NCERT Solutions for Class 8 Maths Chapter 14 Factorization PDF

 Chapter – 14: Factorization Exercise 14.1 Q1: Find the common factors of the terms (i) 12 x, 36 (ii) 2 y, 22 xy (iii) 14 pq, 28 p 2 q 2 (iv) 2 x, 3 x 2 , 4 (iv) 6abc, 24ab 2 , 12a 2b (vi) 16 x 3 , − 4 x...

 Chapter – 14: Factorization Exercise 14.1 Q1: Find the common factors of the terms (i) 12 x, 36 (ii) 2 y, 22 xy (iii) 14 pq, 28 p 2 q 2 (iv) 2 x, 3 x 2 , 4 (iv) 6abc, 24ab 2 , 12a 2b (vi) 16 x 3 , − 4 x 2 , 32 x (vii) 10 pq, 20qr , 30rp (viii) 3 x 2 y 3 , 10 x 3 y 2 , 6 x 2 y 2 z Difficulty level: Easy What is known: Terms. What is unknown: Common factors of given terms. Reasoning: First, we will find factors of each terms then find out which factors are common in each term. Solution: (i) 12 x = 2  2  3  x 36 = 2  2  3  3 The common factors are 2, 2, 3. And, 2  2  3 = 12 ( ii ) 2y = 2 y 22 xy = 2 11 x  y The common factors are 2, y. And, 2  y = 2 y (iii) 14 pq = 2× 7× p× q 28p 2 q 2 = 2× 2× 7× p× p× q× q The common factors are 2, 7, p, q. And, 2  7  p  q = 14 pq ( iv ) 2x = 2× x 3x 2 = 3× x× x 4 = 2× 2 The common factor is 1. (v) 6abc = 2  3  a  b  c 24ab 2 = 2  2  2  3  a  b  b 12a 2b = 2  2  3  a  a  b The common factors are 2, 3, a, b. And, 2  3  a  b = 6ab (vi) 16 x3 = 2  2  2  2  x  x  x − 4 x 2 = −1 2  2  x  x 32 x = 2  2  2  2  2  x The common factors are 2, 2, x. And, 2  2  x = 4 x (vii) 10 pq = 2  5  p  q 20 qr = 2  2  5  q  r 30 rp = 2  3  5  r  p The common factors are 2, 5. And, 2  5 = 10 (viii) 3x 2 y 3 = 3  x  x  y  y  y 10 x3 y 2 = 2  5  x  x  x  y  y 6 x2 y 2 z = 2  3  x  x  y  y  z The common factors are x, x, y, y. And, x  x  y  y = x y 2 2 Q2: Factorise the following expressions (i) 7 x − 42 (ii) 6 p − 12q (iii) 7 a 2 + 14a (iv) − 16 z + 20 z 3 (v) 20 l 2 m + 30alm (vi) 5 x 2 y − 15 xy 2 (vii) 10a 2 − 15b 2 + 20c 2 (viii) − 4a 2 + 4ab − 4ca (ix) x 2 yz + xy 2 z + xyz 2 (x) ax 2 y + bxy 2 + cxyz Difficulty level: Easy What is known: Algebraic expression. What is unknown: Factorisation of given algebraic expression. Reasoning: First, we will find factors of each terms then find out which factors are common in each term and take out that common factor from expression. Solution: (i) 7x = 7  x 42 = 2  3  7 The common factor is 7.  7 x − 42 = (7  x) − ( 2  3 7) = 7 ( x − 6) (ii) 6 p = 2  3  p 12q = 2  2  3  q The common factors are 2 and 3.  6 p − 12q = (2  3  p) − (2  2  3  q) = 2  3[ p − (2  q)] = 6 ( p − 2q ) (iii) 7 a 2 = 7  a  a 14a = 2  7  a The common factors are 7 and a.  7a 2 + 14a = (7  a  a) + (2  7  a) = 7  a [a + 2] = 7a (a + 2) (iv) 16 z = 2  2  2  2  z 20 z 3 = 2  2  5  z  z  z The common factors are 2, 2, and z. −16 z + 20 z 3 = −(2  2  2  2  z ) + (2  2  5  z  z  z) = (2  2  z )[−(2  2) + (5  z  z )] = 4 z ( −4 + 5 z 2 ) (v) 20 l 2 m = 2  2  5  l  l  m 30 alm = 2  3  5  a  l  m The common factors are 2, 5, l and m.  20l 2 m + 30alm = (2  2  5  l  l  m) + (2  3  5  a  l  m) = (2  5  l  m)[(2  l ) + (3  a)] = 10lm (2 l + 3a) ( vi ) 5x2 y = 5  x  x  y 15 xy 2 = 3  5  x  y  y The common factors are 5, x, and y.  5 x 2 y − 15 xy 2 = (5  x  x  y ) − (3  5  x  y  y) = 5  x  y [ x − (3  y)] = 5 xy ( x − 3 y) (vii) 10a 2 = 2  5  a  a 15b2 = 3  5  b  b 20c 2 = 2  2  5  c  c The common factor is 5. 