Class 8 Maths - Algebraic Expressions and Identities NCERT Solutions

Chapter 9: Algebraic Expressions and Identities (NCERT Solutions)

Exercise 9.1

Q1. 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 - p
(v) x/2 + y/2 - xy
(vi) 0.3a - 0.6ab + 0.5b

(i) 5xyz² - 3zy
Terms: 5xyz² (coeff: 5), -3zy (coeff: -3)

(ii) 1 + x + x²
Terms: 1 (coeff: 1), x (coeff: 1), x² (coeff: 1)

(iii) 4x²y² - 4x²y²z² + z²
Terms: 4x²y² (coeff: 4), -4x²y²z² (coeff: -4), z² (coeff: 1)

(iv) 3 - pq + qr - p
Terms: 3 (coeff: 3), -pq (coeff: -1), qr (coeff: 1), -p (coeff: -1)

(v) x/2 + y/2 - xy
Terms: x/2 (coeff: 1/2), y/2 (coeff: 1/2), -xy (coeff: -1)

(vi) 0.3a - 0.6ab + 0.5b
Terms: 0.3a (coeff: 0.3), -0.6ab (coeff: -0.6), 0.5b (coeff: 0.5)

Q2. 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

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: x+x²+x³+x⁴, ab+bc+cd+da (Since they have 4 terms)

Q3. 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

(i) Add: (ab - bc) + (bc - ca) + (ca - ab)
= ab - ab - bc + bc - ca + ca = 0.

(ii) Add: (a - b + ab) + (b - c + bc) + (c - a + ac)
= a - a - b + b - c + c + ab + bc + ac = ab + bc + ac.

(iii) Add: (2p²q² - 3pq + 4) + (5 + 7pq - 3p²q²)
= (2 - 3)p²q² + (-3 + 7)pq + (4 + 5) = -p²q² + 4pq + 9.

(iv) Add: (l² + m²) + (m² + n²) + (n² + l²) + (2lm + 2mn + 2nl)
= 2l² + 2m² + 2n² + 2lm + 2mn + 2nl
= 2(l² + m² + n² + lm + mn + nl).

Q4.
(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

(a) (12a - 9ab + 5b - 3) - (4a - 7ab + 3b + 12)
= 12a - 9ab + 5b - 3 - 4a + 7ab - 3b - 12
= (12a - 4a) + (-9ab + 7ab) + (5b - 3b) + (-3 - 12)
= 8a - 2ab + 2b - 15.

(b) (5xy - 2yz - 2zx + 10xyz) - (3xy + 5yz - 7zx)
= (5xy - 3xy) + (-2yz - 5yz) + (-2zx + 7zx) + 10xyz
= 2xy - 7yz + 5zx + 10xyz.

(c) (18 - 3p - 11q + 5pq - 2pq² + 5p²q) - (4p²q - 3pq + 5pq² - 8p + 7q - 10)
= 18 - 3p - 11q + 5pq - 2pq² + 5p²q - 4p²q + 3pq - 5pq² + 8p - 7q + 10
= (5 - 4)p²q + (-2 - 5)pq² + (5 + 3)pq + (-3 + 8)p + (-11 - 7)q + (18 + 10)
= p²q - 7pq² + 8pq + 5p - 18q + 28.

Exercise 9.2

Q1. Find the product of the following pairs of monomials:
(i) 4, 7p (ii) -4p, 7p (iii) -4p, 7pq (iv) 4p³, -3p (v) 4p, 0

(i) 4 × 7p = 28p

(ii) -4p × 7p = -28p²

(iii) -4p × 7pq = -28p²q

(iv) 4p³ × (-3p) = -12p⁴

(v) 4p × 0 = 0

Q2. 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)

Area = l × b

(i) p × q = pq

(ii) 10m × 5n = 50mn

(iii) 20x² × 5y² = 100x²y²

(iv) 4x × 3x² = 12x³

(v) 3mn × 4np = 12mn²p

Q3. Complete the table of products:

First Monomial → Second Monomial ↓	2x	-5y	3x²	-4xy	7x²y	-9x²y²
2x	4x²	-10xy	6x³	-8x²y	14x³y	-18x³y²
-5y	-10xy	25y²	-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	45x²y³	-27x⁴y²	36x³y³	-63x⁴y³	81x⁴y⁴

