NCERT Solutions Class 8 Maths Chapter 14 Factorisation Ex 14.1

NCERT Solutions Class 8 Mathematics 
Chapter – 14 (Factorisation) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 14 Factorisation  Exercise 14.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter 14: Factorisation

• NCERT Solution Class 8 Maths Ex – 14.2
• NCERT Solution Class 8 Maths Ex – 14.3
• NCERT Solution Class 8 Maths Ex – 14.4

Exercise – 14.1

1. Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14 pq, 28p²q²
(iv) 2x, 3x², 4
(v) 6 abc, 24ab², 12a²b
(vi) 16 x³, – 4x² , 32 x
(vii) 10 pq, 20qr, 30 rp
(viii) 3x²y³ , 10x³y² , 6x²y²z

Solution –

(i) 12x, 36
Factors of 12x and 36
12x = 2×2×3×x
36 = 2×2×3×3
Common factors of 12x and 36 are 2, 2, 3
∴ Required common factors = 2×2×3 = 12

(ii) 2y, 22xy
Factors of 2y and 22xy
2y = 2 × y
22xy = 2 × 11 × x × y
Common factors of 2y and 22xy are 2, y
∴ Required common factors =  2 × y = 2y

(iii) 14 pq, 28p²q²
Factors of 14pq and 28p²q²
14pq = 2 x 7 x p x q
28p²q² = 2 x 2 x 7 x p x p x q x q
Common factors of 14 pq and 28 p²q² are 2, 7 , p , q
∴ Required common factors = 2 x 7 x p x q = 14pq

(iv) 2x, 3x², 4
Factors of 2x, 3x²and 4
2x = 2×x
3x²= 3×x×x
4 = 2×2
Common factors of 2x, 3x² and 4 is 1.

(v) 6 abc, 24ab², 12a²b
Factors of 6abc, 24ab² and 12a²b
6abc = 2×3×a×b×c
24ab² = 2×2×2×3×a×b×b
12 a² b = 2×2×3×a×a×b
Common factors of 6 abc, 24ab² and 12a²b are 2, 3, a, b
∴ Required common factors = 2×3×a×b = 6ab

(vi) 16 x³, – 4x² , 32 x
Factors of 16x³ , -4x²and 32x
16 x³ = 2×2×2×2×x×x×x
– 4x² = -1×2×2×x×x
32x = 2×2×2×2×2×x
Common factors of 16 x³ , – 4x² and 32x are 2,2, x
∴ Required common factors = 2×2×x = 4x

(vii) 10 pq, 20qr, 30 rp
Factors of 10 pq, 20qr and 30rp
10 pq = 2×5×p×q
20qr = 2×2×5×q×r
30rp= 2×3×5×r×p
Common factors of 10 pq, 20qr and 30rp are 2, 5
∴ Required common factors = 2×5 = 10

(viii) 3x²y³ , 10x³y² , 6x²y²z
Factors of 3x²y³ , 10x³y² and 6x²y²z
3x²y³ = 3×x×x×y×y×y
10x³ y² = 2×5×x×x×x×y×y
6x²y²z = 3×2×x×x×y×y×z
Common factors of 3x²y³, 10x³y² and 6x²y²z are x², y²
∴ Required common factors = x²×y² = x²y²

2. Factorise the following expressions.
(i) 7x–42
(ii) 6p–12q
(iii) 7a²+ 14a
(iv) -16z+20 z³
(v) 20l²m+30alm
(vi) 5x²y-15xy²
(vii) 10a²-15b²+20c²
(viii) -4a²+4ab–4 ca
(ix) x²yz+xy²z +xyz²
(x) ax²y+bxy²+cxyz

Solution –
(i) 7x = 7 × x
42 = 2 × 3 × 7
The common factor is 7.
Therefore, 7x – 42 = (7 × x) – (2 × 3 × 7 )
= 7(x – 6)

