{"id":121558,"date":"2021-02-15T11:40:35","date_gmt":"2021-02-15T11:40:35","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=121558"},"modified":"2023-09-16T01:23:16","modified_gmt":"2023-09-16T01:23:16","slug":"ncert-solutions-for-8th-class-maths-chapter-14-factorisation","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-8th-class-maths-chapter-14-factorisation\/","title":{"rendered":"NCERT Solutions for 8th Class Maths: Chapter 14-Factorisation"},"content":{"rendered":"\n

NCERT Solutions for 8th Class Maths: Chapter 14-Factorisation<\/strong><\/h2>\n\n\n\n

Page No: 220<\/p>\n\n\n\n

Exercise 14.1<\/h2>\n\n\n\n

1. Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p2<\/sup>q2<\/sup>
(iv) 2x, 3x2<\/sup>, 4 
(v) 6abc, 24ab2<\/sup>, 12a2<\/sup>b 
(vi) 16x3<\/sup>, -4x2<\/sup>, 32x 
(vii) 10 pq, 20qr, 30rp
(viii) 3x2<\/sup>y3<\/sup>, 10x3<\/sup>y2<\/sup>, 6x2<\/sup>y2<\/sup>z <\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) 12x = 2\u00d72\u00d73\u00d7x
36 = 2\u00d72\u00d73\u00d73
Hence, the common factors are 2, 2 and 3 = 2\u00d72\u00d73 = 12<\/p>\n\n\n\n

(ii) 2y = 2\u00d7y
22xy = 2\u00d711\u00d7x\u00d7y
Hence, the common factors are 2 and y = 2\u00d7y = 2y<\/p>\n\n\n\n

(iii) 14pq = 2\u00d77\u00d7p\u00d7q
28p2<\/sup>q2<\/sup> = 2\u00d72\u00d77\u00d7p\u00d7p\u00d7q\u00d7q
Hence, the common factors are 2\u00d77\u00d7p\u00d7q = 14pq<\/p>\n\n\n\n

(iv) 2x = 2\u00d7x\u00d71
3x2<\/sup> = 3\u00d7x\u00d7x\u00d71
4 = 2\u00d72\u00d71
Hence, the common factor is 1.<\/p>\n\n\n\n

(v) 6abc = 2\u00d73\u00d7a\u00d7b\u00d7c
24ab2<\/sup> = 2\u00d72\u00d72\u00d73\u00d7a\u00d7b\u00d7b
12a2<\/sup>b = 2\u00d72\u00d73\u00d7a\u00d7a\u00d7b
Hence, the common factors are 2\u00d73\u00d7a\u00d7b = 6ab<\/p>\n\n\n\n

(vi) 16x3<\/sup> = 2\u00d72\u00d72\u00d7x\u00d7x\u00d7x
-4x2<\/sup> = (-1)\u00d72\u00d72\u00d7x\u00d7x
32x = 2\u00d72\u00d72\u00d72\u00d72\u00d7x
Hence the common factors are 2\u00d72\u00d7x = 4x<\/p>\n\n\n\n

(vii) 10pq = 2\u00d75\u00d7p\u00d7q
20qr = 2\u00d72\u00d75\u00d7q\u00d7r
30rp = 2\u00d73\u00d75\u00d7r\u00d7p
Hence the common factors are 2\u00d75 = 10<\/p>\n\n\n\n

(viii) 3x2<\/sup>y3<\/sup> = 3\u00d7x\u00d7x\u00d7y\u00d7y\u00d7y
10x3<\/sup>y2<\/sup> = 2\u00d75\u00d7x\u00d7x\u00d7x\u00d7y\u00d7y
6x2<\/sup>y2<\/sup>z = 2\u00d73\u00d7x\u00d7x\u00d7y\u00d7y\u00d7z
Hence the common factors are x\u00d7x\u00d7y\u00d7y = x2<\/sup>y2<\/sup><\/p>\n\n\n\n


2. Factorize the following expressions.
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a2<\/sup> + 14a 
(iv) -16z + 20z3<\/sup> 
(v)20l2<\/sup>m + 30alm 
(vi) 5x2<\/sup>y – 15xy2<\/sup> 
(vii) 10a2<\/sup> – 15b2<\/sup> + 20c2<\/sup> 
(viii) -4a2<\/sup> + 4ab – 4ca 
(ix) x2<\/sup>yz + xy2<\/sup>z + xyz2<\/sup> 
(x) ax2<\/sup>y + bxy2<\/sup> + cxyz <\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) 7x – 42 = 7\u00d7x – 2\u00d73\u00d77
Taking common factors from each term,
= 7(x – 2\u00d73)
= 7(x – 6)<\/p>\n\n\n\n

(ii) 6p – 12q = 2\u00d73\u00d7p – 2\u00d72\u00d73\u00d7q
Taking common factors from each term,
= 2\u00d73(p – 2q)
= 6(p – 2q)<\/p>\n\n\n\n

(iii) 7a2<\/sup> + 14a = 7\u00d7a\u00d7a + 2\u00d77\u00d7a
Taking common factors from each term,
= 7\u00d7a(a + 2)
= 7a(a + 2)<\/p>\n\n\n\n

(iv) -16z + 20z3<\/sup>
= (-1)\u00d72\u00d72\u00d72\u00d72\u00d7z + 2\u00d72\u00d75\u00d7z\u00d7z\u00d7 z
Taking common factors from each term,
= 2\u00d72\u00d7z (-2\u00d72 + 5\u00d7z\u00d7z)
= 4z (-4 + 5z2<\/sup>)<\/p>\n\n\n\n

(v) 20l2<\/sup>m + 30alm
= 2\u00d72\u00d75\u00d7l\u00d7l\u00d7m + 2\u00d73\u00d75\u00d7a\u00d7l\u00d7m
Taking common factors from each term,
= 2\u00d75\u00d7l\u00d7m(2\u00d7l + 3\u00d7a)
= 10 lm(2l +3a)<\/p>\n\n\n\n

(vi) 5x2<\/sup>y – 15xy2<\/sup>
= 5\u00d7x\u00d7x\u00d7y – 3\u00d75\u00d7x\u00d7y\u00d7y (Taking common factors from each term)
= 5\u00d7x\u00d7y(x – 3y)
= 5xy(x – 3y)<\/p>\n\n\n\n

