7) (2a − 7) = (2a − 7)2
= (2a)2 − 2(2a) (7) + (7)2 [(a − b)2 = a2 − 2ab + b2]
= 4a2 − 28a + 49
(iv)
[(a
− b)2 = a2 − 2ab
+ b2]
(v) (1.1m − 0.4) (1.1 m + 0.4)
= (1.1m)2 − (0.4)2 [(a + b) (a − b) = a2 − b2]
= 1.21m2 − 0.16
(vi) (a2 + b2) (− a2 + b2) = (b2 + a2) (b2 − a2)
= (b2)2 − (a2)2 [(a + b) (a − b) = a2 − b2]
= b4 − a4
(vii) (6x − 7) (6x + 7) = (6x)2 − (7)2 [(a + b) (a − b) = a2 − b2]
= 36x2 − 49
(viii) (− a + c) (− a + c) = (− a + c)2
= (− a)2 + 2(− a) (c) + (c)2 [(a + b)2 = a2 + 2ab + b2]
= a2 − 2ac + c2
(ix)
[(a
+ b)2 = a2 + 2ab + b2]
(x) (7a − 9b) (7a − 9b) = (7a − 9b)2
= (7a)2 − 2(7a)(9b) + (9b)2 [(a − b)2 = a2 − 2ab + b2]
= 49a2 − 126ab + 81b2
Page No 151:
Question 2:
Use the identity (x + a) (x + b) = x2 + (a + b)x + ab to find the following products.
(i) (x + 3) (x + 7) (ii) (4x +5) (4x + 1)
(iii) (4x − 5) (4x − 1) (iv) (4x + 5) (4x − 1)
(v) (2x +5y) (2x + 3y) (vi) (2a2 +9) (2a2 + 5)
(vii) (xyz − 4) (xyz − 2)
Answer:
The products will be as follows.
(i) (x + 3) (x + 7) = x2 + (3 + 7) x + (3) (7)
= x2 + 10x + 21
(ii) (4x + 5) (4x + 1) = (4x)2 + (5 + 1) (4x) + (5) (1)
= 16x2 + 24x + 5
(iii)
= 16x2 − 24x + 5
(iv)
= 16x2 + 16x − 5
(v) (2x +5y) (2x + 3y) = (2x)2 + (5y + 3y) (2x) + (5y) (3y)
= 4x2 + 16xy + 15y2
(vi) (2a2 +9) (2a2 + 5) = (2a2)2 + (9 + 5) (2a2) + (9) (5)
= 4a4 + 28a2 + 45
(vii) (xyz − 4) (xyz − 2)
=
= x2y2z2 − 6xyz + 8
Page No 151:
Question 3:
Find the following squares by suing the identities.
(i) (b − 7)2 (ii) (xy + 3z)2 (iii) (6x2 − 5y)2
(iv) (v) (0.4p
− 0.5q)2 (vi) (2xy + 5y)2
Answer:
(i) (b − 7)2 = (b)2 − 2(b) (7) + (7)2 [(a − b)2 = a2 − 2ab + b2]
= b2 − 14b + 49
(ii) (xy + 3z)2 = (xy)2 + 2(xy) (3z) + (3z)2 [(a + b)2 = a2 + 2ab + b2]
= x2y2 + 6xyz + 9z2
(iii) (6x2 − 5y)2 = (6x2)2 − 2(6x2) (5y) + (5y)2 [(a − b)2 = a2 − 2ab + b2]
= 36x4 − 60x2y + 25y2
(iv) [(a
+ b)2 = a2 + 2ab + b2]
(v) (0.4p − 0.5q)2 = (0.4p)2 − 2 (0.4p) (0.5q) + (0.5q)2
[(a − b)2 = a2 − 2ab + b2]
= 0.16p2 − 0.4pq + 0.25q2
(vi) (2xy + 5y)2 = (2xy)2 + 2(2xy) (5y) + (5y)2
[(a + b)2 = a2 + 2ab + b2]
= 4x2y2 + 20xy2 + 25y2
Page No 151:
Question 4:
Simplify.
