NCERT Solutions for Class 8 Maths Chapter 14 Factorisation| PDF Download

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NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

Page No: 220

Exercise 14.1

1. Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p²q²
(iv) 2x, 3x², 4
(v) 6abc, 24ab², 12a²b
(vi) 16x³, -4x², 32x
(vii) 10 pq, 20qr, 30rp
(viii) 3x²y³, 10x³y², 6x²y²z

Answer

(i) 12x = 2×2×3×x
36 = 2×2×3×3
Hence, the common factors are 2, 2 and 3 = 2×2×3 = 12

(ii) 2y = 2×y
22xy = 2×11×x×y
Hence, the common factors are 2 and y = 2×y = 2y

(iii) 14pq = 2×7×p×q
28p²q² = 2×2×7×p×p×q×q
Hence, the common factors are 2×7×p×q = 14pq

(iv) 2x = 2×x×1
3x² = 3×x×x×1
4 = 2×2×1
Hence, the common factor is 1.

(v) 6abc = 2×3×a×b×c
24ab² = 2×2×2×3×a×b×b
12a²b = 2×2×3×a×a×b
Hence, the common factors are 2×3×a×b = 6ab

(vi) 16x³ = 2×2×2×x×x×x
-4x² = (-1)×2×2×x×x
32x = 2×2×2×2×2×x
Hence the common factors are 2×2×x = 4x

(vii) 10pq = 2×5×p×q
20qr = 2×2×5×q×r
30rp = 2×3×5×r×p
Hence the common factors are 2×5 = 10

(viii) 3x²y³ = 3×x×x×y×y×y
10x³y² = 2×5×x×x×x×y×y
6x²y²z = 2×3×x×x×y×y×z
Hence the common factors are x×x×y×y = x²y²

2. Factorize the following expressions.
(i) 7x - 42
(ii) 6p - 12q
(iii) 7a² + 14a
(iv) -16z + 20z³
(v)20l²m + 30alm
(vi) 5x²y - 15xy²
(vii) 10a² - 15b² + 20c²
(viii) -4a² + 4ab - 4ca
(ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz

Answer

(i) 7x - 42 = 7×x - 2×3×7
Taking common factors from each term,
= 7(x - 2×3)
= 7(x - 6)

(ii) 6p - 12q = 2×3×p - 2×2×3×q
Taking common factors from each term,
= 2×3(p - 2q)
= 6(p - 2q)

(iii) 7a² + 14a = 7×a×a + 2×7×a
Taking common factors from each term,
= 7×a(a + 2)
= 7a(a + 2)

(iv) -16z + 20z³
= (-1)×2×2×2×2×z + 2×2×5×z×z× z
Taking common factors from each term,
= 2×2×z (-2×2 + 5×z×z)
= 4z (-4 + 5z²)

(v) 20l²m + 30alm
= 2×2×5×l×l×m + 2×3×5×a×l×m
Taking common factors from each term,
= 2×5×l×m(2×l + 3×a)
= 10 lm(2l +3a)

(vi) 5x²y - 15xy²
= 5×x×x×y - 3×5×x×y×y (Taking common factors from each term)
= 5×x×y(x - 3y)
= 5xy(x - 3y)

(vii) 10a² - 15b² + 20c²
= 2×5×a×a - 3×5×b×b + 2×2×5×c×c
Taking common factors from each term,
= 5(2×a×a - 3×b×b + 2×2×c×c)
= 5(2a² - 3b² + 4c²)

(viii) -4a² + 4ab - 4ca
= (-1)×2×2×a×a + 2×2×a×b - 2×2×c×a
Taking common factors from each term,
= 2×2×a(-a + b -c)
= 4a (-a + b - c)

(ix) x²yz + xy²z + xyz²
= x×x×y×z + x×y×y×z + z×y×z×z
Taking common factors from each term,
= x×y×z( x + y + z)
= xyz(x + y +z)

(x) ax²y + bxy² + cxyz
= a×x×x×y + b×x×y×y + c×x×y×z
Taking common factors from each term,
= x×y(a×x + b×y + c×z)
= xy(ax + by +cz)

3. Factorize:
(i) x² + xy + 8x + 8y
(ii) 15xy - 6x + 5y -2
(iii) ax + bx - ay - by
(iv) 15pq + 15 + 9q + 25p
(v) z - 7 + 7xy -xyz

Answer

(i) x² + xy + 8x + 8y
= x(x + y) + 8(x + y)
= (x + y)(x + 8)