(vii) 10a2<\/sup> – 15b2<\/sup> + 20c2<\/sup>
= 2\u00d75\u00d7a\u00d7a – 3\u00d75\u00d7b\u00d7b + 2\u00d72\u00d75\u00d7c\u00d7c
Taking common factors from each term,
= 5(2\u00d7a\u00d7a – 3\u00d7b\u00d7b + 2\u00d72\u00d7c\u00d7c)
= 5(2a2<\/sup> – 3b2<\/sup> + 4c2<\/sup>)<\/p>\n\n\n\n

(viii) -4a2<\/sup> + 4ab – 4ca
= (-1)\u00d72\u00d72\u00d7a\u00d7a + 2\u00d72\u00d7a\u00d7b – 2\u00d72\u00d7c\u00d7a
Taking common factors from each term,
= 2\u00d72\u00d7a(-a + b -c)
= 4a (-a + b – c)<\/p>\n\n\n\n

(ix) x2<\/sup>yz + xy2<\/sup>z + xyz2<\/sup>
= x\u00d7x\u00d7y\u00d7z + x\u00d7y\u00d7y\u00d7z + z\u00d7y\u00d7z\u00d7z
Taking common factors from each term,
= x\u00d7y\u00d7z( x + y + z)
= xyz(x + y +z)<\/p>\n\n\n\n

(x) ax2<\/sup>y + bxy2<\/sup> + cxyz
= a\u00d7x\u00d7x\u00d7y + b\u00d7x\u00d7y\u00d7y + c\u00d7x\u00d7y\u00d7z
Taking common factors from each term,
= x\u00d7y(a\u00d7x + b\u00d7y + c\u00d7z)
= xy(ax + by +cz)<\/p>\n\n\n\n


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

Answer<\/strong><\/p>\n\n\n\n

(i) x2<\/sup> + xy + 8x + 8y
= x(x + y) + 8(x + y)
= (x + y)(x + 8)<\/p>\n\n\n\n

(ii) 15xy – 6x + 5y – 2
= 3x(5y – 2) + 1(5y – 2)
= (5y -2)(3x + 1)<\/p>\n\n\n\n

(iii) ax + bx – ay – by
= (ax + bx) – (ay + by)
= x(a + b) – y(a + b)
= (a + b)(x – y)<\/p>\n\n\n\n

(iv) 15pq + 15 + 9q + 25p
= 15pq + 25p + 9q + 15
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5)(5p + 3)<\/p>\n\n\n\n

(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)<\/p>\n\n\n\n

Page No. 223<\/p>\n\n\n\n

Exercise 14.2<\/h2>\n\n\n\n

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


Answer<\/strong><\/p>\n\n\n\n


(i) a2<\/sup> + 8a + 16 = a2<\/sup> + (4 + 4)a + 4 \u00d7 4
Using identity x2<\/sup> + (a + b)x + ab = (x + a)(x + b),
Here x = a, a = 4 and b = 4
a2<\/sup> + 8a + 16 = (a + 4)(a + 4) = (a + 4)2<\/sup><\/p>\n\n\n\n

(ii) p2<\/sup> – 10p + 25 = p2<\/sup> +(-5-5)p + (-5)(-5)
Using identity x2<\/sup> + (a +b)x + ab = ( x + a)(x + b),
Here x = p, a = -5 and b = -5
p2<\/sup> – 10p + 25 = (p -5)(p- 5) = (p – 5)2<\/sup><\/p>\n\n\n\n

(iii) 25m2<\/sup> + 30m + 9 = (5m)2<\/sup> + 2 \u00d7 5m \u00d7 3 + (3)2<\/sup>
Using identity a2<\/sup> + 2ab + b2<\/sup> = (a + b)2<\/sup> , here a= 5m, b = 3
 25m2<\/sup> + 30m + 9 = (5m + 3)2<\/sup><\/p>\n\n\n\n

(iv) 49y2<\/sup> + 84yz + 36z2<\/sup> = (7y)2<\/sup> + 2 \u00d7 7y \u00d7 6z + (6z)2<\/sup>
Using identity a2<\/sup> + 2ab + b2<\/sup> = (a + b)2<\/sup> , here a = 7y, b = 6z
49y2<\/sup> + 84yz + 36z2<\/sup> = (7y + 6z)2<\/sup><\/p>\n\n\n\n

(v) 4x2<\/sup> – 8x + 4 = (2x)2<\/sup> – 2 \u00d7 2x \u00d72 + (2)2<\/sup>
Using identity a2 – 2ab + b2 = (a – b)2 , here a = 2x, b = 2
4x2<\/sup> – 8x + 4 = (2x – 2)2<\/sup>
= (2)2<\/sup> (x – 1)2<\/sup> = 4( x – 1)2<\/sup><\/p>\n\n\n\n

(vi) 121b2<\/sup> – 88bc + 16c2<\/sup> = (11b)2<\/sup> – 2 \u00d7 11b \u00d7 4c + (4c)2<\/sup>
Using identity a2 – 2ab + b2 = (a – b)2 , here a = 11b, b = 4c
121b2<\/sup> – 88bc + 16c2<\/sup> = (11b – 4c)2<\/sup><\/p>\n\n\n\n

(vii) (l + m)2<\/sup> – 4lm
= l2<\/sup> + 2 \u00d7 l \u00d7m + m2<\/sup> – 4lm [ \u2235 (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]
= l2<\/sup> + 2lm + m2<\/sup> – 4lm
= l2<\/sup> – 2lm + m2<\/sup>
= (l – m)2<\/sup> [ \u2235 (a- b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup> ]<\/p>\n\n\n\n

(viii) a4<\/sup> + 2a2<\/sup>b2<\/sup> + b4<\/sup> = (a2<\/sup>)2<\/sup> + 2 \u00d7 a2<\/sup> \u00d7 b2<\/sup> + (b2<\/sup>)2<\/sup>
= (a2<\/sup> + b2<\/sup>)2<\/sup> [\u2235 (a + b)2 = a2 + 2ab + b ]<\/p>\n\n\n\n