(i) (a2 − b2)2 (ii) (2x +5)2 − (2x − 5)2
(iii) (7m − 8n)2 + (7m + 8n)2 (iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2
(vi) (ab + bc)2 − 2ab2c (vii) (m2 − n2m)2 + 2m3n2
Answer:
(i) (a2 − b2)2 = (a2)2 − 2(a2) (b2) + (b2)2 [(a − b)2 = a2 − 2ab + b2 ]
= a4 − 2a2b2 + b4
(ii) (2x +5)2 − (2x − 5)2 = (2x)2 + 2(2x) (5) + (5)2 − [(2x)2 − 2(2x) (5) + (5)2]
[(a − b)2 = a2 − 2ab + b2]
[(a + b)2 = a2 + 2ab + b2]
= 4x2 + 20x + 25 − [4x2 − 20x + 25]
= 4x2 + 20x + 25 − 4x2 + 20x − 25 = 40x
(iii) (7m − 8n)2 + (7m + 8n)2
= (7m)2 − 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2
[(a − b)2 = a2 − 2ab + b2 and (a + b)2 = a2 + 2ab + b2]
= 49m2 − 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2
(iv) (4m + 5n)2 + (5m + 4n)2
= (4m)2 + 2(4m) (5n) + (5n)2 + (5m)2 + 2(5m) (4n) + (4n)2
[ (a + b)2 = a2 + 2ab + b2]
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2
(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2
= (2.5p)2 − 2(2.5p) (1.5q) + (1.5q)2 − [(1.5p)2 − 2(1.5p)(2.5q) + (2.5q)2]
[(a − b)2 = a2 − 2ab + b2 ]
= 6.25p2 − 7.5pq + 2.25q2 − [2.25p2 − 7.5pq + 6.25q2]
= 6.25p2 − 7.5pq + 2.25q2 − 2.25p2 + 7.5pq − 6.25q2]
= 4p2 − 4q2
(vi) (ab + bc)2 − 2ab2c
= (ab)2 + 2(ab)(bc) + (bc)2 − 2ab2c [(a + b)2 = a2 + 2ab + b2 ]
= a2b2 + 2ab2c + b2c2 − 2ab2c
= a2b2 + b2c2
(vii) (m2 − n2m)2 + 2m3n2
= (m2)2 − 2(m2) (n2m) + (n2m)2 + 2m3n2 [(a − b)2 = a2 − 2ab + b2 ]
= m4 − 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2
Page No 151:
Question 5:
Show that
(i) (3x + 7)2 − 84x = (3x − 7)2 (ii) (9p − 5q)2 + 180pq = (9p + 5q)2
(iii)
(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0
Answer:
(i) L.H.S = (3x + 7)2 − 84x
= (3x)2 + 2(3x)(7) + (7)2 − 84x
= 9x2 + 42x + 49 − 84x
= 9x2 − 42x + 49
R.H.S = (3x − 7)2 = (3x)2 − 2(3x)(7) +(7)2
= 9x2 − 42x + 49
L.H.S = R.H.S
(ii) L.H.S = (9p − 5q)2 + 180pq
= (9p)2 − 2(9p)(5q) + (5q)2 − 180pq
= 81p2 − 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
R.H.S = (9p + 5q)2
= (9p)2 + 2(9p)(5q) + (5q)2
= 81p2 + 90pq + 25q2
L.H.S = R.H.S
(iii) L.H.S =
(iv) L.H.S = (4pq + 3q)2 − (4pq − 3q)2
= (4pq)2 + 2(4pq)(3q) + (3q)2 − [(4pq)2 − 2(4pq) (3q) + (3q)2]
= 16p2q2 + 24pq2 + 9q2 − [16p2q2 − 24pq2 + 9q2]
= 16p2q2 + 24pq2 + 9q2 −16p2q2 + 24pq2 − 9q2
= 48pq2 = R.H.S
(v) L.H.S = (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)
= (a2 − b2) + (b2 − c2) + (c2 − a2) = 0 = R.H.S.
Page No 152:
Question 6:
Using identities, evaluate.