10a 2 − 15b2 + 20c 2 = (2  5  a  a) − (3  5  b  b) + (2  2  5  c  c) = 5[(2  a  a) − (3  b  b) + (2  2  c  c)] = 5 ( 2a 2 − 3b 2 + 4c 2 ) (viii) 4a 2 = 2  2  a  a 4ab = 2  2  a  b 4ca = 2  2  c  a The common factors are 2, 2, and a. − 4a 2 + 4ab − 4ca = − (2  2  a  a) + (2  2  a  b) − (2  2  c  a) = 2  2  a [ −( a ) + b − c ] = 4a ( − a + b − c ) (ix) x 2 yz = x  x  y  z xy 2 z = x  y  y  z xyz 2 = x  y  z  z The common factors are x, y, and z.  x 2 yz + xy 2 z + xyz 2 = ( x  x  y  z ) + ( x  y  y  z ) + ( x  y  z  z ) = x  y  z [ x + y + z] = xyz ( x + y + z ) (x) ax 2 y = a  x  x  y bxy 2 = b  x  y  y cxyz = c  x  y  z The common factors are x and y. ax 2 y + bxy 2 + cxyz = (a  x  x  y ) + (b  x  y  y ) + (c  x  y  z ) = ( x  y)[(a  x) + (b  y) + (c  z )] = xy (ax + by + cz ) Q3: Factorize: (i) x 2 + xy + 8 x + 8 y (ii) 15 xy − 6 x + 5 y − 2 (iii) ax + bx − ay − by (iv) 15 pq + 15 + 9q + 25 p (v) z − 7 + 7 xy − xyz Difficulty level: Medium What is known: Algebraic expression. What is unknown: Factorisation of given algebraic expression. Reasoning: There are 4 terms in each expression. First, we will make pair of two terms from which we can take out common factors and convert the expression of 4 terms into 2 terms expression then take out common factors from remaining 2 terms. Solution: (i) x 2 + xy + 8 x + 8 y = x  x + x  y + 8  x + 8  y = x( x + y) + 8( x + y) = ( x + y)( x + 8) (ii) 15 xy − 6 x + 5 y − 2 = 3  5  x  y − 3  2  x + 5  y − 2 = 3x(5 y − 2) + 1(5 y − 2) = (5 y − 2)(3x + 1) (iii) ax + bx − ay − by = a  x + b  x − a  y − b  y = x ( a + b) − y ( a + b) = (a + b)( x − y ) (iv) 15 pq + 15 + 9q + 25 p = 15 pq + 9q + 25 p + 15 = 3 5  p  q + 3 3 q + 5  5  p + 3 5 = 3q(5 p + 3) + 5(5 p + 3) = (5 p + 3)(3q + 5) (v ) z − 7 + 7 xy − xyz = z − xyz − 7 + 7 xy = z − x y  z − 7 + 7 x y = z (1 − xy ) − 7(1 − xy ) = (1 − xy )( z − 7)  Chapter – 14: Factorization Exercise 14.2 Q1: Factorize the following expressions. (i) a 2 + 8a + 16 (ii) p 2 − 10 p + 25 (iii) 25m2 + 30m + 9 (iv) 49 y 2 + 84 yz + 36 z 2 (v) 4 x2 − 8x + 4 (vi) 121b 2 − 88bc + 16c 2 (vii) (l + m) 2 − 4lm (Hint: Expand ( l + m ) first ) 2 (viii) a 4 + 2a 2b 2 + b 4 Difficulty level: Medium What is known: Algebraic expression. What is unknown: Factorisation of the algebraic expression. Reasoning: Use identity: ( x + y ) = x 2 + 2 xy + y 2 2 ( x − y ) = x2 − 2 xy + y 2 2 Solution: (i) a 2 + 8a + 16 = (a) 2 + 2  a  4 + (4) 2  Using identity ( x + y )2 = x 2 + 2 xy + y 2 ,  = (a + 4)2   considering x = a and y = 4  (ii) p 2 − 10 p + 25 = ( p) 2 − 2  p  5 + (5) 2  Using identity (a − b) 2 = a 2 − 2ab + b2 ,  = ( p − 5)2   considering a = p and y = 5  (iii) 25m2 + 30m + 9 = (5m) 2 + 2  5m  3 + (3) 2  Using identity (a + b) 2 = a 2 + 2ab + b 2 ,  = (5m + 3)2   considering a = 5m and b = 3  (iv) 49 y 2 + 84 yz + 36 z 2 = (7 y ) 2 + 2  (7 y )  (6 z ) + (6 z) 2  Using identity (a + b) 2 = a 2 + 2ab + b 2 ,  = (7 y + 6 z ) 2   considering a = 7 y and b = 6 z  (v ) 4 x 2 − 8 x + 4 = (2 x) 2 − 2(2 x)(2) + (2) 2  Using identity (a − b) 2 = a 2 − 2ab + b 2 ,  = (2 x − 2)2   Considering a = 2 x and y = 2  = [(2)( x − 1)]2 = 4( x − 1)2 (vi) 121b 2 − 88bc + 16c 2 = (11b) 2 − 2(11b)(4c) + (4c) 2  Using identity (a − b) 2 = a 2 − 2ab + b 2 ,  = (11b − 4c) 2   Considering a = 11b and b = 4c  (vii) (l + m)2 − 4lm = l 2 + 2lm + m2 − 4lm = l 2 − 2lm + m2  Using identity (a − b) 2 = a 2 − 2ab + b 2 ,  = (l − m)2   considering a = l and b = m  a 4 + 2a 2b 2 + b 4 = ( a 2 ) + 2 ( a 2 )( b 2 ) + ( b 2 ) 2 2 (viii)  Using identity ( x + y ) 2 = x 2 + 2 xy + y 2 ,  = (a + b 2 ) 2 2   considering x = a and y = b 2 2  Q2: Factorize (i) 4 p 2 − 9q 2 (ii) 63a 2 − 112b 2 (iii) 49 x 2 − 36 (iv) 16 x5 − 144 x3 (v) (l + m) 2 − (l − m) 2 (vi) 9 x 2 y 2 − 16 (vii) (x 2 − 2 xy + y 2 ) − z 2 (viii) 25a 2 − 4b 2 + 28bc − 49c 2 Difficulty level: Easy What is known: Algebraic expression. What is unknown: Factorisation of the algebraic expression. Reasoning: Use identity: ( a + b ) = a 2 + 2ab + b2 2 ( a − b ) = a 2 − 2ab + b2 2 a 2 − b 2 = ( a − b )( a + b ) Solution: (i) 4 p 2 − 9q 2 = (2 p) 2 − (3q) 2  Using identity a 2 − b 2 = (a − b)(a + b),  = (2 p + 3q) (2 p − 3q)   considering a = 2 p and b = 3q  (ii) 63a 2 − 112b2 = 7 ( 9a 2 − 16b2 ) = 7 (3a)2 − (4b)2   Using identity x 2 − y 2 = ( x − y )( x + y ),  = 7  (3a + 4b)(3a − 4b)   considering x = 3a and y = 4b  (iii) 49 x 2 − 36 = (7 x) 2 − (6) 2  Using identity a 2 − b 2 = (a − b)(a + b),  = (7 x − 6)(7 x + 6)   considering a = 7 x and b = 6.  (iv) 16 x5 − 144 x3 = 16 x3 ( x 2 − 9 ) = 16 x3 ( x)2 − (3)2   Using identity a 2 − b 2 = (a − b)(a + b)  = 16 x3  ( x − 3)( x + 3)   Considering a = x and b = 3.  (l + m) − (l − m) = ( l + m ) − ( l − m )  ( l + m ) + ( l − m )  2 2 (v)  Using identity a 2 − b 2 = (a - b)(a+b),  = ( l + m − l + m )( l + m + l − m )   considering a = (l + m) and b = (l − m)  = 2m  2l = 4ml = 4lm (vi) 9 x 2 y 2 − 16 = (3xy ) 2 − (4) 2  Using the identity a 2 − b 2 = (a − b)(a + b),    = (3xy − 4)(3xy + 4) considering a = 3xy and b = 4.   Using identity ( a − b )2 = a 2 − 2ab + b 2    ( vii ) (x 2 − 2 xy + y 2 ) − z 2 = ( x − y ) − ( z ) 2 2 for ( x − y )2 = x 2 − 2 xy + y 2   Using identity a 2 − b 2 = ( a − b )( a + b )    = ( x − y − z )( x − y + z )  considering a = x − y and b = z.  (viii) 25a 2 − 4b 2 + 28bc − 49c 2 = 25a 2 − ( 4b 2 − 28bc + 49c 2 ) = (5a ) 2 − (2b) 2 − 2  2b  7c + (7c) 2   Using identity ( x − y ) 2 = x 2 − 2xy + y 2    considering x = 2b and y = 7c.  = (5a ) 2 − (2b − 7c) 2  Using identity x 2 − y 2 = ( x − y )( x + y )    considering x = 5a and y = 2b − 7c.  = [5a + (2b − 7c)][5a − (2b − 7c)] = (5a + 2b − 7c)(5a − 2b + 7c) Q3: Factorise the expressions (i) ax 2 + bx (ii) 7 p 2 + 21q 2 (iii) 2 x3 + 2 xy 2 + 2 xz 2 (iv) am2 + bm 2 + bn 2 + an 2 (v) (lm + l ) + m + 1 (vi) y ( y + z ) + 9( y + z ) (vii) 5 y 2 − 20 y − 8 z + 2 yz (viii) 10ab + 4a + 5b + 2 (ix) 6 xy − 4 y + 6 − 9 x Difficulty level: Medium What is known: Algebraic expression. What is unknown: Factorisation of given algebraic expression. Reasoning: For part (i), (ii), (iii) and (vi) - First we will find factors of each terms then find out which factors are common in each term and take out that common factor from expression. For part (iv), (v), (vii), (viii), (ix) - There are 4 terms in each expression. First, we will make pair of two terms from which we can take out common factors and convert the expression of 4 terms into 2 terms expression then take out common factors from remaining 2 terms. Solution: (i) ax 2 + bx = a  x  x + b  x = x(ax + b) ( ii ) 7 p 2 + 21q 2 = 7  p  p + 3  7  q  q = 7 ( p 2 + 3q 2 ) ( iii ) 2 x 3 + 2 xy 2 + 2 xz 2 = 2 x ( x 2 + y 2 + z 2 ) (iv) am2 +bm2 +bn 2 + an 2 = am2 +bm 2 + an 2 +bn 2 = m 2 ( a + b) + n 2 ( a + b ) = ( a + b) ( m 2 + n 2 ) (v) ( lm + l ) + m + 1 = lm + m + l + 1 = m ( l + 1) + 1( l + 1) = ( l + 1)( m + 1) ( vi ) y (y + z ) + 9(y + z ) = (y + z )(y + 9) (vii) 5 y 2 − 20 y − 8 z + 2 yz = 5 y 2 − 20 y + 2 yz − 8 z = 5 y( y − 4) + 2 z ( y − 4) = ( y − 4)(5 y + 2 z ) (viii) 10ab + 4a + 5b + 2 = 10ab + 5b + 4a + 2 = 5b(2a + 1) + 2(2a + 1) = (2a + 1)(5b + 2) (ix) 6 xy − 4 y + 6 − 9 x = 6 xy − 9 x − 4 y + 6 = 3x(2 y − 3) − 2(2 y − 3) = (2 y − 3)(3x − 2) Q4: Factorise (i) a 4 − b4 (ii) p 4 − 81 (iii) x 4 − ( y + z )4 (iv) x 4 − ( x − z )4 (v) a 4 − 2a 2 b 2 + b 4 Difficulty level: Easy What is known: Algebraic expression. What is unknown: Factorisation of the algebraic expression. Reasoning: Use identity: ( a − b) = a 2 − 2ab + b 2 2 a 2 − b 2 = ( a − b )( a + b ) Solution: a 4 − b4 = ( a 2 ) − ( b2 ) 2 2 (i) = ( a 2 − b 2 )( a 2 + b 2 ) = (a − b)(a + b) ( a 2 + b 2 ) p 4 − 81 = ( p 2 ) − (9) 2 2 (ii) = ( p 2 − 9 )( p 2 + 9 ) = ( p) 2 − (3) 2  ( p 2 + 9 ) = ( p − 3)( p + 3) ( p 2 + 9 ) ( iii ) x 4 − ( y + z ) 4 = ( x 2 ) − ( y + z ) 2  2 2 =  x 2 − ( y + z ) 2   x 2 + ( y + z ) 2  = [ x − ( y + z )][ x + ( y + z )]  x 2 + ( y + z ) 2  = ( x − y − z )( x + y + z )  x 2 + ( y + z ) 2  x 4 − ( x − z ) = ( x 2 ) − ( x − z )  2 ( iv ) 4 2 2   =  x2 − ( x − z )   x2 + ( x − z )  2 2    =  x − ( x − z )   x + ( x − z )   x + ( x − z )  2 2   = z ( 2x − z )  x 2 + x 2 − 2xz + z 2  = z ( 2x − z ) ( 2x 2 − 2xz + z 2 ) a 4 − 2a 2b 2 + b 4 = ( a 2 ) − 2 ( a 2 )( b 2 ) + ( b 2 ) 2 2 (v) = ( a 2 − b2 ) 2 =  (a − b)(a + b)  2 = ( a − b) 2 ( a + b) 2 Q5: Factorise the following expressions (i) p 2 + 6 p + 8 (ii) q 2 − 10q + 21 (iii) p 2 + 6 p − 16 Difficulty level: Medium What is known: Algebraic expression. What is unknown: Factorisation of the algebraic expression. Reasoning: In general, for factorising an algebraic expression of the type x + px + q , we find two 2 factors a and b of q (i.e., the constant term) such that ab = q and a + b = p. Solution: (i) p 2 + 6 p + 8 It can be observed that, 8 = 4  2 and 4 + 2 = 6  p2 + 6 p + 8 = p2 + 2 p + 4 p + 8 = p( p + 2) + 4( p + 2) = ( p + 2)( p + 4) ( ii ) q 2 − 10q + 21 It can be observed that, 21 = ( −7 )  ( −3) and ( −7 ) + ( −3) = −10  q 2 − 10q + 21 = q 2 − 7q − 3q + 21 = q(q − 7) − 3(q − 7) = (q − 7)(q − 3) ( iii ) p 2 + 6 p − 16 It can be observed that, −16 = ( −2 )  8 and 8 + ( −2 ) = 6 p 2 + 6 p − 16 = p 2 + 8 p − 2 p − 16 = p( p + 8) − 2( p + 8) = ( p + 8)( p − 2)  Chapter – 14: Factorization Exercise 14.3 Q1: Carry out the following divisions. (i) 28 x 4  56 x (ii) − 36 y 3  9 y 2 (iii) 66 pq 2 r 3  11qr 2 (iv) 34 x3 y 3 z 3  51xy 2 z 3 (v) 12a8b8  ( −6a 6b 4 ) (i) 28x 4  56 x Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of 28x4 and 56x then cancel out common factors of 28x4 and 56x. Solution: 28 x 4 can be written as 2  2  7  x  x  x  x and 56 x can be written as 2  2  2  7  x Then, 2 2 7 x x x x 28 x 4  56 x = 2 2 2 7 x 3 x = 2 1 = x3 2 (ii) − 36 y 3  9 y 2 Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of −36y and 9y then cancel out common factors of −36y and 9y. 3 2 3 2 Solution: −36 y 3 can be written as − 2  2  3  3  y  y  y and 9 y 2 can be written as 3  3  y  y Then, −2  2  3  3  y  y  y −36 y 3  9 y 2 = 3 3 y  y = −4 y (iii) 66 pq 2 r 3 11qr 2 Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: 2 3 2 2 3 Find out factors of 66 pq r and 11qr then cancel out common factors of 66 pq r and 11qr 2. Solution: 66 pq 2 r 3 can be written as 2  3 11 p  q  q  r  r  r and 11qr 2 can be written as 11 q  r  r Then, 2  3 11 p  q  q  r  r  r 66 pq 2 r 3  11qr 2 = 11 q  r  r = 6 pqr (iv) 34 x3 y 3 z 3  51xy 2 z 3 Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: 3 3 3 2 3 3 3 3 Find out factors of 34x y z and 51xy z then cancel out common factors of 34x y z and 51xy 2 z 3. Solution: 34 x3 y 3 z 3 can be written as 2 17  x  x  x  y  y  y  z  z  z and 51xy 2 z 3 can be written as 3 17  x  y  y  z  z  z Then, 2 17  x  x  x  y  y  y  z  z  z 34 x3 y 3 z 3  51xy 2 z 3 = 3 17  x  y  y  z  z  z 2 = x2 y 3 (v) 12a 8b8  ( −6a 6b 4 ) Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of 12a8b8 and −6a 6b4 then cancel out common factors of 12a8b8 and −6a 6b4. Solution: 12a8b8 can be written as 2  2  3  a 8  b8 and −6a 6b 4 can be written as − 2  3  a 6  b 4 Then, 2  2  3  a 8  b8 12a8b8  ( −6a 6b 4 ) = −2  3  a 6  b 4 = −2a 2b 4 Q2: Divide the