2. Factorize:
(i) 4p2<\/sup> – 9q2<\/sup> (ii) 63a2<\/sup> – 112b2<\/sup> 
(iii) 49x2<\/sup> – 36
(iv) 16x5<\/sup> – 144x2<\/sup> 
(v) (l + m)2<\/sup> – (l -m)2<\/sup> 
(vi) 9x2<\/sup>y2<\/sup> – 16 
(vii) (x2<\/sup> – 2xy + y2<\/sup>) – z2<\/sup> 
(viii) 25a2<\/sup> – 4b2<\/sup> + 28bc – 49c2<\/sup> <\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n


(i) 4p2<\/sup> – 9q2<\/sup> = (2p)2<\/sup> – (3q)2<\/sup>
= (2p -3q)(2p + 3q) [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]<\/p>\n\n\n\n

(ii) 63a2<\/sup> – 112b2<\/sup> = 7(9a2<\/sup> – 16b2<\/sup>)
= 7 [ (3a)2<\/sup> – (4b)2<\/sup>]
= 7(3a – 4b)(3a + 4b)  [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]<\/p>\n\n\n\n

(iii) 49x2<\/sup> – 36 = (7x)2<\/sup> – (6)2<\/sup>
= (7x – 6)(7x + 6)  [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]<\/p>\n\n\n\n

(iv) 16x5<\/sup> – 144x3<\/sup> = 16x3<\/sup>(x2<\/sup> – 9)
= 16x3<\/sup> [(x)2<\/sup> – (3)2<\/sup>]
= 16x3<\/sup> (x – 3)(x + 3) [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]<\/p>\n\n\n\n

(v) (l + m)2<\/sup> – (l – m)2<\/sup>
= [(l + m) + ( l – m)][(l + m)- (l – m)] [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= (l + m + l – m)(l + m – l +m)
= (2l) (2m) = 4lm<\/p>\n\n\n\n

(vi) 9x2<\/sup>y2<\/sup> – 16 = (3xy)2<\/sup> – (4)2<\/sup>
= (3xy – 4)(3xy + 4) [ \u2235 a2 – b2 = (a – b)(a +b)]<\/p>\n\n\n\n

(vii) ( x2<\/sup> – 2xy + y2<\/sup>) – z2<\/sup> = ( x – y)2<\/sup> – z2<\/sup>   [\u2235 (a -b)2<\/sup> = a2<\/sup> -2ab + b2<\/sup>]
 = ( x – y – z)( x – y + z) [ \u2235 a2 – b2 = (a – b)(a +b)]
(viii) 25a2<\/sup> – 4b2<\/sup> + 28bs – 49c2<\/sup>
= 25a2<\/sup> – (4b2<\/sup> – 28bc + 49c2<\/sup>)
= 25a2<\/sup> – [ (2b)2<\/sup> – 2 \u00d7 2b \u00d7 7c + (7c)2<\/sup>]
= 25a2<\/sup> – (2b – 7c)2<\/sup> [ \u2235 (a -b)2 = a2 -2ab + b2]
= (5a)2<\/sup> – (2b – 7c)2<\/sup>
= [5a – (2b – 7c)][5a + (2b – 7c)] [ \u2235 a2 – b2 = (a – b)(a +b)]
= (5a – 2b + 7c)(5a + 2b – 7c)<\/p>\n\n\n\n

3. Factorize the expressions:<\/strong><\/p>\n\n\n\n

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


Answer<\/strong><\/p>\n\n\n\n

(i) ax2<\/sup> + bx = x(ax + b)<\/p>\n\n\n\n

(ii) 7p2<\/sup> + 21q2<\/sup> = 7(p2<\/sup> + 3q2<\/sup>)<\/p>\n\n\n\n

(iii) 2x3<\/sup> + 2xy2<\/sup> + 2xz2<\/sup> = 2x( x2<\/sup> + y2<\/sup> + z2<\/sup>)<\/p>\n\n\n\n

(iv) am2<\/sup> + bm2<\/sup> + bn2<\/sup> + an2<\/sup>
= m2<\/sup>( a + b) + n2<\/sup>(a + b)
= (a + b )(m2<\/sup> + n2<\/sup>)<\/p>\n\n\n\n

(v) (lm + l) + m + 1
= l(m + 1) + 1(m + 1)
= (m + 1)( l + 1)<\/p>\n\n\n\n

(vi) y(y + z) + 9(y + z)
= (y + z)(y + 9)<\/p>\n\n\n\n

(vii) 5y2<\/sup> – 20y – 8z + 2yz
= 5y2<\/sup> – 20y + 2yz – 8z
= 5y(y – 4) + 2z(y – 4)
= (y – 4)(5y + 2z)<\/p>\n\n\n\n

(viii) 10ab + 4a + 5b + 2
= 2a(5b + 2) + 1 (5b + 2)
= (5b + 2)(2a + 1)<\/p>\n\n\n\n

(ix) 6xy – 4y + 6 – 9x
= 6xy – 9x – 4y + 6
= 3x(2y – 3) – 2(2y – 3)
= (2y – 3) (3x – 2)<\/p>\n\n\n\n


4. Factorize:
(i) a4<\/sup> – b4<\/sup>
(ii) p4<\/sup> – 81 
(iii) x4<\/sup> – (y + z)4<\/sup>
(iv)x4<\/sup> – (x -z)4<\/sup> 
(v) a4<\/sup> – 2a2<\/sup>b2<\/sup> + b4<\/sup> <\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) a4<\/sup> – b4<\/sup> = (a2<\/sup>)2<\/sup> – (b2<\/sup>)2<\/sup>
= (a2<\/sup> – b2<\/sup>)( a2<\/sup> + b2<\/sup>) [ \u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= (a – b)(a + b)(a2<\/sup> + b2<\/sup>) [ \u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]<\/p>\n\n\n\n

(ii) p4<\/sup> – 81 = (p2<\/sup>)2<\/sup> – (9)2<\/sup>
= (p2<\/sup> – 9)(p2<\/sup> + 9) [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= (p2<\/sup> – 32<\/sup>)(p2<\/sup> + 9)
= ( p – 3)(p + 3)(p2<\/sup> + 9) [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]<\/p>\n\n\n\n

(iii) x4<\/sup> – (y + z)4<\/sup> = (x2<\/sup>)2<\/sup> – [(y + z)2<\/sup>]2<\/sup>
= [x2<\/sup> – (y + z)2<\/sup>][ x2<\/sup> + (y + z)2<\/sup>] [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= [x -(y +z)][x + (y + z)][x2<\/sup> + (y + z)2<\/sup>] [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= (x – y – z) (x + y + z) [x2<\/sup> + (y + z)2<\/sup>]<\/p>\n\n\n\n

