Ncert Solutions For Class 8 Maths Ex 14.1

Ncert Solutions For Class 8 Maths Chapter 14 Ex 14.1

Question 1.  Calculate the common factors of the following:

( i )16a and  28                                  ( i i )26x and  13ya

( iii )20pq, 30rp and 20pqr            ( iv )12a3y2 and  5a2y3z2

( v ) 8abc and  18ab2                       ( vi )6pqr, 24pq2 and 12p2q

 

Sol.

( i ) 16a = 2 * 2 *2 * 2 * a

28 = 2 * 2 * 7

Thus, the common factors are 2 and 2

 

( ii ) 26a = 2 * 13 * a

13ya = 13 * y * a

Thus, the common factors are 13 and a

 

( iii ) 20pq = 2 * 2 * 5 * p * q

30rp = 2 * 3  * 5 * r * p

20qr = 2 * 2 * 5 *p* q * r

Thus, the common factors are 2, 5 and p

 

( iv ) 12a3y2 = 2 * 2 * 3 * a * a * a * y * y

5a2y3z2 = 5 * a * a * y * y * y *z *z

Thus, the common factors are a * a * y * y

 

( v ) 8abc   =  2 * 2 * 2 * a * b * c

18ab2 = 2 * 3 * 3 * a * b*b                       

        Thus, the common factors are 2, a and b

 

( vi ) 6pqr   =2 * 3* p*q*r

24pq2 =2 * 2 * 2 *3*p* q*q

12p2q = 2*2*3*p*p*q

Thus, the common factors are 2, 3, p and q.

 

Question 2. Factorize the given expressions.

                ( i ) 7a – 56           ( ii ) 6a -30b

                ( iii ) 3a2+18a     ( iv ) -12a + 20b2

                ( v ) 4c2+4ab – 8ca           ( vi ) a2bc + ab2c + abc2 

                ( vii ) ap2q + bpq2+ cpqw  ( viii ) 20a2b + 30abc

 

sol.

( i )7a – 56   = 7 * a – (2 * 2 * 2 * 7)

Taking the common factors,

=7(a – 8)

 

( ii ) 6a – 30b = (2 * 3 *a)– (2 * 3* 5 *b)

Taking the common factors,

=2*3(a – 5*b)

=6(a-5b)

 

( iii ) 3a2 + 18a = 3 * a * a + (2 * 3 * 3 *a)

Taking the common factors,

= 3 * a (a + 2 * 3 )

= 3a(a + 6)

 

( iv ) -12a + 20b2= -(2 * 2* 3*a) + (2*2*5*b*b)

Taking the common factors,

= 2*2(-3*a  + 5*b*b)

=  4(-3a +5*b*b)

=-4(3a – 5b2)

 

( v ) 4c2+4ab -8ca= (2 * 2 * c * c) + (2*2*a*b) – (2*2*2*c*a)

Taking the common factors,

=2*2(c*c  + a*b – 2*c*a)

=4(c2+ab – 2ca)

 

( vi ) a2bc + ab2c + abc2 = a*a*b*c + a*b*b*c + a*b*c*c

Taking the common factors,

= abc(a+b+c)

 

( vii ) ap2q + bpq2+ cpqw  = a * p* p *q + b*p*q*q + c*p*q*w

Taking the common factors,

= p*q(a*p + b*q + c*w)

=pq(ap + bq + cw)

 

( viii ) 20a2b + 30abc = 2*2*5*a*a*b + 2*3*5*a*b*c

Taking the common factors,

=2*5*a*b(2*a + 3*c)

=10ab(2a + 3c)

 

Question 3. Factorize the following expressions:

                ( i ) a 2+ ab + 19a  + 19b                 ( ii )20ab -8a +5a-2

                ( iii ) pa + pb – qa – qb                    ( iv ) 18ab + 15 + 30b + 9b

                ( v ) 8ab + c – 8- abc

sol.

( i ) a2 + ab + 19a + 19b =  a(a + b) + 19(a + b)

= (a + 19)(a + b)

 

( ii ) 20ab – 8a + 5b – 2     = 4a(5b – 2) + 1(5b – 2)

= (4a + 1)(5b – 2)

 

( iii ) pa + pb – qa –qb     = p(a + b) –q(a + b)

= (a + b)(p – q)

 

( iv ) 18ab + 15 + 30a + 9b = 18ab + 30a + 15 + 9b

=  6a(3b + 5) + 3(5 + 3b)

= (6a +3)(3b + 5)

 

( v ) 8ab + c -8 –abc          =8ab -8 +c –abc

= 8(ab -1) – c(ab – 1)

=(8 – c)(ab – 1)

Or, (-1)( c – 8)(-1)(1-ab)

Thus, we have : (1 – ab) (c – 8)