Q4. 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

Volume = l × b × h

(i) 5a × 3a² × 7a⁴ = (5 × 3 × 7) × a⁷ = 105a⁷

(ii) 2p × 4q × 8r = (2 × 4 × 8) × pqr = 64pqr

(iii) xy × 2x²y × 2xy² = (1 × 2 × 2) × x⁴y⁴ = 4x⁴y⁴

(iv) a × 2b × 3c = (1 × 2 × 3) × abc = 6abc

Q5. 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

(i) xy × yz × zx = x²y²z²

(ii) a × (-a²) × a³ = -a⁶

(iii) 2 × 4y × 8y² × 16y³ = 1024y⁶

(iv) a × 2b × 3c × 6abc = 36a²b²c²

(v) m × (-mn) × mnp = -m³n²p

Exercise 9.3

Q1. 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

(i) 4p × (q+r) = 4pq + 4pr

(ii) ab × (a-b) = a²b - ab²

(iii) (a+b) × 7a²b² = 7a³b² + 7a²b³

(iv) (a²-9) × 4a = 4a³ - 36a

(v) (pq+qr+rp) × 0 = 0

Q2. Complete the table (First expression, Second expression, Product):

First Expression	Second Expression	Product
a	b+c+d	ab+ac+ad
x+y-5	5xy	5x²y+5xy²-25xy
p	6p²-7p+5	6p³-7p²+5p
4p²q²	p²-q²	4p⁴q²-4p²q⁴
a+b+c	abc	a²bc+ab²c+abc²

Q3. Find the product:
(i) (a²) × (2a²²) × (4a²⁶)
(ii) (2/3 xy) × (-9/10 x²y²)
(iii) (-10/3 pq³) × (6/5 p³q)
(iv) x × x² × x³ × x⁴

(i) (1 × 2 × 4) × a⁵⁰ = 8a⁵⁰

(ii) (2/3 × -9/10) × (x³y³) = -3/5 x³y³

(iii) (-10/3 × 6/5) × (p⁴q⁴) = -4p⁴q⁴

(iv) x^1+2+3+4 = x¹⁰

Q4.
(a) Simplify 3x(4x-5) + 3 and find its values for (i) x=3 (ii) x=1/2
(b) Simplify a(a²+a+1) + 5 and find its value for (i) a=0 (ii) a=1 (iii) a=-1

(a) Simplify: 12x² - 15x + 3
(i) For x=3: 12(3)² - 15(3) + 3 = 108 - 45 + 3 = 66
(ii) For x=1/2: 12(1/2)² - 15(1/2) + 3 = 12(1/4) - 15/2 + 3 = 3 - 7.5 + 3 = -1.5

(b) Simplify: a³ + a² + a + 5
(i) For a=0: 0 + 0 + 0 + 5 = 5
(ii) For a=1: 1 + 1 + 1 + 5 = 8
(iii) For a=-1: (-1)³ + (-1)² + (-1) + 5 = -1 + 1 - 1 + 5 = 4

Q5.
(a) Add: p(p-q), q(q-r) and r(r-p)
(b) Add: 2x(z-x-y) and 2y(z-y-x)
(c) Subtract: 3l(l-4m+5n) from 4l(10n-3m+2l)
(d) Subtract: 3a(a+b+c) - 2b(a-b+c) from 4c(-a+b+c)

(a) (p² - pq) + (q² - qr) + (r² - rp)
= p² + q² + r² - pq - qr - rp

(b) (2xz - 2x² - 2xy) + (2yz - 2y² - 2xy)
= -2x² - 2y² - 4xy + 2xz + 2yz

(c) (40ln - 12lm + 8l²) - (3l² - 12lm + 15ln)
= 40ln - 12lm + 8l² - 3l² + 12lm - 15ln
= 5l² + 25ln