(iv) x4<\/sup> – (x – z)4<\/sup> = (x2<\/sup>)2<\/sup> – [(x – z)2<\/sup>]2<\/sup>
= [x2<\/sup> -(x – z)2<\/sup>][x2<\/sup> + (x – z)2<\/sup>] [ \u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= [x – (x – z)][x + (x – z)] [x2<\/sup> + (x – z)2<\/sup>] [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= [x – x + z] [x + x – z] [x2<\/sup> + x2<\/sup> – 2xz + z2<\/sup>] [\u2235 (a -b)2<\/sup> = a2<\/sup> -2ab + b2<\/sup>]
= z(2x – z) (2x2<\/sup> – 2xz + z2<\/sup>)<\/p>\n\n\n\n

(v) a4<\/sup> – 2a2<\/sup>b2<\/sup> + b4<\/sup> = (a2<\/sup>)2<\/sup> – 2a2<\/sup>b2<\/sup> + (b2<\/sup>)2<\/sup>
= (a2<\/sup> – b2<\/sup>)2<\/sup> [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= [(a – b)(a + b)]2<\/sup> [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a +b)]
= (a -b)2<\/sup> (a + b)2<\/sup> [ (xy)m<\/sup> = xm<\/sup>ym<\/sup>]<\/p>\n\n\n\n


5. Factorize the following expressions:
(i) p2<\/sup> + 6p + 8
(ii) q2<\/sup> – 10q + 21 
(iii) p2<\/sup> + 6p – 16 <\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) p2<\/sup> + 6p + 8 = p2<\/sup> + ( 4 + 2)p + 4 \u00d7 2
= p2<\/sup> + 4p + 2p + 4 \u00d72
= p(p + 4) + 2 ( p + 4)
= (p + 4)(p + 2)<\/p>\n\n\n\n

(ii) q2<\/sup> – 10q + 21 = q2<\/sup> – ( 7 + 3)q + 7 \u00d7 3
= q2<\/sup> – 7q – 3q + 7 \u00d7 3
= q(q – 7) – 3(q – 7)
= (q – 7)( q – 3)<\/p>\n\n\n\n

(iii) p2<\/sup> + 6p – 16
= p2<\/sup> + (8 – 2)p – 8\u00d72
= p2<\/sup> + 8p – 2p – 8\u00d72
= p(p + 8) – 2(p + 8)
= ( p + 8)(p -2)<\/p>\n\n\n\n

Page No. 227<\/p>\n\n\n\n

Exercise 14.3<\/h2>\n\n\n\n

1. Carry out the following divisions:
(i) 2x4<\/sup> \u00f7 56x
(ii) -36y3<\/sup> \u00f7 9y2<\/sup>
(iii) 66pq2<\/sup>r3<\/sup> \u00f7 11 qr2<\/sup>
(iv) 34x3<\/sup>y3<\/sup>x3<\/sup> \u00f7 51xy2<\/sup>z3<\/sup>
(v) 12a8<\/sup>b8<\/sup> \u00f7 (-6a6<\/sup>b4<\/sup>)<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n


 (i) 2x4<\/sup> \u00f7 56x
= 28x4<\/sup>\/56x
= 28\/56 \u00d7 x4<\/sup>\/x
= 1\/2 x3<\/sup> [xm<\/sup> \u00f7 xn<\/sup> = xm-n<\/sup>]<\/p>\n\n\n\n

(ii) -36y3<\/sup> \u00f7 9y2<\/sup> = -36y3<\/sup>\/9y2<\/sup>
= -36\/9 \u00d7 y3<\/sup>\/y2<\/sup>
= -4y [xm<\/sup> \u00f7 xn<\/sup> = xm-n<\/sup>]<\/p>\n\n\n\n

(iii) 66pq2<\/sup>r3<\/sup> \u00f7 11qr2<\/sup>
= 66pq2<\/sup>r3<\/sup>\/11qr2<\/sup>
= 66\/11 \u00d7 pq2<\/sup>r3<\/sup>\/qr2<\/sup>
= 6pqr [xm<\/sup> \u00f7 xn<\/sup> = xm-n<\/sup>]<\/p>\n\n\n\n

(iv) 34x3<\/sup>y3<\/sup>z3<\/sup> \u00f7 51xy2<\/sup>z3<\/sup>
= 34x3<\/sup>y3<\/sup>z3<\/sup>\/51xy2<\/sup>z3<\/sup>
= 34\/51 \u00d7x3<\/sup>y3<\/sup>z3<\/sup>\/xy2<\/sup>z3<\/sup>
= 2\/3x2<\/sup>y [xm<\/sup> \u00f7 xn<\/sup> = xm-n<\/sup>]<\/p>\n\n\n\n

(v) 12a8<\/sup>b8<\/sup> \u00f7 (- 6a6<\/sup>b4<\/sup>)
= 12a8<\/sup>b8<\/sup>\/- 6a6<\/sup>b4<\/sup>
= 12\/-6 \u00d7 a8<\/sup>b8<\/sup>\/a6<\/sup>b4<\/sup>
= -2a2<\/sup>b4<\/sup> [xm<\/sup> \u00f7 xn<\/sup> = xm-n<\/sup>]<\/p>\n\n\n\n


2. Divide the given polynomial by the given monomial:
(i) (5x2<\/sup> – 6x) \u00f7 3x
(ii) (3y8<\/sup> – 4y6<\/sup> + 5y4<\/sup>) \u00f7 y4<\/sup>
(iii) 8(x3<\/sup>y2<\/sup>z2<\/sup> + x2<\/sup>y3<\/sup>z2<\/sup> + x2<\/sup>y2<\/sup>z3<\/sup>) \u00f7 4x2<\/sup>y2<\/sup>z2<\/sup>
(iv) (x3<\/sup> + 2x2<\/sup> + 3x) \u00f72x
(v) (p3<\/sup>q6<\/sup> – p6<\/sup>q3<\/sup>) \u00f7 p3<\/sup>q3<\/sup><\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) (5x2<\/sup> – 6x) \u00f73x
= (5x2<\/sup> – 6x)\/3x
= 5x2<\/sup>\/3x – 6x\/3x = (5\/3)x – 2 = 1\/3 (5x – 6)<\/p>\n\n\n\n

