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Ncert Solutions for Factorization Class 8 CBSE Part 1



In this page we have NCERT book Solutions for Class 8th Maths:Factorization for EXERCISE 1 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Question 1

 Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14 pq, 28 p2q2

(iv) 2x, 3x², 4

(v) 6 abc, 24ab², 12 a²b

(vi) 16 x³, – 4x², 32x

(vii) 10 pq, 20qr, 30rp

(viii) 3x² y³, 10x³ y²,6 x² y²z

Answer:

1) We need to convert each of these terms in simple factors forms and then choose the common factors

12x =2×2×3×x

36=2×2×3×3

So common Factor are 2,2,3

And  2×2×3=12

 

2) 2y=2×y

22xy=2×11×y×x

So common factors are 2,y

And 2×y=2y

 

3) 14 pq= 2×7×p×q

 28 p2q2=2×2×7×p×p×q×q

So common factors are  2,7,p,q

=14pq

 

4) 2x=2×x

3x2=3×x×x

4=2×2

There is no common factor other than unity, so common factor= 1

 

5) 6abc=2×3×a×b×c

24ab2=2×2×2×3×a×b×b

12a2b=2×2×3×a×a×b

Common factors are 2, 3, a, b

=6ab

 

6) 16x3=2×2×2×2×x×x×x

-4x2=-2×2×x×x

32x=-2×2×2×2×2×x

Common factors are 2, 2, x

=4x

7) 10pq=-2×5×p×q

20qr=-2×2×5×q×r

30rp=-2×3×5×p×r

Common factors are 2,5

=10

 

8) 3x2y3=3×x×x×y×y×y

10x3y3=2×5×x×x×x×y×y×y

6x2y2z=-2×3×x×x×y×y×z

Common factors are x, x, y, y

=x2y2

 
 

Question 2

 Factorize the following expressions.

(i) 7x – 42

(ii) 6p – 12q

(iii) 7a² + 14a

(iv) – 16 z + 20 z³

(v) 20 l² m + 30 a l m

(vi) 5 x² y – 15 xy²

(vii) 10 a² – 15 b² + 20 c²

(viii) – 4 a² + 4 ab – 4 ca

(ix) x² y z + x y²z + x y z²

(x) a x² y + b x y² + c x y z²

Answer

In these problem, we need to factorize each term and then find common factors to factorize the expression

 

1) 7x-42

=  (7×x) – (7×6)

=7(x-6)
 

2) 6p-12q

=  (6×p) – (6×2q)

=6(p-2q)
 

3) 7a2+14a

=(7×a×a) +  (2×7×a)

=7a(a+2)

4) -16z+ 20z3

=4z(5z2-4)

 

5) 20 l² m + 30 a l m

=(2×2×5×l×m×l)  + (2×3×5×a×l×m)

=2lm(10l+15a)

 

6) 5 x²y -15 xy2

=(5×x×x×y)  + (5×3×x×y×y)

=5x(xy-3y2)

 

7) 10a² -15 b2+20c2

=(2×5×a×a)  - (5×3×b×b) + (5×2×2×c×c)

=5(2a2-3b2+4c2)

 

8) -4a² +4a b-4ca

=(-4×a×a)  +(4×b×a) - (4×c×a)

=-4a(a-b+c)

 

9)  x²yz + xy2z  +xyz2

=(x×x×y×z)  +(x×y×y×z) (x×y×z×z)

=xyz(x+y+z)

10) ax²yz + bxy2z  +cxyz2

=(x×x×y×z×a)  +(x×y×y×z×b) (x×y×z×z×c)

=xyz(ax+by+cz)

 

Question 3

Factorize

(i) x² + x y + 8x + 8y

(ii) 15 xy – 6x + 5y – 2

(iii) ax + bx – ay – by

(iv) 15 pq + 15 + 9q + 25p

(v) z – 7 + 7 x y – x y z

Answer

1) x² + x y + 8x + 8y

= x(x+y) + 8(x+y)

=(x+y)(x+8)

2) 15 xy – 6x + 5y – 2

= 3x(5y-2) +1(5y-2)

=(3x+1)(5y-2)

3) ax + bx – ay – by

= x(a+b) –y(a+b)

=(a+b)(x-y)

4) 15 pq + 15 + 9q + 25p

=15pq+9q +25p+15

= 3q(5p+3) + 5(5p+3)

=(3q+5)(5p+3)

5) z – 7 + 7 x y – x y z

=z-xyz-7+7xy

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

=(z-7)(1-xy)

 


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