(i) 712 (ii) 992 (iii) 1022 (iv) 9982
(v) (5.2)2 (vi) 297 × 303 (vii) 78 × 82
(viii) 8.92 (ix) 1.05 × 9.5
Answer:
(i) 712 = (70 + 1)2
= (70)2 + 2(70) (1) + (1)2 [(a + b)2 = a2 + 2ab + b2 ]
= 4900 + 140 + 1 = 5041
(ii) 992 = (100 − 1)2
= (100)2 − 2(100) (1) + (1)2 [(a − b)2 = a2 − 2ab + b2 ]
= 10000 − 200 + 1 = 9801
(iii) 1022 = (100 + 2)2
= (100)2 + 2(100)(2) + (2)2 [(a + b)2 = a2 + 2ab + b2 ]
= 10000 + 400 + 4 = 10404
(iv) 9982 = (1000 − 2)2
= (1000)2 − 2(1000)(2) + (2)2 [(a − b)2 = a2 − 2ab + b2 ]
= 1000000 − 4000 + 4 = 996004
(v) (5.2)2 = (5.0 + 0.2)2
= (5.0)2 + 2(5.0) (0.2) + (0.2)2 [(a + b)2 = a2 + 2ab + b2 ]
= 25 + 2 + 0.04 = 27.04
(vi) 297 × 303 = (300 − 3) × (300 + 3)
= (300)2 − (3)2 [(a + b) (a − b) = a2 − b2]
= 90000 − 9 = 89991
(vii) 78 × 82 = (80 − 2) (80 + 2)
= (80)2 − (2)2 [(a + b) (a − b) = a2 − b2]
= 6400 − 4 = 6396
(viii) 8.92 = (9.0 − 0.1)2
= (9.0)2 − 2(9.0) (0.1) + (0.1)2 [(a − b)2 = a2 − 2ab + b2 ]
= 81 − 1.8 + 0.01 = 79.21
(ix) 1.05 × 9.5 = 1.05 × 0.95 × 10
= (1 + 0.05) (1− 0.05) ×10
= [(1)2 − (0.05)2] × 10
= [1 − 0.0025] × 10 [(a + b) (a − b) = a2 − b2]
= 0.9975 × 10 = 9.975
Video Solution for algebraic expressions and identities (Page: 152 , Q.No.: 6)
NCERT Solution for Class 8 math - algebraic expressions and identities 152 , Question 6
Page No 152:
Question 7:
Using a2 − b2 = (a + b) (a − b), find
(i) 512 − 492 (ii) (1.02)2 − (0.98)2 (iii) 1532 − 1472
(iv) 12.12 − 7.92
Answer:
(i) 512 − 492 = (51 + 49) (51 − 49)
= (100) (2) = 200
(ii) (1.02)2 − (0.98)2 = (1.02 + 0.98) (1.02 − 0.98)
= (2) (0.04) = 0.08
(iii) 1532 − 1472 = (153 + 147) (153 − 147)
= (300) (6) = 1800
(iv) 12.12 − 7.92 = (12.1 + 7.9) (12.1 − 7.9)
= (20.0) (4.2) = 84
Video Solution for algebraic expressions and identities (Page: 152 , Q.No.: 7)
NCERT Solution for Class 8 math - algebraic expressions and identities 152 , Question 7
Page No 152:
Question 8:
Using (x + a) (x + b) = x2 + (a + b) x + ab, find
(i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8
Answer:
(i) 103 × 104 = (100 + 3) (100 + 4)
= (100)2 + (3 + 4) (100) + (3) (4)
= 10000 + 700 + 12 = 10712
(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2)
= (5)2 + (0.1 + 0.2) (5) + (0.1) (0.2)
= 25 + 1.5 + 0.02 = 26.52
(iii) 103 × 98 = (100 + 3) (100 − 2)
= (100)2 + [3 + (− 2)] (100) + (3) (− 2)
= 10000 + 100 − 6
= 10094
(iv) 9.7 × 9.8 = (10 − 0.3) (10 − 0.2)
= (10)2 + [(− 0.3) + (− 0.2)] (10) + (− 0.3) (− 0.2)
= 100 + (− 0.5)10 + 0.06 = 100.06 − 5 = 95.06
Video Solution for algebraic expressions and identities (Page: 152 , Q.No.: 8)
NCERT Solution for Class 8 math - algebraic expressions and identities 152 , Question 8
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