(ii) 15xy - 6x + 5y - 2
= 3x(5y - 2) + 1(5y - 2)
= (5y -2)(3x + 1)

(iii) ax + bx - ay - by
= (ax + bx) - (ay + by)
= x(a + b) - y(a + b)
= (a + b)(x - y)

(iv) 15pq + 15 + 9q + 25p
= 15pq + 25p + 9q + 15
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5)(5p + 3)

(v) z -7 + 7xy - xyz = 7xy - 7 - xyz + z
= 7(xy - 1) - z(xy - 1)
= (xy -1)(7 - z) = (-1)(1 - xy)(-1)(z - 7)
= (1 - xy)(z - 7)

Page No. 223

Exercise 14.2

1. Factorize the following expressions:(i) a² + 8a + 16
(ii) p² - 10p + 25
(iii) 25m² + 30m + 9
(iv) 49y² + 84yz + 36z²
(v) 4x² - 8x + 4
(vi) 121b² - 88bc + 16c²
(vii) (l + m)² - 4lm [Hint: Expand (l + m)² first]
(viii) a⁴ + 2a²b² + b⁴

Answer

(i) a² + 8a + 16 = a² + (4 + 4)a + 4 × 4
Using identity x² + (a + b)x + ab = (x + a)(x + b),
Here x = a, a = 4 and b = 4
a² + 8a + 16 = (a + 4)(a + 4) = (a + 4)²

(ii) p² - 10p + 25 = p² +(-5-5)p + (-5)(-5)
Using identity x² + (a +b)x + ab = ( x + a)(x + b),
Here x = p, a = -5 and b = -5
p² - 10p + 25 = (p -5)(p- 5) = (p - 5)²

(iii) 25m² + 30m + 9 = (5m)² + 2 × 5m × 3 + (3)²
Using identity a² + 2ab + b² = (a + b)² , here a= 5m, b = 3
25m² + 30m + 9 = (5m + 3)²

(iv) 49y² + 84yz + 36z² = (7y)² + 2 × 7y × 6z + (6z)²
Using identity a² + 2ab + b² = (a + b)² , here a = 7y, b = 6z
49y² + 84yz + 36z² = (7y + 6z)²

(v) 4x² - 8x + 4 = (2x)² - 2 × 2x ×2 + (2)²
Using identity a2 - 2ab + b2 = (a - b)2 , here a = 2x, b = 2
4x² - 8x + 4 = (2x - 2)²
= (2)² (x - 1)² = 4( x - 1)²

(vi) 121b² - 88bc + 16c² = (11b)² - 2 × 11b × 4c + (4c)²
Using identity a2 - 2ab + b2 = (a - b)2 , here a = 11b, b = 4c
121b² - 88bc + 16c² = (11b - 4c)²

(vii) (l + m)² - 4lm
= l² + 2 × l ×m + m² - 4lm [ ∵ (a + b)² = a² + 2ab + b² ]
= l² + 2lm + m² - 4lm
= l² - 2lm + m²
= (l - m)² [ ∵ (a- b)² = a² - 2ab + b² ]

(viii) a⁴ + 2a²b² + b⁴ = (a²)² + 2 × a² × b² + (b²)²
= (a² + b²)² [∵ (a + b)2 = a2 + 2ab + b ]

2. Factorize:
(i) 4p² - 9q² (ii) 63a² - 112b²
(iii) 49x² - 36
(iv) 16x⁵ - 144x²
(v) (l + m)² - (l -m)²
(vi) 9x²y² - 16
(vii) (x² - 2xy + y²) - z²
(viii) 25a² - 4b² + 28bc - 49c²

Answer

(i) 4p² - 9q² = (2p)² - (3q)²
= (2p -3q)(2p + 3q) [∵ a² - b² = (a - b)(a +b)]

(ii) 63a² - 112b² = 7(9a² - 16b²)
= 7 [ (3a)² - (4b)²]
= 7(3a - 4b)(3a + 4b) [∵ a² - b² = (a - b)(a +b)]

(iii) 49x² - 36 = (7x)² - (6)²
= (7x - 6)(7x + 6) [∵ a² - b² = (a - b)(a +b)]

(iv) 16x⁵ - 144x³ = 16x³(x² - 9)
= 16x³ [(x)² - (3)²]
= 16x³ (x - 3)(x + 3) [∵ a² - b² = (a - b)(a +b)]

