Class 8 Exam  >  Class 8 Notes  >  NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8 PDF Download

Exercise 14.1 

Question 1: 

Find the common factors of the given terms. 

(i)    12 x , 36 
 (ii)    2y,22 x y
 (iii)    14pq,28p2q
 (iv)    2 x ,3 x 2,4
 (v)    6abc, 24ab2 , 12a2
 (vi)    16 x 3 ,-4 x 2, 32 x
 (vii)  10 pq, 20qr , 30rp
 (viii)    3 x 2y3 ,10 x 3y2 , 6 x 2y2z

 

Answer 1:

(i)

12 x = 2 x 2 x 3 x x
36 = 2 x 2 x 3 x 3
Hence, the common factors are 2,2 and 3 = 2 x 2 x 3 = 12

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

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

(iv) 2 x = 2 x x x 1

3 x 2 = 3 x x x x x 1 

 4 = 2 x 2 x 1
Hence, the common factor is 1.

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

(vi)    16 x = 2 x 2 x 2 x 2 x x x x x x
-4 x 2 = (-l) x 2 x 2 x x x x
32 x = 2 x 2 x 2 x 2 x 2 x x
Hence, the common factors are 2 x 2 x x = 4 x

(vii) 10pq =2 x5 x p x q
20qr =2x 2 x5 x qx r
30rp =2x3x5x r x p
Hence.the common factors are 2x 5 = 10

(viii) 3x2y3 =3x x x x x y x y x y
10x3y=2x 5x x x x x x x y x y
6x2y2z = 2x3 x x x x x y x y x z
Hence, the common factors are x x x x y x y =x2y2

Question 2: 

Factorize the following expressions. 

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

Answer 2: 

(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) 7a+ 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)
= 4=(-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 * / +3* a)
= 10lm( 2l+3a)

(vi)    5*2y-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)

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(viii)

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(ix)

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(x)

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8  

Question 3: 

Factorize: 

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

Answer 3:

(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)- =( xy -1 )
= (xy- 1)(7-z)=(-1)(1-xy)(- 1)(z-7)
= (1-xy)(z -7)

Exercise 14.2 

Question 1: 

Factorize the following expressions: 
(i) a2+8a +16
 (ii) p-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)first]
 (viii) a+ 2a2b2+b4

Answer 1: 

(i) 
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(ii)
 NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(iii) 
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(ivI)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(v)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(vi)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(vii)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(viii)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

Question 2: 

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

Answer 2: 

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

 

Question 3: 

Factorize the expressions: 

(i) ax2 +bx    
 (ii)    7p + 2lq2
 (iii)    2x3+ 2xy2 + 2xz2
 (iv)    am2'+ bm2 + bn2 + an2
 (v)    ( lm+l)+m+ I    
 (vi)    y( y+z)+ 9( y+ z)
 (vii)    5y2 -20y - 8z+2yz    
 (viii)    10ab+4a +5b+2
 (ix)    6xy - 4y + 6 -9x

Answer 3:

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

 

Question 4: 

Factorize: 

(i)    a4-b4
 (ii)    p4-81
 (iii)    x4-(y+z)4
 (iv)    x4-(x -z) 4
 (v)    a4-2a2b2+b2

Answer 4: 

(i) 
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(ii) 
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(iii)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(iv)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

(v)
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

Question 5: 

Factorize the following expressions:

NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8
NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

Answer 5: 

(i)    p 2 +6p+8 = p 2 + (4+ 2) p +4 x 2
= p 2 +4p +2p+4 x 2
= p(p + 4) +2(p + 4)
= ( p +4)( p + 2)

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

(iii)    p+6p -16 = p2 + (8-2)p-8x2
=  p 2 + 8p-2p-8x2
= p (p+8) -2(p+ 8)
=  (p+ 8)(p-2) 

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FAQs on NCERT Solutions (Ex: 14.1 - 14.2) - Factorisation - Class 8

1. What is factorisation?
Ans. Factorisation is the process of finding the factors of a given algebraic expression. It is the reverse process of multiplication, where we find the product of given factors. In factorisation, we try to express the given expression as a product of two or more expressions.
2. What are the methods of factorisation?
Ans. There are several methods of factorisation, including: 1. Common factor method 2. Factorisation by regrouping terms 3. Factorisation of quadratic expressions 4. Factorisation using the identity (a+b)^2 = a^2 + 2ab + b^2 5. Factorisation using the identity (a-b)^2 = a^2 - 2ab + b^2
3. How do I factorise quadratic expressions?
Ans. To factorise quadratic expressions, we need to find two factors of the quadratic expression that when multiplied, give the original quadratic expression. The general form of a quadratic expression is ax^2 + bx + c. We can use the following methods to factorise quadratic expressions: 1. Factorisation by splitting the middle term 2. Factorisation by using the quadratic formula 3. Factorisation by completing the square
4. How important is factorisation in mathematics?
Ans. Factorisation is an essential concept in mathematics that has many real-life applications. It is used in algebra, geometry, calculus, and many other branches of mathematics. Factorisation is used in simplifying algebraic expressions, solving equations, finding roots of polynomials, and many more. It is also used in cryptography to encode and decode messages.
5. What are the applications of factorisation in daily life?
Ans. Factorisation has various applications in daily life, including: 1. Finding the total cost of items sold in a store 2. Calculating the area and perimeter of a rectangle 3. Solving problems related to speed, distance, and time 4. Estimating the amount of paint needed to paint a room 5. Calculating the interest rate on loans or investments.
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