In this page we have NCERT solutions for class 8 maths chapter 14 Exercise 14.1 for Factorisation.This exercise has questions on finding common factors, Factorisation by regrouping terms. Hope you like them and do not forget to like , social share
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NCERT solutions for class 8 maths chapter 14 Exercise 14.1
Question 1
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14 pq, 28 p
2q
2
(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:
(i) 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
(ii) 2y=2×y
22xy=2×11×y×x
So common factors are 2,y
And 2×y=2y
(iii) 14 pq= 2×7×p×q
28 p
2q
2=2×2×7×p×p×q×q
So common factors are 2,7,p,q
=14pq
(iv) 2x=2×x
3x
2=3×x×x
4=2×2
There is no common factor other than unity, so common factor= 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
Common factors are 2, 3, a, b
=6ab
(vi) 16x
3=2×2×2×2×x×x×x
-4x
2=-2×2×x×x
32x=-2×2×2×2×2×x
Common factors are 2, 2, x
=4x
(vii) 10pq=-2×5×p×q
20qr=-2×2×5×q×r
30rp=-2×3×5×p×r
Common factors are 2,5
=10
(viii) 3x
2y
3=3×x×x×y×y×y
10x
3y
3=2×5×x×x×x×y×y×y
6x
2y
2z=-2×3×x×x×y×y×z
Common factors are x, x, y, y
=x
2y
2
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
(i) 7x-42
= (7×x) – (7×6)
=7(x-6)
(ii) 6p-12q
= (6×p) – (6×2q)
=6(p-2q)
(iii) 7a
2+14a
=(7×a×a) + (2×7×a)
=7a(a+2)
(iv) -16z+ 20z
3
=4z(5z
2-4)
(v) 20 l² m + 30 a l m
=(2×2×5×l×m×l) + (2×3×5×a×l×m)
=2lm(10l+15a)
(vi) 5 x²y -15 xy
2
=(5×x×x×y) + (5×3×x×y×y)
=5x(xy-3y
2)
(vii) 10a² -15 b
2+20c
2
=(2×5×a×a) - (5×3×b×b) + (5×2×2×c×c)
=5(2a
2-3b
2+4c
2)
(viii) -4a² +4a b-4ca
=(-4×a×a) +(4×b×a) - (4×c×a)
=-4a(a-b+c)
(ix) x²yz + xy
2z +xyz
2
=(x×x×y×z) +(x×y×y×z) +(x×y×z×z)
=xyz(x+y+z)
(x) ax²yz + bxy
2z +cxyz
2
=(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
(i) x² + x y + 8x + 8y
= x(x+y) + 8(x+y)
=(x+y)(x+8)
(ii) 15 xy – 6x + 5y – 2
= 3x(5y-2) +1(5y-2)
=(3x+1)(5y-2)
(iii) ax + bx – ay – by
= x(a+b) –y(a+b)
=(a+b)(x-y)
(iv) 15 pq + 15 + 9q + 25p
=15pq+9q +25p+15
= 3q(5p+3) + 5(5p+3)
=(3q+5)(5p+3)
(v) z – 7 + 7 x y – x y z
=z-xyz-7+7xy
=z(1-xy)-7(1-xy)
=(z-7)(1-xy)
Summary
- NCERT solutions for class 8 maths chapter 14 Exercise 14.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
Download NCERT solutions Factorisation Exercise 14.1 as pdf
- This chapter 14 has total 4 Exercise 14.1,14.2,14.3 and 14.4. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below
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