(v) (l + m)² - (l - m)²
= [(l + m) + ( l - m)][(l + m)- (l - m)] [∵ a² - b² = (a - b)(a +b)]
= (l + m + l - m)(l + m - l +m)
= (2l) (2m) = 4lm

(vi) 9x²y² - 16 = (3xy)² - (4)²
= (3xy - 4)(3xy + 4) [ ∵ a2 - b2 = (a - b)(a +b)]

(vii) ( x² - 2xy + y²) - z² = ( x - y)² - z² [∵ (a -b)² = a² -2ab + b²]
= ( x - y - z)( x - y + z) [ ∵ a2 - b2 = (a - b)(a +b)]
(viii) 25a² - 4b² + 28bs - 49c²
= 25a² - (4b² - 28bc + 49c²)
= 25a² - [ (2b)² - 2 × 2b × 7c + (7c)²]
= 25a² - (2b - 7c)² [ ∵ (a -b)2 = a2 -2ab + b2]
= (5a)² - (2b - 7c)²
= [5a - (2b - 7c)][5a + (2b - 7c)] [ ∵ a2 - b2 = (a - b)(a +b)]
= (5a - 2b + 7c)(5a + 2b - 7c)

3. Factorize the expressions:

(i) ax² + bx(ii) 7p² + 21q²
(iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an²
(v) (lm + l ) + m + 1
(vi) y( y + z) + 9 ( y + z)
(vii) 5y² - 20y - 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy - 4y + 6 - 9x

Answer

(i) ax² + bx = x(ax + b)

(ii) 7p² + 21q² = 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz² = 2x( x² + y² + z²)

(iv) am² + bm² + bn² + an²
= m²( a + b) + n²(a + b)
= (a + b )(m² + n²)

(v) (lm + l) + m + 1
= l(m + 1) + 1(m + 1)
= (m + 1)( l + 1)

(vi) y(y + z) + 9(y + z)
= (y + z)(y + 9)

(vii) 5y² - 20y - 8z + 2yz
= 5y² - 20y + 2yz - 8z
= 5y(y - 4) + 2z(y - 4)
= (y - 4)(5y + 2z)

(viii) 10ab + 4a + 5b + 2
= 2a(5b + 2) + 1 (5b + 2)
= (5b + 2)(2a + 1)

(ix) 6xy - 4y + 6 - 9x
= 6xy - 9x - 4y + 6
= 3x(2y - 3) - 2(2y - 3)
= (2y - 3) (3x - 2)

4. Factorize:
(i) a⁴ - b⁴
(ii) p⁴ - 81
(iii) x⁴ - (y + z)⁴
(iv)x⁴ - (x -z)⁴
(v) a⁴ - 2a²b² + b⁴

Answer

(i) a⁴ - b⁴ = (a²)² - (b²)²
= (a² - b²)( a² + b²) [ ∵ a² - b² = (a - b)(a +b)]
= (a - b)(a + b)(a² + b²) [ ∵ a² - b² = (a - b)(a +b)]

(ii) p⁴ - 81 = (p²)² - (9)²
= (p² - 9)(p² + 9) [∵ a² - b² = (a - b)(a +b)]
= (p² - 3²)(p² + 9)
= ( p - 3)(p + 3)(p² + 9) [∵ a² - b² = (a - b)(a +b)]

(iii) x⁴ - (y + z)⁴ = (x²)² - [(y + z)²]²
= [x² - (y + z)²][ x² + (y + z)²] [∵ a² - b² = (a - b)(a +b)]
= [x -(y +z)][x + (y + z)][x² + (y + z)²] [∵ a² - b² = (a - b)(a +b)]
= (x - y - z) (x + y + z) [x² + (y + z)²]

(iv) x⁴ - (x - z)⁴ = (x²)² - [(x - z)²]²
= [x² -(x - z)²][x² + (x - z)²] [ ∵ a² - b² = (a - b)(a +b)]
= [x - (x - z)][x + (x - z)] [x² + (x - z)²] [∵ a² - b² = (a - b)(a +b)]
= [x - x + z] [x + x - z] [x² + x² - 2xz + z²] [∵ (a -b)² = a² -2ab + b²]
= z(2x - z) (2x² - 2xz + z²)

(v) a⁴ - 2a²b² + b⁴ = (a²)² - 2a²b² + (b²)²
= (a² - b²)² [∵ a² - b² = (a - b)(a +b)]
= [(a - b)(a + b)]² [∵ a² - b² = (a - b)(a +b)]
= (a -b)² (a + b)² [ (xy)^m = x^my^m]