(ii) (3y8<\/sup> – 4y6<\/sup> + 5y4<\/sup>) \u00f7 y4<\/sup>
= (3y8<\/sup> – 4y6<\/sup> + 5y4<\/sup>)\/ y4<\/sup>
= 3y8<\/sup>\/y4<\/sup> – 4y6<\/sup>\/y4<\/sup> + 5y4<\/sup>\/y4<\/sup> = 3y4<\/sup> – 4y2<\/sup> + 5<\/p>\n\n\n\n

(iii) 8(x3<\/sup>y2<\/sup>z2<\/sup> + x2<\/sup>y3<\/sup>z2<\/sup> + x2<\/sup>y2<\/sup>z3<\/sup>) \u00f7 4x2<\/sup>y2<\/sup>z2<\/sup>
= {8(x3<\/sup>y2<\/sup>z2<\/sup> + x2<\/sup>y3<\/sup>z2<\/sup> + x2<\/sup>y2<\/sup>z3<\/sup>)}\/4 x2<\/sup>y2<\/sup>z2<\/sup>
= 8 x3<\/sup>y2<\/sup>z2<\/sup>\/4 x2<\/sup>y2<\/sup>z2<\/sup> + 8 x2<\/sup>y3<\/sup>z2<\/sup>\/4x2<\/sup>y2<\/sup>z2<\/sup> + 8 x2<\/sup>y2<\/sup>z3<\/sup>\/4x2<\/sup>y2<\/sup>z2<\/sup>
= 2x + 2y + 2z
= 2(x + y + z)<\/p>\n\n\n\n

(iv) (x3<\/sup> + 2x2<\/sup> + 3x) \u00f7 2x
= (x3<\/sup> + 2x2<\/sup> + 3x)\/2x<\/p>\n\n\n\n

= x3<\/sup>\/2x + 2x2<\/sup>\/2x + 3x\/2x = x2<\/sup>\/2 + 2x\/2 + 3\/2
= 1\/2( x2<\/sup> + 2x + 3)
(v) (p3<\/sup>q6<\/sup> – p6<\/sup>q3<\/sup>) \u00f7 p3<\/sup>q3<\/sup>
= (p3<\/sup>q6<\/sup> – p6<\/sup>q3<\/sup>)\/p3<\/sup>q3<\/sup>
= p3<\/sup>q6<\/sup>\/p3<\/sup>q3<\/sup> – p6<\/sup>q3<\/sup>\/p3<\/sup>q3<\/sup> = q3<\/sup> – p3<\/sup><\/p>\n\n\n\n


3. Work out the following divisions:
(i) (10x – 25) \u00f7 5
(ii) (10x – 25) \u00f7 (2x – 5)
(iii) 10y (6y + 21) \u00f7 5(2y + 7)
(iv) 9x2<\/sup>y2<\/sup>(3z – 24) \u00f7 27xy(z – 8)
(v) 96abc(3a – 12)(5b – 30) \u00f7 144(a -4)(b – 6)<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) (10x – 25) \u00f7 5
= (10x – 25)\/5
= {5(2x – 5)}\/5
= 2x -5<\/p>\n\n\n\n

(ii) (10x – 25) \u00f7 (2x – 5)
= (10x – 25)\/(2x – 5)
= {5(2x – 5)\/(2x – 5)
= 5<\/p>\n\n\n\n

(iii) 10y(6y + 21) \u00f7 5(2y + 7)
= {10y(6y + 21)}\/5(2y + 7)
= {2\u00d75\u00d7y\u00d7 3(2y + 7)}\/5(2y + 7)
= 2\u00d7y\u00d73
= 6y<\/p>\n\n\n\n

(iv) 9x2<\/sup>y2<\/sup>(3z – 24) \u00f7 27xy(z – 8)
= {9x2<\/sup>y2<\/sup>(3z – 24)}\/27xy(z – 8)
= 9\/27 \u00d7 {xy \u00d7 xy \u00d7 3(z – 8)}\/xy(z – 8)
= xy<\/p>\n\n\n\n

(v) 96abc(3a – 12)(5b – 30) \u00f7 144(a- 4)(b – 6)
= {96abc(3a – 12)(5b – 30)}\/144(a – 4)(b – 6)
= {12\u00d74\u00d72\u00d7abc\u00d7 3(a-4) \u00d7 5(b-6)}\/{12\u00d74\u00d73 (a – 4)(b – 6)
= 10abc<\/p>\n\n\n\n


4. Divide as directed:
(i) 5(2x + 1)(3x + 5) \u00f7 (2x + 1)
(ii) 26xy(x + 5)(y – 4) \u00f7 13x(y – 4)
(iii) 52pqr(p + q)(q + r)(r + p) \u00f7 104pq(q + r)(r + p)
(iv) 20(y + 4)(y2<\/sup> + 5y + 3) \u00f7 5(y + 4)
(v) x(x + 1)(x + 2)(x + 3) \u00f7 x(x + 1)<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) 5(2x + 1)(3x + 5) \u00f7 (2x + 1)
= {5(2x + 1)(3x +5)}\/(2x + 1)
= 5(3x + 5)<\/p>\n\n\n\n

(ii) 26xy( x + 5)(y – 4) \u00f7 13x(y – 4)
26xy( x + 5)(y -4) \u00f7 13x(y – 4)
= {26xy(x + 5)(y – 4)}\/13x(y – 4)
= {13\u00d72\u00d7xy(x + 5)(y – 4)}\/13x(y – 4)
= 2y(x + 5)<\/p>\n\n\n\n

(iii) 52pqr( p + q)(q + r)( r + p) \u00f7 104pq(q + r)(r + p)
= {52pqr(p + q)(q + r)( r + p)}\/{52 \u00d7 2 \u00d7 pq(q + r)(r + p)}
= (1\/2)r (p + q)<\/p>\n\n\n\n

(iv) 20( y + 4)(y2<\/sup> + 5y + 3) \u00f7 5(y + 4)
= {20(y + 4)(y2<\/sup> + 5y + 3)}\/5(y + 4)
= 4(y2<\/sup> + 5y + 3)<\/p>\n\n\n\n