(d) First term = 3a² + 3ab + 3ac - 2ab + 2b² - 2bc = 3a² + 2b² + ab + 3ac - 2bc
Second term = -4ac + 4bc + 4c²
Subtract: (-4ac + 4bc + 4c²) - (3a² + 2b² + ab + 3ac - 2bc)
= -4ac + 4bc + 4c² - 3a² - 2b² - ab - 3ac + 2bc
= -3a² - 2b² + 4c² - ab + 6bc - 7ac

Exercise 9.4

Q1. 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) (3/4 a² + 3b²) and 4(a² - 2/3 b²)

(i) 2x(4x-3) + 5(4x-3) = 8x² - 6x + 20x - 15 = 8x² + 14x - 15

(ii) y(3y-4) - 8(3y-4) = 3y² - 4y - 24y + 32 = 3y² - 28y + 32

(iii) (2.5l)² - (0.5m)² = 6.25l² - 0.25m²

(iv) a(x+5) + 3b(x+5) = ax + 5a + 3bx + 15b

(v) 2pq(3pq - 2q²) + 3q²(3pq - 2q²) = 6p²q² - 4pq³ + 9pq³ - 6q⁴ = 6p²q² + 5pq³ - 6q⁴

(vi) (3/4 a² + 3b²)(4a² - 8/3 b²)
= 3/4 a²(4a² - 8/3 b²) + 3b²(4a² - 8/3 b²)
= 3a⁴ - 2a²b² + 12a²b² - 8b⁴ = 3a⁴ + 10a²b² - 8b⁴

Q2. Find the product:
(i) (5 - 2x)(3 + x)
(ii) (x + 7y)(7x - y)
(iii) (a² + b)(a + b²)
(iv) (p² - q²)(2p + q)

(i) 5(3+x) - 2x(3+x) = 15 + 5x - 6x - 2x² = 15 - x - 2x²

(ii) x(7x-y) + 7y(7x-y) = 7x² - xy + 49xy - 7y² = 7x² + 48xy - 7y²

(iii) a²(a+b²) + b(a+b²) = a³ + a²b² + ab + b³

(iv) p²(2p+q) - q²(2p+q) = 2p³ + p²q - 2pq² - q³

Q3. 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)

(i) x²(x+5) - 5(x+5) + 25 = x³ + 5x² - 25x - 25 + 25 = x³ + 5x² - 25x

(ii) a²b³ + 3a² + 5b³ + 15 + 5 = a²b³ + 3a² + 5b³ + 20

(iii) t³ - ts + s²t² - s³

(iv) (ac-ad+bc-bd) + (ac+ad-bc-bd) + 2ac + 2bd = 2ac - 2bd + 2ac + 2bd = 4ac

(v) (2x²+xy+2xy+y²) + (x²-xy+2xy-2y²) = (2x²+3xy+y²) + (x²+xy-2y²) = 3x² + 4xy - y²

(vi) x(x²-xy+y²) + y(x²-xy+y²) = x³-x²y+xy² + x²y-xy²+y³ = x³ + y³

(vii) (2.25x² + 6xy + 4.5x - 6xy - 16y² - 12y) - 4.5x + 12y = 2.25x² - 16y² + 4.5x - 12y - 4.5x + 12y = 2.25x² - 16y²

(viii) ((a+b)+c)((a+b)-c) = (a+b)² - c² = a² + 2ab + b² - c²

Exercise 9.5

Q1. 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) (3a - 1/2)(3a - 1/2) (v) (1.1m-0.4)(1.1m+0.4) (vi) (a²+b²)(-a²+b²) (vii) (6x-7)(6x+7) (viii) (-a+c)(-a+c) (ix) (x/2+3y/4)(x/2+3y/4) (x) (7a-9b)(7a-9b)

(i) (x+3)² = x² + 6x + 9 [(a+b)² = a²+2ab+b²]

(ii) (2y+5)² = 4y² + 20y + 25

(iii) (2a-7)² = 4a² - 28a + 49

(iv) (3a - 1/2)² = 9a² - 3a + 1/4

(v) (1.1m)² - (0.4)² = 1.21m² - 0.16 [(a-b)(a+b) = a²-b²]