5. Factorize the following expressions:
(i) p² + 6p + 8
(ii) q² - 10q + 21
(iii) p² + 6p - 16

Answer

(i) p² + 6p + 8 = p² + ( 4 + 2)p + 4 × 2
= p² + 4p + 2p + 4 ×2
= p(p + 4) + 2 ( p + 4)
= (p + 4)(p + 2)

(ii) q² - 10q + 21 = q² - ( 7 + 3)q + 7 × 3
= q² - 7q - 3q + 7 × 3
= q(q - 7) - 3(q - 7)
= (q - 7)( q - 3)

(iii) p² + 6p - 16
= p² + (8 - 2)p - 8×2
= p² + 8p - 2p - 8×2
= p(p + 8) - 2(p + 8)
= ( p + 8)(p -2)

Page No. 227

Exercise 14.3

1. Carry out the following divisions:
(i) 2x⁴ ÷ 56x
(ii) -36y³ ÷ 9y²
(iii) 66pq²r³ ÷ 11 qr²
(iv) 34x³y³x³ ÷ 51xy²z³
(v) 12a⁸b⁸ ÷ (-6a⁶b⁴)

Answer

(i) 2x⁴ ÷ 56x
= 28x⁴/56x
= 28/56 × x⁴/x
= 1/2 x³ [x^m ÷ xⁿ = x^m-n]

(ii) -36y³ ÷ 9y² = -36y³/9y²
= -36/9 × y³/y²
= -4y [x^m ÷ xⁿ = x^m-n]

(iii) 66pq²r³ ÷ 11qr²
= 66pq²r³/11qr²
= 66/11 × pq²r³/qr²
= 6pqr [x^m ÷ xⁿ = x^m-n]

(iv) 34x³y³z³ ÷ 51xy²z³
= 34x³y³z³/51xy²z³
= 34/51 ×x³y³z³/xy²z³
= 2/3x²y [x^m ÷ xⁿ = x^m-n]

(v) 12a⁸b⁸ ÷ (- 6a⁶b⁴)
= 12a⁸b⁸/- 6a⁶b⁴
= 12/-6 × a⁸b⁸/a⁶b⁴
= -2a²b⁴ [x^m ÷ xⁿ = x^m-n]

2. Divide the given polynomial by the given monomial:
(i) (5x² - 6x) ÷ 3x
(ii) (3y⁸ - 4y⁶ + 5y⁴) ÷ y⁴
(iii) 8(x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
(iv) (x³ + 2x² + 3x) ÷2x
(v) (p³q⁶ - p⁶q³) ÷ p³q³

Answer

(i) (5x² - 6x) ÷3x
= (5x² - 6x)/3x
= 5x²/3x - 6x/3x = (5/3)x - 2 = 1/3 (5x - 6)

(ii) (3y⁸ - 4y⁶ + 5y⁴) ÷ y⁴
= (3y⁸ - 4y⁶ + 5y⁴)/ y⁴
= 3y⁸/y⁴ - 4y⁶/y⁴ + 5y⁴/y⁴ = 3y⁴ - 4y² + 5

(iii) 8(x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
= {8(x³y²z² + x²y³z² + x²y²z³)}/4 x²y²z²
= 8 x³y²z²/4 x²y²z² + 8 x²y³z²/4x²y²z² + 8 x²y²z³/4x²y²z²
= 2x + 2y + 2z
= 2(x + y + z)

(iv) (x³ + 2x² + 3x) ÷ 2x
= (x³ + 2x² + 3x)/2x
= x³/2x + 2x²/2x + 3x/2x = x²/2 + 2x/2 + 3/2
= 1/2( x² + 2x + 3)
(v) (p³q⁶ - p⁶q³) ÷ p³q³
= (p³q⁶ - p⁶q³)/p³q³
= p³q⁶/p³q³ - p⁶q³/p³q³ = q³ - p³

3. Work out the following divisions:
(i) (10x - 25) ÷ 5
(ii) (10x - 25) ÷ (2x - 5)
(iii) 10y (6y + 21) ÷ 5(2y + 7)
(iv) 9x²y²(3z - 24) ÷ 27xy(z - 8)
(v) 96abc(3a - 12)(5b - 30) ÷ 144(a -4)(b - 6)