(v) x( x + 1)(x + 2)(x + 3) \u00f7 x(x + 1)
= {x(x + 1)(x + 2)(x + 3)}\/x(x + 1)
= (x + 2)(x + 3)<\/p>\n\n\n\n


5. Factorize the expressions and divide them as directed:
(i) (y2<\/sup> + 7y + 10) \u00f7 (y + 5)
(ii) (m2<\/sup> – 14m – 32) \u00f7 (m + 2)
(iii) (5p2<\/sup> – 25p + 20) \u00f7 (p – 1)
(iv) 4yz(z2<\/sup> + 6z – 16) \u00f7 2y( z + 8)
(v) 5pq(p2<\/sup> – q2<\/sup>) \u00f7 2p(p + q)
(vi) 12xy(9x2<\/sup> – 16y2<\/sup>) \u00f7 4xy(3x + 4y)
(vii) 39y3<\/sup>(50y2<\/sup> – 98) \u00f7 26y2<\/sup>(5y + 7)<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(i) (y2<\/sup> + 7y + 10) \u00f7 (y + 5)
= (y2<\/sup> + 7y + 10)\/(y + 5)
= {y2<\/sup> + ( 2 + 5)y + 2 \u00d7 5}\/(y +5)
= (y2<\/sup> + 2y + 5y + 2 \u00d7 5)\/(y + 5)
= {(y + 2)(y + 5)}\/(y + 5) [\u2235 x2<\/sup> + (a+b)x + ab = (x +a)(x+b)]
= y + 2<\/p>\n\n\n\n

(ii) (m2<\/sup> – 14m + 32) \u00f7 (m + 2)
= (m2<\/sup> – 14m + 32)\/(m +2)
= { m2<\/sup> + (-16 + 2)m + (-16) \u00d7 2}\/(m + 2)
= {(m – 16)(m + 2)}\/(m +2) [\u2235 x2<\/sup> + (a+b)x + ab = (x +a)(x+b)]
= (m – 16)<\/p>\n\n\n\n

(iii) (5p2<\/sup> – 25p + 20) \u00f7 (p -1)
= (5p2<\/sup> – 25p + 20)\/(p -1)
= (5p2<\/sup> – 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)<\/p>\n\n\n\n

(iv) 4yz (z2<\/sup> + 6z – 16) \u00f7 2y(z + 8)
= {4yz(z2<\/sup> + 6z – 16)}\/2y(z + 8)
= [4yz{z2<\/sup> + (8 – 2)z + 8 \u00d7 (-2)}]\/2y(z + 8)
= {4yz(z – 2)(z + 8)}\/2y(z + 8) [\u2235 x2<\/sup> + (a+b)x + ab = (x +a)(x+b)]
= 2z ( z -2)<\/p>\n\n\n\n

(v) 5pq(p2<\/sup> – q2<\/sup>) \u00f7 2p( p + q)
= {5pq(p2<\/sup> – q2<\/sup>)}\/2p(p + q)
= {5pq(p – q)(p + q)}\/2p( p + q) [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a + b)]
= (5\/2)q (p – q)<\/p>\n\n\n\n

(vi) 12xy(9x2<\/sup> – 16y2<\/sup>) \u00f7 4xy(3x + 4y)
= {12xy (9x2<\/sup> – 16y2<\/sup>)}\/4xy(3x + 4y)
= {12xy[(3x)2<\/sup> – (4y)2<\/sup>]}\/4xy(3x + 4y)
= {12xy(3x – 4y)(3x + 4y)}\/4xy(3x + 4y) [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a + b)]
= 3(3x – 4y)<\/p>\n\n\n\n

(vii) 39y3<\/sup>(50y2<\/sup> – 98) \u00f7 26y2<\/sup>(5y + 7)
= {39y3<\/sup>(50y2<\/sup> – 98)}\/26y2<\/sup>(5y + 7)
= {39y3<\/sup> \u00d7 2(25y2<\/sup> – 49)}\/26y2<\/sup>(5y + 7)
= {39y2<\/sup> \u00d7 2[(5y)2<\/sup> – (7)2<\/sup>]}\/26y2<\/sup>(5y + 7)
= {39y2<\/sup> \u00d7 2(5y – 7)(5y + 7)}\/26y2<\/sup>(5y + 7) [\u2235 a2<\/sup> – b2<\/sup> = (a – b)(a + b)]
= 3y(5y – 7)<\/p>\n\n\n\n

Page No. 228<\/p>\n\n\n\n

Exercise 14.4<\/h2>\n\n\n\n

1. Find and correct the errors in the following mathematical statements:
4(x-5) = 4x-5<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = 4(x-5) = 4x- 20 \u2260R.H.S.
Hence, the correct mathematical statement is 4(x-5) = 4x- 20.<\/p>\n\n\n\n


2. x(3x+2) = 3x2<\/sup>+ 2<\/strong>
<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

L.H.S. = x(3x+2) = 3×2+ 2 \u2260 R.H.S.
Hence, the correct mathematical statement is x(3x+2) = 3×2+ 2<\/p>\n\n\n\n


3. 2x + 3y = 5xy<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = 2x + 3y \u2260 R.H.S.
Hence, the correct mathematical statement is 2x+ 3y = 2x+ 3y<\/p>\n\n\n\n


4. x+ 2x +3x = 5x<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = x+ 2x + 3x = 6x \u2260R.H.S.
Hence, the correct mathematical statement is x+ 2x + 3x = 6x.<\/p>\n\n\n\n


5. 5y + 2y+ y-7y = 0<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = 5y + 2y+ y – 7y = 8y-7y = y \u2260 R.H.S.
Hence, the correct mathematical statement is 5y+ 2y+y- 7y = 4<\/p>\n\n\n\n


6. 3x+2x = 5 x2<\/sup><\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

L.H.S. = 3x+ 2x = 5x \u2260 R.H.S.
Hence the correct mathematical statement is 3x+ 2x = 5x<\/p>\n\n\n\n


7. (2x)2<\/sup>+ 4(2x) + 7 = 2x2<\/sup>+ 8x+ 7<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (2x)2<\/sup> + 4(2x) + 7 = 4x2<\/sup> + 8x+ 7 \u2260 R.H.S.
Hence, the correct mathematical statement is (2x)2<\/sup> + 4(2x) + 7 = 4x2<\/sup> + 8x+ 7<\/p>\n\n\n\n