(vi) (b²+a²)(b²-a²) = b⁴ - a⁴

(vii) (6x)² - (7)² = 36x² - 49

(viii) (-a+c)² = (-a)² - 2ac + c² = a² - 2ac + c²

(ix) (x/2+3y/4)² = x²/4 + 2(x/2)(3y/4) + 9y²/16 = x²/4 + 3xy/4 + 9y²/16

(x) (7a-9b)² = 49a² - 126ab + 81b²

Q2. Use the identity (x+a)(x+b) = x² + (a+b)x + ab to find 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)

(i) x² + (3+7)x + (3 × 7) = x² + 10x + 21

(ii) (4x)² + (5+1)(4x) + (5 × 1) = 16x² + 24x + 5

(iii) (4x)² + (-5-1)(4x) + (-5 × -1) = 16x² - 24x + 5

(iv) (4x)² + (5-1)(4x) + (5 × -1) = 16x² + 16x - 5

(v) (2x)² + (5y+3y)(2x) + (5y × 3y) = 4x² + 16xy + 15y²

(vi) (2a²)² + (9+5)(2a²) + (9 × 5) = 4a⁴ + 28a² + 45

(vii) (xyz)² + (-4-2)(xyz) + (-4 × -2) = x²y²z² - 6xyz + 8

Q3. Find the following squares by using the identities:
(i) (b-7)² (ii) (xy+3z)² (iii) (6x²-5y)² (iv) (2/3 m + 3/2 n)² (v) (0.4p-0.5q)² (vi) (2xy+5y)²

(i) b² - 14b + 49

(ii) x²y² + 6xyz + 9z²

(iii) 36x⁴ - 60x²y + 25y²

(iv) 4/9 m² + 2(2/3 m)(3/2 n) + 9/4 n² = 4/9 m² + 2mn + 9/4 n²

(v) 0.16p² - 0.4pq + 0.25q²

(vi) 4x²y² + 20xy² + 25y²

Q4. 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

(i) (a²)² - 2a²b² + (b²)² = a⁴ - 2a²b² + b⁴

(ii) ((2x+5)-(2x-5))((2x+5)+(2x-5)) = (10)(4x) = 40x

(iii) 49m² - 112mn + 64n² + 49m² + 112mn + 64n² = 98m² + 128n²

(iv) 16m²+40mn+25n² + 25m²+40mn+16n² = 41m² + 80mn + 41n²

(v) 6.25p² - 7.5pq + 2.25q² - (2.25p² - 7.5pq + 6.25q²) = 4p² - 4q²

(vi) a²b² + 2ab²c + b²c² - 2ab²c = a²b² + b²c²

(vii) m⁴ - 2m³n² + n⁴m² + 2m³n² = m⁴ + n⁴m²

Q5. Show that:
(i) (3x+7)² - 84x = (3x-7)²
(ii) (9p-5q)² + 180pq = (9p+5q)²
(iii) (4/3 m - 3/4 n)² + 2mn = 16/9 m² + 9/16 n²
(iv) (4pq+3q)² - (4pq-3q)² = 48pq²
(v) (a-b)(a+b) + (b-c)(b+c) + (c-a)(c+a) = 0

(i) LHS = 9x² + 42x + 49 - 84x = 9x² - 42x + 49 = (3x-7)² = RHS.

(ii) LHS = 81p² - 90pq + 25q² + 180pq = 81p² + 90pq + 25q² = (9p+5q)² = RHS.

(iii) LHS = 16/9 m² - 2mn + 9/16 n² + 2mn = 16/9 m² + 9/16 n² = RHS.

(iv) LHS = 2 × (2 × 4pq × 3q) or RHS using a²-b²: = ((4pq+3q)-(4pq-3q))((4pq+3q)+(4pq-3q)) = (6q)(8pq) = 48pq² = RHS.

(v) LHS = (a²-b²) + (b²-c²) + (c²-a²) = 0 = RHS.