Answer

(i) (10x - 25) ÷ 5
= (10x - 25)/5
= {5(2x - 5)}/5
= 2x -5

(ii) (10x - 25) ÷ (2x - 5)
= (10x - 25)/(2x - 5)
= {5(2x - 5)/(2x - 5)
= 5

(iii) 10y(6y + 21) ÷ 5(2y + 7)
= {10y(6y + 21)}/5(2y + 7)
= {2×5×y× 3(2y + 7)}/5(2y + 7)
= 2×y×3
= 6y

(iv) 9x²y²(3z - 24) ÷ 27xy(z - 8)
= {9x²y²(3z - 24)}/27xy(z - 8)
= 9/27 × {xy × xy × 3(z - 8)}/xy(z - 8)
= xy

(v) 96abc(3a - 12)(5b - 30) ÷ 144(a- 4)(b - 6)
= {96abc(3a - 12)(5b - 30)}/144(a - 4)(b - 6)
= {12×4×2×abc× 3(a-4) × 5(b-6)}/{12×4×3 (a - 4)(b - 6)
= 10abc

4. Divide as directed:
(i) 5(2x + 1)(3x + 5) ÷ (2x + 1)
(ii) 26xy(x + 5)(y - 4) ÷ 13x(y - 4)
(iii) 52pqr(p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)
(iv) 20(y + 4)(y² + 5y + 3) ÷ 5(y + 4)
(v) x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)

Answer

(i) 5(2x + 1)(3x + 5) ÷ (2x + 1)
= {5(2x + 1)(3x +5)}/(2x + 1)
= 5(3x + 5)

(ii) 26xy( x + 5)(y - 4) ÷ 13x(y - 4)
26xy( x + 5)(y -4) ÷ 13x(y - 4)
= {26xy(x + 5)(y - 4)}/13x(y - 4)
= {13×2×xy(x + 5)(y - 4)}/13x(y - 4)
= 2y(x + 5)

(iii) 52pqr( p + q)(q + r)( r + p) ÷ 104pq(q + r)(r + p)
= {52pqr(p + q)(q + r)( r + p)}/{52 × 2 × pq(q + r)(r + p)}
= (1/2)r (p + q)

(iv) 20( y + 4)(y² + 5y + 3) ÷ 5(y + 4)
= {20(y + 4)(y² + 5y + 3)}/5(y + 4)
= 4(y² + 5y + 3)

(v) x( x + 1)(x + 2)(x + 3) ÷ x(x + 1)
= {x(x + 1)(x + 2)(x + 3)}/x(x + 1)
= (x + 2)(x + 3)

5. Factorize the expressions and divide them as directed:
(i) (y² + 7y + 10) ÷ (y + 5)
(ii) (m² - 14m - 32) ÷ (m + 2)
(iii) (5p² - 25p + 20) ÷ (p - 1)
(iv) 4yz(z² + 6z - 16) ÷ 2y( z + 8)
(v) 5pq(p² - q²) ÷ 2p(p + q)
(vi) 12xy(9x² - 16y²) ÷ 4xy(3x + 4y)
(vii) 39y³(50y² - 98) ÷ 26y²(5y + 7)

Answer

(i) (y² + 7y + 10) ÷ (y + 5)
= (y² + 7y + 10)/(y + 5)
= {y² + ( 2 + 5)y + 2 × 5}/(y +5)
= (y² + 2y + 5y + 2 × 5)/(y + 5)
= {(y + 2)(y + 5)}/(y + 5) [∵ x² + (a+b)x + ab = (x +a)(x+b)]
= y + 2

(ii) (m² - 14m + 32) ÷ (m + 2)
= (m² - 14m + 32)/(m +2)
= { m² + (-16 + 2)m + (-16) × 2}/(m + 2)
= {(m - 16)(m + 2)}/(m +2) [∵ x² + (a+b)x + ab = (x +a)(x+b)]
= (m - 16)

(iii) (5p² - 25p + 20) ÷ (p -1)
= (5p² - 25p + 20)/(p -1)
= (5p² - 20p -5p + 20)/(p -1)
= {5p(p - 4) -5 (p - 4)}/(p -1)
= {(5p - 5)(p - 4)}/(p -1) = {5(p -1)(p -4)}/(p - 1)
= 5 (p - 4)