given polynomial by the given monomial. (i) ( 5 x − 6 x )  3x 2 (ii) ( 3 y − 4 y + 5 y )  y 8 6 4 4 (iii) 8 ( x y z + x y z + x y z )  4 x y z 3 2 2 2 3 2 2 2 3 2 2 2 ( iv ) ( x + 2 x + 3x )  2 x 3 2 (v) ( p q − p q )  p q 3 6 6 3 3 3 (i) (5x 2 − 6 x )  3x Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of ( 5 x 2 − 6 x ) and 3x then cancel out common factors of ( 5 x 2 − 6 x ) and 3x Solution: 5 x 2 − 6 x can be written as x(5 x − 6) Then, ( 5x2 − 6 x )  3x = x(53xx− 6) 1 = (5 x − 6) 3 (ii) (3 y 8 − 4 y6 + 5 y4 )  y4 Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of ( 3 y 8 − 4 y 6 + 5 y 4 ) and y then cancel out common factors of 4 (3 y 8 − 4 y 6 + 5 y 4 ) and y4. Solution: 3 y 8 − 4 y 6 + 5 y 4 can be written as y 4 ( 3 y 4 − 4 y 2 + 5 ) Then, y 4 ( 3 y 4 − 4 y 2 + 5) (3 y 8 − 4y + 5y 6 4 ) y 4 = y4 = 3y4 − 4 y2 + 5 (iii) 8 ( x3 y 2 z 2 + x 2 y 3 z 2 + x 2 y 2 z 3 )  4 x 2 y 2 z 2 Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of 8 ( x3 y 2 z 2 + x 2 y 3 z 2 + x 2 y 2 z 3 ) and 4x y z then cancel out common 2 2 2 factors of 8 ( x3 y 2 z 2 + x 2 y 3 z 2 + x 2 y 2 z 3 ) and 4x y z. 2 2 2 Solution: 8 ( x3 y 2 z 2 + x 2 y 3 z 2 + x 2 y 2 z 3 ) can be written as 8 x 2 y 2 z 2 ( x + y + z ) Then, 8x2 y 2 z 2 ( x + y + z) 8 ( x y z + x y z + x y z )  4x y z = 3 2 2 2 3 2 2 2 3 2 2 2 4x2 y 2 z 2 = 2( x + y + z ) ( iv ) (x 3 + 2 x 2 + 3x )  2 x Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of ( x3 + 2 x 2 + 3x ) and 2x then cancel out common factors of ( x3 + 2 x 2 + 3x ) and 2x. Solution: x3 + 2 x 2 + 3 x can be written as x ( x 2 + 2 x + 3) Then, x ( x 2 + 2 x + 3) (x 3 + 2 x + 3x )  2 x = 2 2x = 2 ( x + 2 x + 3) 1 2 (v) (p q 3 6 − p 6 q3 )  p3q3 Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of ( p 3q 6 − p 6 q 3 ) and p q then cancel out common factors of 3 3 (p q 3 6 − p 6 q 3 ) and p3q3. Solution: p3q 6 − p 6 q 3 can be written as p 3q 3 ( q 3 − p 3 ) Then, p3q3 ( q3 − p3 ) (p q 3 6 − p q ) p q = 6 3 3 3 p3q3 = q3 − p3 Q3: Work out the following divisions. (i) (10 x − 25)  5 (ii) (10 x − 25)  (2 x − 5) (iii) 10 y (6 y + 21)  5(2 y + 7) (iv) 9 x 2 y 2 (3z − 24)  27 xy( z − 8) (v) 96abc(3a − 12)(5b − 30)  144(a − 4)(b − 6) (i) (10 x − 25)  5 Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of (10 x − 25) then cancel out common factors of (10 x − 25) and 5. Solution: Factorising (10 x − 25) , we get (10 x − 25) = 5  2  x − 5  5 = 5 ( 2 x − 5) 5(2 x − 5) (10 x − 25)  5 = 5 = 2x − 5 (ii) (10 x − 25)  (2 x − 5) Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of (10 x − 25) then cancel out common factors of (10 x − 25) and (2 x − 5) Solution: Factorising (10 x − 25) , we get (10 x − 25) = 5  2  x − 5  5 = 5 ( 2 x − 5) 5(2 x − 5) (10 x − 25)  (2 x − 5) = 2x − 5 =5 (iii) 10 y(6 y + 21)  5(2 y + 7) Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of 10 y(6 y + 21) then cancel out common factors of 10 y(6 y + 21) and 5(2 y + 7). Solution: Factorising 10 y(6 y + 21) , we get, 10 y (6 y + 21) = 5  2  y  ( 2  3  y + 3  7 ) = 5 2  y  3( 2  y + 7) = 30 y ( 2 y + 7 ) 30 y (2 y + 7) 10 y (6 y + 21)  5(2 y + 7) = 5(2 y + 7) = 6y (iv) 9 x 2 y 2 (3z − 24)  27 xy( z − 8) Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of 9 x y (3z − 24) then cancel out common factors of 9 x y (3z − 24) and 2 2 2 2 27 xy( z − 8). Solution: Factorising 9 x 2 y 2 (3z − 24) , we get, 9 x 2 y 2 (3z − 24) = 3  3  x  x  y  y  ( 3  z − 2  2  2  3) = 3 3 x  x  y  y  3( z − 2  2  2) = 27 x 2 y 2 ( z − 8 ) 27 x 2 y 2  ( z − 8) 9 x 2 y 2 (3z − 24)  27 xy ( z − 8) = 27 xy ( z − 8) = xy (v) 96abc(3a − 12)(5b − 30) 144(a − 4)(b − 6) Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Find out factors of 96abc(3a − 12)(5b − 30) then cancel out common factors of 96abc(3a − 12)(5b − 30) and 144(a − 4)(b − 6). Solution: Factorising 96abc(3a − 12)(5b − 30) , we get, 96abc(3a − 12)(5b − 30) = 96abc  ( 3  a − 2  2  3)  ( 5  b − 5  2  3) = 96abc  3 ( a − 2  2 )  5 ( b − 2  3) = 1440abc ( a − 4 )( b − 6 ) 96abc(3a − 12)(5b − 30)  144(a − 4)(b − 6) 1440abc ( a − 4 )( b − 6 ) = 144(a − 4)(b − 6) = 10abc Q4: Divide as directed. (i) 5(2 x + 1)(3 x + 5)  (2 x + 1) (ii) 26 xy ( x + 5)( y − 4)  13x( y − 4) (iii) 52 pqr( p + q)(q + r )(r + p )  104 pq (q + r )(r + p ) (iv) 20( y + 4) ( y 2 + 5 y + 3)  5( y + 4) (v) x( x + 1)( x + 2)( x + 3)  x( x + 1) Difficulty level: Easy What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Cancel out common factors of the following. Solution: 5(2 x + 1)(3 x + 1) (i ) 5(2 x + 1)(3 x + 5)  (2 x + 1) = (2 x + 1) = 5(3 x + 1) 2 13  xy ( x + 5)( y − 4) ( ii ) 26 xy ( x + 5)( y − 4)  13x( y − 4) = 13x( y − 4) = 2 y ( x + 5) (iii) 52 pqr ( p + q)(q + r )(r + p)  104 pq(q + r )(r + p) 2  2 13  p  q  r  ( p + q)  (q + r )  (r + p) = 2  2  2 13  p  q  (q + r )  (r + p) 1 = r ( p + q) 2 (iv) 20( y + 4) ( y 2 + 5 y + 3)  5( y + 4) 2  2  5  ( y + 4)  ( y 2 + 5 y + 3) = 5  ( y + 4) = 4 ( y + 5 y + 3) 2 x( x + 1)( x + 2)( x + 3) ( v) x( x + 1)( x + 2)( x + 3)  x( x + 1) = x( x + 1) = ( x + 2)( x + 3) Q5: Factorize the expressions and divide them as directed. (i) ( y + 7 y + 10)  ( y + 5) 2 (ii) ( m − 14m − 32 )  (m + 2) 2 (iii) ( 5 p − 25 p + 20 )  ( p − 1) 2 (iv) 4 yz ( z + 6 z − 16 )  2 y ( z + 8) 2 (v) 5 pq ( p − q )  2 p( p + q ) 2 2 (vi) 12 xy ( 9 x − 16 y )  4 xy (3x + 4 y ) 2 2 (vii) 39 y ( 50 y − 98 )  26 y (5 y + 7) 3 2 2 (i) (y 2 + 7 y + 10 )  ( y + 5) Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Factorise ( y 2 + 7 y + 10 ) then cancel out common factors of ( y 2 + 7 y + 10 ) and ( y + 5). Solution: (y 2 + 7 y + 10 ) can be written as, y 2 + 2 y + 5 y + 10 = y ( y + 2) + 5( y + 2) = ( y + 2)( y + 5) Then, ( y + 2)( y + 5) (y 2 + 7 y + 10 )  ( y + 5) = ( y + 5) = y+2 (ii) (m 2 − 14m − 32 )  ( m + 2) Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Factorise ( m 2 − 14m − 32 ) then cancel out common factors of ( m 2 − 14m − 32 ) and Solution: m2 − 14m − 32 can be written as, m2 + 2m − 16m − 32 = m(m + 2) − 16(m + 2) = (m + 2)(m − 16) Then, (m + 2)(m − 16) (m 2 − 14m − 32 )  (m + 2) = (m + 2) = m − 16 (iii) (5 p 2 − 25 p + 20 )  ( p − 1) Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Factorise ( 5 p 2 − 25 p + 20 ) then cancel out common factors of (5 p 2 − 25 p + 20 ) and ( p − 1). Solution: 5 p 2 − 25 p + 20 can be written as, 5 ( p2 − 5 p + 4) = 5 ( p2 − p − 4 p + 4) = 5  p( p − 1) − 4( p − 1)  = 5( p − 1)( p − 4) Then, 5( p − 1)( p − 4) (5 p 2 − 25 p + 20 )  ( p − 1) = ( p − 1) = 5( p − 4) (iv) 4 yz ( z 2 + 6 z − 16 )  2 y ( z + 8) Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Factorise 4 yz ( z 2 + 6 z − 16 ) then cancel out common factors of 4 yz ( z 2 + 6 z − 16 ) and 2 y( z + 8). Solution: 4 yz ( z 2 + 6 z − 16 ) can be written as, 4 yz ( z 2 − 2 z + 8z − 16 ) = 4 yz  z ( z − 2) + 8( z − 2) = 4 yz ( z − 2)( z + 8) Then, 4 yz ( z − 2)( z + 8) 4 yz ( z 2 + 6 z − 16 )  2 y ( z + 8) = 2 y ( z + 8) = 2 z ( z − 2) (v) 5 pq ( p 2 − q 2 )  2 p ( p + q ) Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Factorise 5 pq ( p 2 − q 2 ) by using identity a 2 − b 2 = ( a − b )( a + b ) then cancel out common factors of 5 pq ( p 2 − q 2 ) and 2 p( p + q). Solution: 5 pq ( p 2 − q 2 ) can be written as 5 pq( p − q)( p + q) Then, 5 pq( p − q )( p + q ) 5 pq ( p 2 − q 2 )  2 p( p + q ) = 2 p( p + q) 5 = q( p − q) 2 (vi) 12 xy ( 9 x 2 − 16 y 2 )  4 xy (3 x + 4 y ) Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Factorise 12 xy ( 9 x 2 − 16 y 2 ) by using identity a 2 − b 2 = ( a − b )( a + b ) then cancel out common factors of 12 xy ( 9 x 2 − 16 y 2 ) and 4 xy(3x + 4 y). Solution: 12 xy ( 9 x 2 − 16 y 2 ) can be written as, 12 xy (3x)2 − (4 y)2  = 12 xy(3x − 4 y)(3x + 4 y) = 2  2  3  x  y  (3x − 4 y)  (3x + 4 y) Then, 2  2  3  x  y  (3 x − 4 y )  (3 x + 4 y ) 12 xy ( 9 x 2 − 16 y 2 )  4 xy (3x + 4 y ) = 2  2  x  y  (3x + 4 y ) = 3(3 x − 4 y ) (vii) 39 y 3 ( 50 y 2 − 98 )  26 y 2 (5 y + 7) Difficulty level: Medium What is known: Algebraic expression. What is unknown: Division of the algebraic expression. Reasoning: Factorise 39 y 3 ( 50 y 2 − 98 ) by using identity a 2 − b 2 = ( a − b )( a + b ) then cancel out common factors of 39 y 3 ( 50 y 2 − 98 ) and 26 y (5 y + 7). 2 Solution: 39 y 3 ( 50 y 2 − 98 ) can be written as, 3 13  y  y  y   2  ( 25 y 2 − 49 ) = 3 13  2  y  y  y  (5 y) 2 − (7) 2  = 3 13  2  y  y  y(5 y − 7)(5 y + 7) and 26 y (5 y + 7) can be written as 2 13  y  y  (5 y + 7) 2 Then, 3 13  2  y  y  y (5 y − 7)(5 y + 7) 39 y 3 ( 50 y 2 − 98 )  26 y 2 (5 y + 7) = 2 13  y  y  (5 y + 7) = 3 y (5 y − 7)  Chapter – 14: Factorization Exercise 14.4 Q1: Find and correct the errors in the statement: 4( x − 5) = 4 x − 5 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: S = 4( x − 5)  R.H.S. The correct statement is 4( x − 5) = 4 x − 20 Q2: Find and correct the errors in the statement: x(3x + 2) = 3x 2 + 2 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S = x(3x + 2) = 3x 2 + 2 x L.H.S  R.H.S. The correct statement is x(3x + 2) = 3x + 2 x 2 Q3: Find and correct the errors in the statement: 2 x + 3 y = 5 xy Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S = 2 x + 3 y = 2 x + 3 y L.H.S  R.H.S. The correct statement is 2 x + 3 y = 2 x + 3 y Q4: Find and correct the errors in the statement: x + 2 x + 3x = 5x Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = x + 2 x + 3x = 1x + 2 x + 3x = x(1 + 2 + 3) = 6 x L.H.S.  R.H.S. The correct statement is x + 2 x + 3x = 6 x Q5: Find and correct the errors in the statement: 5 y + 2 y + y − 7 y = 0 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = 5 y + 2 y + y − 7 y = 8 y − 7 y = y  R.H.S. The correct statement is 5 y + 2 y + y − 7 y = y Q6: Find and correct the errors in the statement: 3x + 2 x = 5x 2 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = 3x + 2 x = 5 x L.H.S.  R.H.S. The correct statement is 3x + 2 x = 5x Q7: Find and correct the errors in the statement: ( 2 x )2 + 4 ( 2 x ) + 7 = 2 x 2 + 8 x + 7 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = (2 x) 2 + 4(2 x) + 7 = 4 x 2 + 8 x + 7 L.H.S.  R.H.S The correct statement is (2 x) + 4(2 x) + 7 = 4 x + 8 x + 7 2 2 Q8: Find and correct the errors in the statement: (2 x)2 + 5 x = 4 x + 5 x = 9 x Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = (2 x) 2 + 5 x = 4 x 2 + 5 x L.H.S.  R.H.S. The correct statement is ( 2 x )2 + 5 x = 4 x 2 + 5 x Q9: Find and correct the errors in the statement : ( 3x + 2 ) = 3 x 2 + 6 x + 4 2 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = (3x + 2)2 = (3x) 2 + 2(3x)(2) + (2) 2  Using identity (a + b) = a + 2ab + b  2 2 2 = 9 x 2 + 12 x + 4 L.H.S.  