Contents
NCERT Solutions for 8th Class 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 = xmym]
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 [xm ÷ xn = xm-n]
(ii) -36y3 ÷ 9y2 = -36y3/9y2
= -36/9 × y3/y2
= -4y [xm ÷ xn = xm-n]
(iii) 66pq2r3 ÷ 11qr2
= 66pq2r3/11qr2
= 66/11 × pq2r3/qr2
= 6pqr [xm ÷ xn = xm-n]
(iv) 34x3y3z3 ÷ 51xy2z3
= 34x3y3z3/51xy2z3
= 34/51 ×x3y3z3/xy2z3
= 2/3x2y [xm ÷ xn = xm-n]
(v) 12a8b8 ÷ (- 6a6b4)
= 12a8b8/- 6a6b4
= 12/-6 × a8b8/a6b4
= -2a2b4 [xm ÷ xn = xm-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) = 3×2+ 2 ≠ R.H.S.
Hence, the correct mathematical statement is x(3x+2) = 3×2+ 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
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.
Indcareer Schools 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.
NCERT Solutions for Class 8 Maths Chapters:
FAQ on Chapter 14 Factorisation
Factorise 9x + 18y + 6xy + 27.
Here, we have a common factor 3 in all the terms.
∴ 9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy + 9]
We find that 3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy + 9)
i.e. a common factor in both the groups does not eist,
Thus, 3x + 6y + 2xy + 9 cannot be factorised.
On regrouping the terms, we have
3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y
= 3(x + 3) + 2y(x + 3)
= (x + 3)(3 + 2y)
Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]
Thus, 9x + 18y + 6xy + 27 = 3(x + 3)(2y + 3)
Write 10y as irreducible factor form.
We have
10 = 2 × 5
xy = x × y
∴ 10xy = 2 × 5 × x × y.
Factorise: x – 9 + 9zy – xyz.
By regrouping, we have
x – 9 + 9zy – xyz = x – 9 – xyz + 9zy
= 1(x – 9) – yz(x – 9)
= (x – 9)(1 – yz)
= (x – 9)(1 – yz).
Factorise: 54x2 – 96y2
We have 54x2 – 96y2 = 6[9x2 – 16y2]
= 6[(3x)2 – (4y)2]
= 6[(3x + 4y)(3x – 4y)] [Using a2 – b2 = (a + b)(a – b)]
Thus, 54x2 – 96y2 = 6 (3x + 4y)(3x –4y).
Chapterwise NCERT Solutions for 8th Class Maths
Chapter 1 – Rational Numbers
Chapter 2 – Linear Equations in One Variable
Chapter 3 – Understanding Quadrilaterals
Chapter 4 – Practical Geometry
Chapter 5 – Data Handling
Chapter 6 – Squares and Square Roots
Chapter 7 – Cubes and Cube Roots
Chapter 8 – Comparing Quantities
Chapter 9 – Algebraic Expressions and Identities
Chapter 10 – Visualizing Solid Shapes
Chapter 11 – Mensuration
Chapter 12 – Exponents and Powers
Chapter 13 – Direct and Inverse Proportions
Chapter 14 – Factorization
Chapter 15 – Introduction to Graphs
Chapter 16 – Playing with Numbers
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