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Ncert Solutions for Class 8 Mathematics Chapter 14 - Factorization CBSE Part 1



In this page we have NCERT book Solutions for Class 8th Mathematics:Factorization Chapter 14 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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