Q6. Using identities, evaluate:
(i) 71² (ii) 99² (iii) 102² (iv) 998² (v) 5.2² (vi) 297 × 303 (vii) 78 × 82 (viii) 8.9² (ix) 1.05 × 9.5

(i) (70+1)² = 4900+140+1 = 5041

(ii) (100-1)² = 10000-200+1 = 9801

(iii) (100+2)² = 10000+400+4 = 10404

(iv) (1000-2)² = 1000000-4000+4 = 996004

(v) (5+0.2)² = 25+2+0.04 = 27.04

(vi) (300-3)(300+3) = 90000-9 = 89991

(vii) (80-2)(80+2) = 6400-4 = 6396

(viii) (9-0.1)² = 81-1.8+0.01 = 79.21

(ix) 1.05 × 9.5 ≈ 10.5 × 0.95 = (10+0.5)(10-0.5)/10 = (100-0.25)/10 = 9.975

Q7. 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

(i) (51-49)(51+49) = 2 × 100 = 200

(ii) (1.02-0.98)(1.02+0.98) = 0.04 × 2.00 = 0.08

(iii) (153-147)(153+147) = 6 × 300 = 1800

(iv) (12.1-7.9)(12.1+7.9) = 4.2 × 20.0 = 84

Q8. 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

(i) (100+3)(100+4) = 100²+(3+4)100+(3×4) = 10000+700+12 = 10712

(ii) (5+0.1)(5+0.2) = 5²+(0.1+0.2)5+(0.1×0.2) = 25+1.5+0.02 = 26.52

(iii) (100+3)(100-2) = 100²+(3-2)100+(3×-2) = 10000+100-6 = 10094

(iv) (10-0.3)(10-0.2) = 10²+(-0.3-0.2)10+(-0.3×-0.2) = 100-5+0.06 = 95.06

Class 8 Maths - Algebra Practice Questions

Chapter 9: Algebraic Expressions and Identities (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. Add: 7x² - 4x + 5 and -3x² + 2x - 1

Sum = (7x² - 3x²) + (-4x + 2x) + (5 - 1)
= 4x² - 2x + 4.

Q2. Subtract 3a - 4b + 5c from 5a + 2b - c.

Difference = (5a + 2b - c) - (3a - 4b + 5c)
= 5a + 2b - c - 3a + 4b - 5c
= (5a - 3a) + (2b + 4b) + (-c - 5c)
= 2a + 6b - 6c.

Q3. Multiply -2x²y by 5xy³.

Product = (-2 × 5) × (x² × x) × (y × y³)
= -10x³y⁴.

Q4. Simplify: 2x(3x - 5) - x(2x - 3)

= (6x² - 10x) - (2x² - 3x)
= 6x² - 10x - 2x² + 3x
= 4x² - 7x.

Q5. Find the product: (a² - b²)(a² + b²)

Using (x - y)(x + y) = x² - y², where x = a² and y = b²
= (a²)² - (b²)²
= a⁴ - b⁴.

Q6. Evaluate 103² using a standard identity.

103² = (100 + 3)²
Using (a + b)² = a² + 2ab + b²
= 100² + 2(100)(3) + 3²
= 10000 + 600 + 9
= 10609.

Q7. Evaluate 98 × 102 using an identity.

98 × 102 = (100 - 2)(100 + 2)
Using (a - b)(a + b) = a² - b²
= 100² - 2²
= 10000 - 4
= 9996.

Q8. Find the value of x if 3x = 24² - 21².

3x = (24 - 21)(24 + 21) [Using a² - b² = (a-b)(a+b)]
3x = (3)(45)
x = (3 × 45)/3
x = 45.

Q9. Simplify: (3x + 2y)² - (3x - 2y)²

Using a² - b² = (a-b)(a+b) where a = 3x+2y and b = 3x-2y
= [(3x+2y) - (3x-2y)] × [(3x+2y) + (3x-2y)]
= (4y) × (6x)
= 24xy.

Q10. If x + 1/x = 5, find the value of x² + 1/x².

Squaring both sides:
(x + 1/x)² = 5²
x² + 2(x)(1/x) + 1/x² = 25
x² + 2 + 1/x² = 25
x² + 1/x² = 25 - 2 = 23.

Q11. Multiply a binomial (2x - 3y) by a trinomial (4x² + 6xy + 9y²).

= 2x(4x² + 6xy + 9y²) - 3y(4x² + 6xy + 9y²)
= 8x³ + 12x²y + 18xy² - 12x²y - 18xy² - 27y³
= 8x³ + (12x²y - 12x²y) + (18xy² - 18xy²) - 27y³
= 8x³ - 27y³.