(ii) 6p -12q
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
The common factors are 2 and 3.
Therefore,
6p -12q = (2 × 3 × p) – (2 × 2 × 3 × q)
= 2 × 3(p – 2 × q)
= 6(p – 2q)

(iii) 7a²+14a
7a²= 7 × a × a
14a = 2 × 7 × a
The common factors are 7 and a .
Therefore,
7a² + 14a = (7 × a × a) + (2 × 7 × a)
= 7 × a(a + 2)
= 7a(a + 2)

(iv) -16z + 20z³
16z = -1 × 2 × 2 × 2 × 2 × z
20z³ = 2 × 2 × 5 × z × z × z
The common factors are 2, 2, and z.
Therefore,
-16z + 20z³ 
= -(2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)
= (2 × 2 × z)[-(2 × 2) + (5 × z × z)]
= 4z(-4 + 5z²)

(v) 20l²m + 30alm
20l²m= 2 × 2 × 5 × l × l × m
30alm = 2 × 3 × 5 × a × l × m
The common factors are 2, 5, l and m.
Therefore,
20l²m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)
= (2 × 5 × l × m)[(2 × l) + (3 × a)]
= 10lm(2l + 3a)

(vi) 5x²y -15xy²
5x²y = 5 × x × x × y
15xy² = 3 × 5 × x × y × y
The common factors are 5, x, and y.
Therefore,
5x²y – 15xy² = (5 × x × x × y) – (3 × 5 × x × y × y)
= 5 × x × y[x – (3 × y)]
= 5xy(x – 3y)

(vii) 10a² -15b² + 20c²
10a² = 2 × 5 × a × a
15b² = 3 × 5 × b × b
20c² = 2 × 2 × 5 × c × c
The common factor is 5.
Therefore,
10a² -15b² + 20c² = (2 × 5 × a × a) – (3 × 5 × b × b) + (2 × 2 × 5 × c × c)
= 5[(2 × a × a) – (3 × b × b) + (2 × 2 × c × c)]
= 5(2a² – 3b² + 4c² )

(viii) -4a² + 4ab – 4ca
4a² = 2 × 2 × a × a
4ab = 2 × 2 × a × b
4ca = 2 × 2 × c × a
The common factors are 2, 2, and a .
Therefore, – 4a² + 4ab – 4ca = -(2 × 2 × a × a) + (2 × 2 × a × b) – (2 × 2 × c × a)
= 2 × 2 × a (-a + b – c)
= 4a(-a + b – c)

(ix) x²yz + xy²z + xyz²
x²yz = x × x × y × z
xy²z= x × y × y × z
xyz² = x × y × z × z
The common factors are x, y, and z.
Therefore, x²yz + xy²z + xyz² = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)
= x × y × z (x + y + z)
= xyz(x + y + z)

(x) ax²y + bxy² + cxyz
ax²y= a × x × x × y
bxy² = b × x × y × y
cxyz= c × x × y × z
The common factors are x and y.
Therefore,
ax²y + bxy² + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)
= (x × y)[(a × x) + (b × y) + (c × z)]
= xy(ax + by + cz)

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

Solution –

(i) x² + xy + 8x + 8y
= x × x + x × y + 8 × x + 8 × y
= x(x + y) + 8(x + y)
= (x + y) (x + 8)

(ii) 15xy – 6x + 5y – 2
= 3 × 5 × x × y – 3 × 2 × x + 5 × y – 2
= 3x (5y – 2) + 1(5y – 2)
= (5y – 2) (3x + 1)

(iii) ax + bx – ay – by
= a × x + b × x – a × y – b × y
= x(a + b) – y(a + b)
= (a + b)(x – y)

(iv) 15pq + 15 + 9q + 25 p
= 15pq + 9q + 25p + 15
= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5
= 3q(5p + 3) + 5(5p + 3)
= (5p + 3)(3q + 5)

(v) z – 7 + 7xy – xyz
= z – xyz – 7 + 7xy
= z – x × y × z – 7 + 7 × x × y
= z(1- xy) – 7(1 – xy)
= (1 – xy)(z – 7)