(iv) 4yz (z² + 6z - 16) ÷ 2y(z + 8)
= {4yz(z² + 6z - 16)}/2y(z + 8)
= [4yz{z² + (8 - 2)z + 8 × (-2)}]/2y(z + 8)
= {4yz(z - 2)(z + 8)}/2y(z + 8) [∵ x² + (a+b)x + ab = (x +a)(x+b)]
= 2z ( z -2)

(v) 5pq(p² - q²) ÷ 2p( p + q)
= {5pq(p² - q²)}/2p(p + q)
= {5pq(p - q)(p + q)}/2p( p + q) [∵ a² - b² = (a - b)(a + b)]
= (5/2)q (p - q)

(vi) 12xy(9x² - 16y²) ÷ 4xy(3x + 4y)
= {12xy (9x² - 16y²)}/4xy(3x + 4y)
= {12xy[(3x)² - (4y)²]}/4xy(3x + 4y)
= {12xy(3x - 4y)(3x + 4y)}/4xy(3x + 4y) [∵ a² - b² = (a - b)(a + b)]
= 3(3x - 4y)

(vii) 39y³(50y² - 98) ÷ 26y²(5y + 7)
= {39y³(50y² - 98)}/26y²(5y + 7)
= {39y³ × 2(25y² - 49)}/26y²(5y + 7)
= {39y² × 2[(5y)² - (7)²]}/26y²(5y + 7)
= {39y² × 2(5y - 7)(5y + 7)}/26y²(5y + 7) [∵ a² - b² = (a - b)(a + b)]
= 3y(5y - 7)

Page No. 228

Exercise 14.4

1. Find and correct the errors in the following mathematical statements:
4(x-5) = 4x-5

Answer

L.H.S. = 4(x-5) = 4x- 20 ≠R.H.S.
Hence, the correct mathematical statement is 4(x-5) = 4x- 20.

2. x(3x+2) = 3x²+ 2

Answer

L.H.S. = x(3x+2) = 3x2+ 2 ≠ R.H.S.
Hence, the correct mathematical statement is x(3x+2) = 3x2+ 2

3. 2x + 3y = 5xy

Answer

L.H.S. = 2x + 3y ≠ R.H.S.
Hence, the correct mathematical statement is 2x+ 3y = 2x+ 3y

4. x+ 2x +3x = 5x

Answer

L.H.S. = x+ 2x + 3x = 6x ≠R.H.S.
Hence, the correct mathematical statement is x+ 2x + 3x = 6x.

5. 5y + 2y+ y-7y = 0

Answer

L.H.S. = 5y + 2y+ y - 7y = 8y-7y = y ≠ R.H.S.
Hence, the correct mathematical statement is 5y+ 2y+y- 7y = 4

6. 3x+2x = 5 x²

Answer

L.H.S. = 3x+ 2x = 5x ≠ R.H.S.
Hence the correct mathematical statement is 3x+ 2x = 5x

7. (2x)²+ 4(2x) + 7 = 2x²+ 8x+ 7

Answer

L.H.S. = (2x)² + 4(2x) + 7 = 4x² + 8x+ 7 ≠ R.H.S.
Hence, the correct mathematical statement is (2x)² + 4(2x) + 7 = 4x² + 8x+ 7

8. (2x)²+ 5x = 4x+ 5x = 9x

Answer

L.H.S. = (2x)² + 5x = 4x²+ 5x ≠ R.H.S.
Hence the correct mathematical statement is (2x)² + 5x = 4x²+ 5x.

9. (3x + 2)²= 3x² + 6x + 4

Answer

L.H.S. = (3x + 2)² = 3x² + 2 × 3x × 2+ (2)²
= 9x² + 12x + 4 ≠ RHS
Hence, the correct mathematical statementsis (3x + 2)² = 9x² + 12X + 4 × 3x

10. Substituting x = -3 in:
(a) x² + 5X + 4 gives (-3)² + 5(-3) + 4 = 9+ 2+4 = 15
(b) x² - 5X + 4 gives (-3)² - 5(-3) + 4 = 9 - 15 + 4 = -2
(c) x² + 5X gives (-3)² + 5(-3) = -9 - 15 = -24

Answer

(a) L.H.S. = x² + 5x + 4
Putting x = -3 in given expression,
= (-3)² + 5(-3) + 4 = 9 - 15 + 4 = -2 R.H.S.
Hence, x² + 5x + 4 gives (-3)² + 5(-3) + 4 = 9 - 15 + 4 = -2