Q12. Show that (a-b)(a+b) + (b-c)(b+c) + (c-a)(c+a) = 0.

LHS = (a² - b²) + (b² - c²) + (c² - a²)
= a² - a² - b² + b² - c² + c²
= 0 = RHS. (Hence proved).

Q13. Simplify: (5x - 2y)(5x + 2y)(25x² + 4y²)

First multiply the first two terms: (5x)² - (2y)² = 25x² - 4y²
Now multiply by the third term: (25x² - 4y²)(25x² + 4y²)
= (25x²)² - (4y²)²
= 625x⁴ - 16y⁴.

Q14. Find the area of a rectangle whose length is (p + 3) and breadth is (p - 2).

Area = length × breadth
= (p + 3)(p - 2)
Using (x+a)(x+b) = x² + (a+b)x + ab
= p² + (3 - 2)p + (3)(-2)
= p² + p - 6 square units.

Q15. Expand using suitable identity: (x/2 + 2/y)²

Using (a + b)² = a² + 2ab + b²
= (x/2)² + 2(x/2)(2/y) + (2/y)²
= x²/4 + 2x/y + 4/y².

Q16. Calculate 104 × 105 using (x+a)(x+b) = x² + (a+b)x + ab.

104 × 105 = (100 + 4)(100 + 5)
Here x=100, a=4, b=5
= 100² + (4+5)(100) + (4 × 5)
= 10000 + 900 + 20
= 10920.

Q17. Simplify: (xy + yz)² - 2xy²z

= [(xy)² + 2(xy)(yz) + (yz)²] - 2xy²z
= x²y² + 2xy²z + y²z² - 2xy²z
= x²y² + y²z².

Q18. If x - 1/x = 8, find the value of x² + 1/x².

Squaring both sides:
(x - 1/x)² = 8²
x² - 2(x)(1/x) + 1/x² = 64
x² - 2 + 1/x² = 64
x² + 1/x² = 64 + 2 = 66.

Q19. Find the coefficient of x² in the product of (3x - 1)(2x - 5).

Product = 3x(2x-5) - 1(2x-5)
= 6x² - 15x - 2x + 5
= 6x² - 17x + 5
The coefficient of x² is 6.

Q20. If a² + b² = 74 and ab = 35, find the value of (a+b).

We know that (a+b)² = a² + b² + 2ab
(a+b)² = 74 + 2(35)
(a+b)² = 74 + 70 = 144
a+b = √144 = ± 12.

Class 8 Maths - Algebraic Expressions and Identities Summary

Chapter 9: Algebraic Expressions and Identities (Concepts & Formulas)

1. Key Terms & Definitions

Algebraic Expression: An expression formed from variables and constants using the operations of addition, subtraction, multiplication, and division.
Terms: Terms are added to form expressions. E.g., in 4x + 5, 4x and 5 are terms.
Factors: A term is a product of its factors. E.g., factors of 4x are 4 and x.
Coefficients: The numerical factor of a term is called its numerical coefficient or simply coefficient. E.g., in 7xy, 7 is the coefficient.

2. Types of Polynomials

Monomial: An expression containing exactly one term. E.g., 3x, -5y, 7x².
Binomial: An expression containing exactly two terms. E.g., x + y, a - b.
Trinomial: An expression containing exactly three terms. E.g., x + y + z, x² + 2x + 1.
Polynomial: An expression containing one or more terms with non-zero coefficients (and variables having non-negative integral exponents).

3. Like and Unlike Terms

Like Terms: Terms which have the same algebraic factors. E.g., 2xy and -5xy.

Unlike Terms: Terms which have different algebraic factors. E.g., 2xy and -5x.

Note: Addition and subtraction can only be performed on like terms.

4. Standard Algebraic Identities

Identity I: (a + b)² = a² + 2ab + b²
Identity II: (a - b)² = a² - 2ab + b²
Identity III: (a + b)(a - b) = a² - b²
Identity IV: (x + a)(x + b) = x² + (a + b)x + ab

Applications: These identities are useful in carrying out squares and products of algebraic expressions. They also help in easy calculation of numeric squares and products.

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