8. (2x)2<\/sup>+ 5x = 4x+ 5x = 9x<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (2x)2<\/sup> + 5x = 4x2<\/sup>+ 5x \u2260 R.H.S.
Hence the correct mathematical statement is (2x)2<\/sup> + 5x = 4x2<\/sup>+ 5x.<\/p>\n\n\n\n


9. (3x + 2)2<\/sup>= 3x2<\/sup> + 6x + 4<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (3x + 2)2<\/sup> = 3x2<\/sup> + 2 \u00d7 3x \u00d7 2+ (2)2<\/sup>
= 9x2<\/sup> + 12x + 4 \u2260 RHS
Hence, the correct mathematical statementsis (3x + 2)2<\/sup> = 9x2<\/sup> + 12X + 4 \u00d7 3x<\/p>\n\n\n\n


10. Substituting x = -3 in:
(a) x2<\/sup> + 5X + 4 gives (-3)2<\/sup> + 5(-3) + 4 = 9+ 2+4 = 15
(b) x2<\/sup> – 5X + 4 gives (-3)2<\/sup> – 5(-3) + 4 = 9 – 15 + 4 = -2
(c) x2<\/sup> + 5X gives (-3)2<\/sup> + 5(-3) = -9 – 15 = -24<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

(a) L.H.S. = x2<\/sup> + 5x + 4
Putting x = -3 in given expression,
 = (-3)2<\/sup> + 5(-3) + 4 = 9 – 15 + 4 = -2 R.H.S.
Hence, x2<\/sup> + 5x + 4 gives (-3)2<\/sup> + 5(-3) + 4 = 9 – 15 + 4 = -2<\/p>\n\n\n\n

(b) L.H.S. = x2<\/sup> – 5X + 4
Putting x = -3 cin given expression,
 = (-3)2<\/sup> – 5(-3) + 4 = 9 + 15 + 4 = 28 \u2260 R.H.S.
Hence x2<\/sup> \u2013 5x + 4 gives (-3)2<\/sup> – 5(-3) + 4 = 9 + 15 + 4 = 28<\/p>\n\n\n\n

(c) L.H.S. = x2<\/sup> + 5X
Putting x= -3 in given expression,
 = (-3)2<\/sup> + 5(-3) = 9 – 15 = -6 \u2260 R.H.S.
Hence, x2 + 5X gives (-3)2<\/sup> + 5(-3) = 9 – 15 = -6<\/p>\n\n\n\n


11. (y-3)2<\/sup>= y2<\/sup> – 9 <\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (y-3)2<\/sup> = y2<\/sup> – 2 \u00d7 y \u00d7 3 +(3)2 <\/sup> [ (a-b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
= y2<\/sup> – 6y + 9 \u2260 R.H.S.
Hence, the correct statement is (y-3)2<\/sup> = y2<\/sup> – 6y + 9<\/p>\n\n\n\n


12. (z+5)2<\/sup> = z2<\/sup> + 25<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (z+5)2<\/sup> = z2<\/sup> + 2 \u00d7 z\u00d75+ (5)2<\/sup>
= z2<\/sup> + 10z +25 [ (a-b)2 <\/sup>= a2<\/sup> – 2ab + b2<\/sup>]
Hence, the correct statement is (z+5)2<\/sup> = z2<\/sup> + 10z + 25<\/p>\n\n\n\n


13. (2a +3b)(a-b) = 2a2<\/sup>– 3b2<\/sup><\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (2a + 3b)(a-b) = 2a(a-b) + 3b(a-b)
= 2a2<\/sup> – 2ab + 3ab – 3b2<\/sup>
= 2a2<\/sup> + ab – 3b2<\/sup> \u2260 R.H.S.
Hence, the correct statement is (2a +3b)(a-b) = 2a2<\/sup> + ab – 3b2<\/sup><\/p>\n\n\n\n


14. (a + 4) (a + 2) = a2<\/sup>+ 8<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (a+4)(a+2) =a(a+2) + 4(a+2)
= a2<\/sup> + 2a + 4a + 8 = a2<\/sup> + 6a + 8 \u2260 R.H.S.
Hence, the correct statement is (a+4)(a+2) = a2<\/sup>+6a+ 8<\/p>\n\n\n\n


15. (a-4)(a-2) = a2<\/sup>– 8<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (a-4)(a-2) = a(a-2)-4(a-2)
 = a2<\/sup> – 2a -4a+8 = a2<\/sup>– 6a + 8 \u2260 R.H.S
Hence, the correct statement is (a-4)(a-2) = a2<\/sup>– 6a + 8
<\/p>\n\n\n\n

16. 3x2<\/sup>\/3x2<\/sup>= 0 
<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

L.H.S. = 3x2<\/sup>\/3x2<\/sup> =1\/1 = 1 \u2260 R.H.S.
Hence, the correct statement is 3x2<\/sup>\/3x2<\/sup> =1
<\/p>\n\n\n\n

17. 3x2<\/sup> + 1 \/ 3x2<\/sup> = 1+ 1 = 2
<\/strong><\/p>\n\n\n\n

Answer<\/strong>
L.H.S. = 3x2<\/sup> + 1 \/ 3x2<\/sup> = 3x2<\/sup>\/ 3x2<\/sup> + 1\/3x2<\/sup>
= 1 + 1 \/ 3x2<\/sup> R.H.S.
Hence, the correct statement is 3x2<\/sup> + 1 \/ 3x2<\/sup> = 1 + 1\/3x2 <\/sup><\/p>\n\n\n\n


18. 3x\/(3x+2) = 1\/2<\/strong><\/p>\n\n\n\n


Answer<\/strong><\/p>\n\n\n\n

L.H.S. = 3x\/(3x+2) \u2260 R.H.S.
Hence, the correct statement is 3x\/(3x+2) = 3x\/(3x+2)<\/p>\n\n\n\n


19. 3\/(4x+3) = 1\/4x<\/strong><\/p>\n\n\n\n


Answer<\/strong>
L.H.S. = 3\/(4x+3) \u2260 R.H.S.
Hence, the correct statement is 3\/(4x+3) = 3\/(4x+3)
<\/p>\n\n\n\n