(b) L.H.S. = x² - 5X + 4
Putting x = -3 cin given expression,
= (-3)² - 5(-3) + 4 = 9 + 15 + 4 = 28 ≠ R.H.S.
Hence x² – 5x + 4 gives (-3)² - 5(-3) + 4 = 9 + 15 + 4 = 28

(c) L.H.S. = x² + 5X
Putting x= -3 in given expression,
= (-3)² + 5(-3) = 9 - 15 = -6 ≠ R.H.S.
Hence, x2 + 5X gives (-3)² + 5(-3) = 9 - 15 = -6

11. (y-3)²= y² - 9

Answer

L.H.S. = (y-3)² = y² - 2 × y × 3 +(3)² [ (a-b)² = a² - 2ab + b²]
= y² - 6y + 9 ≠ R.H.S.
Hence, the correct statement is (y-3)² = y² - 6y + 9

12. (z+5)² = z² + 25

Answer

L.H.S. = (z+5)² = z² + 2 × z×5+ (5)²
= z² + 10z +25 [ (a-b)²= a² - 2ab + b²]
Hence, the correct statement is (z+5)² = z² + 10z + 25

13. (2a +3b)(a-b) = 2a²- 3b²

Answer

L.H.S. = (2a + 3b)(a-b) = 2a(a-b) + 3b(a-b)
= 2a² - 2ab + 3ab - 3b²
= 2a² + ab - 3b² ≠ R.H.S.
Hence, the correct statement is (2a +3b)(a-b) = 2a² + ab - 3b²

14. (a + 4) (a + 2) = a²+ 8

Answer

L.H.S. = (a+4)(a+2) =a(a+2) + 4(a+2)
= a² + 2a + 4a + 8 = a² + 6a + 8 ≠ R.H.S.
Hence, the correct statement is (a+4)(a+2) = a²+6a+ 8

15. (a-4)(a-2) = a²- 8

Answer

L.H.S. = (a-4)(a-2) = a(a-2)-4(a-2)
= a² - 2a -4a+8 = a²- 6a + 8 ≠ R.H.S
Hence, the correct statement is (a-4)(a-2) = a²- 6a + 8

16. 3x²/3x²= 0

Answer

L.H.S. = 3x²/3x² =1/1 = 1 ≠ R.H.S.
Hence, the correct statement is 3x²/3x² =1

17. 3x² + 1 / 3x² = 1+ 1 = 2

Answer

L.H.S. = 3x² + 1 / 3x² = 3x²/ 3x² + 1/3x²
= 1 + 1 / 3x² R.H.S.
Hence, the correct statement is 3x² + 1 / 3x² = 1 + 1/3x²

18. 3x/(3x+2) = 1/2

Answer

L.H.S. = 3x/(3x+2) ≠ R.H.S.
Hence, the correct statement is 3x/(3x+2) = 3x/(3x+2)

19. 3/(4x+3) = 1/4x

Answer

L.H.S. = 3/(4x+3) ≠ R.H.S.
Hence, the correct statement is 3/(4x+3) = 3/(4x+3)

20. (4x+5)/4x = 5

Answer

L.H.S. = (4x+5)/4x = 4x/4x + 5/4x = 1 + 5/4x ≠R.H.S.
Hence the correct statement is (4x+ 5)/4x = 1 + 5/4x

21. (7x+5)/5 = 7x

Answer

L.H.S. = (7x+5)/5 = 7x/5 + 5/5 = 7x/5 + 1 ≠ R.H.S.
Hence, the correct statement is (7x+5)/5 = 7x/5 +1

Go Back To NCERT Solutions of Class 8 Maths

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation

Chapter 14 Factorisation NCERT Solutions are accurate and detailed which will increase concentration and you can solve questions of supplementary books easily. Factorisation means write an expression as a product of its factors. When we factorise an expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.

• Some expression can easily be factorised using these identities:

(i) a² + 2ab + b² = (a + b)²

(ii) a² – 2ab + b² = (a – b)²

(iii) a² – b² = (a – b)(a + b)

(iv) x² + (a + b)x + ab = (x + a)( x+ b)

• The number 1 is a factor of every algebraic term also, but it is shown only when needed.

Below are exercisewise Class 8 Maths NCERT Solutions by which you can understand the concepts behind the questions and easily solve them.

Studyrankers experts have prepared these NCERT Solutions with the sole intention of helping students in better manner. These NCERT Solutions for Class 8are updated as per the latest marking scheme released by CBSE.