R.H.S The correct statement is ( 3x + 2 )2 = 9 x 2 + 12 x + 4 Q10: Find and correct the errors in the following mathematical statement. Substituting x = −3 in (a) x 2 + 5 x + 4 gives (−3)2 + 5(−3) + 4 = 9 + 2 + 4 = 15 (b) x 2 − 5 x + 4 gives (−3)2 − 5(−3) + 4 = 9 − 15 + 4 = −2 (c) x 2 + 5 x gives (−3)2 + 5(−3) = −9 − 15 = −24 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Put value of x in L.H.S and find correct solution. Solution: a) For x = −3 L.H.S = x 2 + 5 x + 4 = (−3) 2 + 5(−3) + 4 = 9 − 15 + 4 = 13 − 15 = −2 L.H.S  R.H.S The correct answer is x 2 + 5x + 4 = −2 b) For x = −3 x 2 − 5 x + 4 = (−3) 2 − 5(−3) + 4 = 9 + 15 + 4 = 28 The correct answer is x − 5 x + 4 = 28 2 c) For x = −3 x 2 + 5 x = ( −3) 2 + 5( −3) = 9 − 15 = −6 The correct answer is x 2 + 5 x = −6 Q11: Find and correct the errors in the statement: ( y − 3)2 = y 2 − 9 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Use identity (a − b) = a − 2ab + b 2 2 2 Solution: L.H.S. = ( y − 3) 2 = ( y ) 2 − 2( y)(3) + (3) 2  Using identity (a − b) = a − 2ab + b  2 2 2 = y2 − 6 y + 9 L.H.S.  R.H.S. The correct statement is ( y − 3) = y − 6 y + 9 | 2 2 Q12: Find and correct the errors in the statement : ( z + 5 ) = z 2 + 25 2 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Use identity (a + b) = a + 2ab + b 2 2 2 Solution: L.H.S. = ( z + 5)2 = ( z )2 + 2( z )(5) + (5)2  Using identity (a + b)2 = a 2 + 2ab + b 2  = z 2 + 10 z + 25 L.H.S.  R.H.S. The correct statement is ( z + 5 ) = z 2 + 10 z + 25 2 Q13: Find and correct the errors in the statement: ( 2a + 3b )( a − b ) = 2a 2 − 3b 2 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = (2a + 3b)(a − b) = 2a  a − 2a  b + 3b  a − 3b  b = 2a 2 − 2ab + 3ab − 3b 2 = 2a 2 + ab − 3b 2 L.H.S.  R.H.S. The correct statement is ( 2a + 3b )( a − b ) = 2a 2 + ab − 3b 2 Q14: Find and correct the errors in the statement : ( a + 4 )( a + 2 ) = a 2 + 8 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = (a + 4)(a + 2) = a  a + 2  a + 4  a + 4  2 = (a) 2 + a(4 + 2) + (4  2) = a 2 + 6a + 8 L.H.S.  R.H.S. The correct statement is ( a + 4 )( a + 2 ) = a 2 + 6a + 8 Q15: Find and correct the errors in the statement : ( a − 4 )( a − 2 ) = a 2 − 8 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: L.H.S. = (a − 4)(a − 2) = a  a + ( −2 )  a + ( −4 )  a + (−4)  (−2)  = a 2 − 2a − 4a + 8 = a 2 − 6a + 8 L.H.S.  R.H.S. The correct statement is ( a − 4 )( a − 2 ) = a − 6a + 8 2 3x 2 Q16: Find and correct the errors in the statement: 2 = 0 3x Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: 3x 2 3  x  x L.H.S. = 2 = =1 3x 3 x  x L.H.S.  R.H.S. 3x 2 The correct statement is 2 = 1 3x 3x 2 + 1 Q17: Find and correct the errors in the statement: = 1+1 = 2 3x 2 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: 3x 2 + 1 3x 2 1 L.H.S. = 2 = 2+ 2 3x 3x 3x 1 = 1+ 2 3x L.H.S.  R.H.S. 3x 2 + 1 1 The correct statement is 2 = 1+ 2 3x 3x 3x 1 Q18: Find and correct the errors in the statement: = 3x + 2 2 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: 3x L.H.S. =  R.H.S. 3x + 2 3x 3x The correct statement is = 3x + 2 3x + 2 3 1 Q19: Find and correct the errors in the statement: = 4x + 3 4x Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: 3 L.H.S =  R.H.S 4x + 3 3 3 The correct statement is = 4x + 3 4x + 3  4x + 5 Q20: Find and correct the errors in the statement: =5 4x Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: 4x + 5 4x 5 L.H.S. = = + 4x 4x 4x 5 = 1+ 4x L.H.S.  R.H.S. 4x + 5 5 The correct statement is = 1+ 4x 4x 7x + 5 Q21: Find and correct the errors in the statement: = 7x 5 Difficulty level: Easy What is known: Incorrect mathematical statement. What is unknown: Correct mathematical statement. Reasoning: Solve L.H.S. Solution: 7x + 5 7x 5 L.H.S. = = + 5 5 5 7x = +1 5 L.H.S.  R.H.S. 7x + 5 7x The correct statement is = +1 5 5 www.cuemath.com