20. (4x+5)\/4x = 5<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (4x+5)\/4x = 4x\/4x + 5\/4x = 1 + 5\/4x \u2260R.H.S.
Hence the correct statement is (4x+ 5)\/4x = 1 + 5\/4x
<\/p>\n\n\n\n

21. (7x+5)\/5 = 7x<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

L.H.S. = (7x+5)\/5 = 7x\/5 + 5\/5 = 7x\/5 + 1 \u2260 R.H.S.
Hence, the correct statement is (7x+5)\/5 = 7x\/5 +1<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

\u2022 Some expression can easily be factorised using these identities:<\/p>\n\n\n\n

(i)  a2<\/sup> + 2ab + b2<\/sup> = (a + b)2<\/sup><\/p>\n\n\n\n

(ii) a2<\/sup> \u2013 2ab + b2<\/sup> = (a \u2013 b)2<\/sup><\/p>\n\n\n\n

(iii) a2<\/sup> \u2013 b2<\/sup> = (a \u2013 b)(a + b)<\/p>\n\n\n\n

(iv) x2<\/sup> + (a + b)x + ab = (x + a)( x+ b)<\/p>\n\n\n\n

\u2022 The number 1 is a factor of every algebraic term also, but it is shown only when needed.<\/p>\n\n\n\n

Below are exercisewise Class 8 Maths NCERT Solutions by which you can understand the concepts behind the questions and easily solve them.<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

NCERT Solutions for Class 8 Maths Chapters:<\/h3>\n\n\n\n

FAQ on Chapter <\/strong>14 Factorisation<\/strong><\/p>\n\n\n\n

Factorise 9x + 18y + 6xy + 27.<\/h4>\n\n\n\n

Here, we have a common factor 3 in all the terms.<\/p>\n\n\n\n

\u2234 9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy + 9]<\/p>\n\n\n\n

We find that 3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy + 9)<\/p>\n\n\n\n

i.e. a common factor in both the groups does not eist,<\/p>\n\n\n\n

Thus, 3x + 6y + 2xy + 9 cannot be factorised.<\/p>\n\n\n\n

On regrouping the terms, we have<\/p>\n\n\n\n

3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y<\/p>\n\n\n\n

= 3(x + 3) + 2y(x + 3)<\/p>\n\n\n\n

= (x + 3)(3 + 2y)<\/p>\n\n\n\n

Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]<\/p>\n\n\n\n

Thus, 9x + 18y + 6xy + 27 = 3(x + 3)(2y + 3)<\/p>\n\n\n\n

Write 10y as irreducible factor form.<\/h4>\n\n\n\n

We have  <\/p>\n\n\n\n

10 = 2 \u00d7 5<\/p>\n\n\n\n

xy = x \u00d7 y<\/p>\n\n\n\n

\u2234  10xy = 2 \u00d7 5 \u00d7 x \u00d7 y.<\/p>\n\n\n\n

Factorise: x \u2013 9 + 9zy \u2013 xyz.<\/h4>\n\n\n\n

By regrouping, we have<\/p>\n\n\n\n

x \u2013 9 + 9zy \u2013 xyz = x \u2013 9 \u2013 xyz + 9zy<\/p>\n\n\n\n

= 1(x \u2013 9) \u2013 yz(x \u2013 9)<\/p>\n\n\n\n

= (x \u2013 9)(1 \u2013 yz)<\/p>\n\n\n\n

= (x \u2013 9)(1 \u2013 yz).<\/p>\n\n\n\n

Factorise: 54x2<\/sup> \u2013 96y2<\/sup><\/h4>\n\n\n\n

We have 54x2<\/sup> \u2013 96y2<\/sup> = 6[9x2<\/sup> \u2013 16y2<\/sup>]<\/p>\n\n\n\n

= 6[(3x)2<\/sup> \u2013 (4y)2<\/sup>]<\/p>\n\n\n\n

= 6[(3x + 4y)(3x \u2013 4y)] [Using a2<\/sup> \u2013 b2<\/sup> = (a + b)(a \u2013 b)]<\/p>\n\n\n\n

Thus, 54x2<\/sup> \u2013 96y2<\/sup> = 6 (3x + 4y)(3x \u20134y).<\/p>\n\n\n\n

Chapterwise NCERT Solutions for 8th Class Maths<\/h2>\n\n\n\n

Chapter 1 \u2013 Rational Numbers<\/a>
Chapter 2 \u2013 Linear Equations in One Variable<\/a>
Chapter 3 \u2013 Understanding Quadrilaterals<\/a>
Chapter 4 \u2013 Practical Geometry<\/a>
Chapter 5 \u2013 Data Handling<\/a>
Chapter 6 \u2013 Squares and Square Roots<\/a>
Chapter 7 \u2013 Cubes and Cube Roots<\/a>
Chapter 8 \u2013 Comparing Quantities<\/a>
Chapter 9 \u2013 Algebraic Expressions and Identities<\/a>
Chapter 10 \u2013 Visualizing Solid Shapes<\/a>
Chapter 11 \u2013 Mensuration<\/a>
Chapter 12 \u2013 Exponents and Powers<\/a>
Chapter 13 \u2013 Direct and Inverse Proportions<\/a>
Chapter 14 \u2013 Factorization<\/a>
Chapter 15 \u2013 Introduction to Graphs<\/a>
Chapter 16 \u2013 Playing with Numbers<\/a><\/p>\n\n\n\n

About<\/h2>\n\n\n\n

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n

\n
NCERT Solutions<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 8<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 8 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

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, 3×2, 4 (v) 6abc, 24ab2, 12a2b (vi) 16×3, -4×2, 32x (vii) 10 pq, 20qr, 30rp(viii) 3x2y3, 10x3y2, 6x2y2z  Answer (i) 12x = 2\u00d72\u00d73\u00d7x36 = 2\u00d72\u00d73\u00d73Hence, the common factors are […]<\/p>\n","protected":false},"author":302,"featured_media":627566,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,58],"tags":[1506],"boards":[1180],"class_list":["post-121558","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-8","tag-ncert-maths-class-8","boards-ncert","entry"],"yoast_head":"\nNCERT Solutions for 8th Class Maths: Chapter 14-Factorisation - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 8th Class Maths: Chapter 14-Factorisation Page No: 220 Exercise 14.1 1. 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