MP Board Class 8th Maths Solutions Chapter 14 Factorization Ex 14.1

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

(ii) The given terms are 2y and 22xy
2y = 2 xy
22xy = 2 × 11 × X × y
Thus, the common factor of the given terms is 2 × y = 2y.

(iii) The given terms are 14pq and 18p²q²
14 pq = 2 × 7 × p × q
18p2q2 = 2 × 2 × 7 × p × p × q × q
Thus, the common factors of the given terms is 2 × 7 × p × q = 14pq

(iv) The given terms are 2x, 3x² and 4
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
Hence, the given three terms have no factor in common except 1.

(v) The given terms are 6abc, 24ab² and 12a²b
6abc= 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b.
12 a²b = 2 × 2 × 3 × a × a × b
Thus, the common factor of the given terms is 2 × 3 × a × b = 6ab.

(vi) The given terms are 16x³, -4x² and 32x
16x³ = 2 × 2 × 2 × 2 × x × x × x
– 4x² = -1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
Thus, the common factor of the given terms is 2 × 2 × x = 4x.

(vii) The given terms are 10pq, 20qr and 30rp
10pq = 2 × 5 × p × q
20 qr = 2 × 2 × 5 × q × r
30 rp.= 2 × 3 × 5 × r × p
Thus, the common factor of the given terms is 2 × 5 = 10.

(viii)The given terms are 3x²y³, 10x³y² and 6x²y²z
3x²y² = 3 × x × x × y × y × y
10x³y³ = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × Z
Thus, the common factor of the given terms is x × x y × y = x²y².

Question 2.
Factorise the following expressions.
(i) 7x – 42
(ii) 6p – 12g
(iii) 7a² + 14a
(iv) -16z + 20x³
(v) 20l²m + 30alm
(vi) 5x²y – 15xy²
(vii) 10a² – 15b² + 20c²
(viii) – 4a² + 4ab – 4ca
(ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz
Solution:
(i) The expression is 7x – 42
Factors of 7x = 7 × x and 42 = 2 × 3 × 7
∴ 7x – 42 = 7 × x – 2 × 3 × 7 = 7(x – 2 × 3) = 7(x – 6)

(ii) The expression is 6p – 12q
Factors of 6p = 2 × 3 × p and
12q = 2 × 2 × 3 × q
∴ 6p – 12 q = 2 × 3 × p – 2 × 2 × 3 × q
= 2 × 3(p – 2 × q) = 6(p – 2 q)

(iii) The expression is 7a² + 14a
Factors of 7a² = 7 × a × a and
14a = 2 × 7 × a
∴ 7a² + 14a = 7 × a × a + 2 × 7 × a
= 7 × a(a + 2) = 7a (a + 2)

(iv) The expression is -16z + 20z³
Factors of -16z = – 1 × 2 × 2 × 2 × 2 × z and
20z³ = 2 × 2 × 5 × z × z × z
∴ -16z + 20z³ = -1 × 2 × 2 × 2 × 2 × z + 2 × 2 × 5 × z × z × z
= 2 × 2 × z(-1 × 2 × 2 + 5 × z × z)
= 4z (- 4 + 5z²)

(v) The expression is 20l²m + 30alm
Factors of 20l²m = 2 × 2 × 5 × l × l × m and
30alm = 2 × 3 × 5 × a × l × m;
∴ 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) The expression is 5x²y – 15xy²
Factors of 5x²y = 5 × x × x × y and
15xy² = 3 × 5 × x × y × y
∴ 5x²y – 15xy² = 5 × x × x × y – 3 × 5 × x × y × y
= 5 × x × y (x – 3 × y) = 5xy (x – 3y).

(vii) The expression is 10a² – 15b² + 20c²
Factors of 10a² = 2 × 5 × a × a
-15b² = (-1) × 3 × 5 × b × b
20c² = 2 × 2 × 5 × c × c
∴ 10a² – 15b² + 20c²
= 2 × 5 × a × a – 3 × 5 × b × b + 2 × 2 × 5 × c × c
= 5(2a² – 3b² + 4c²)

(viii) The expression is – 4a² + 4ab – 4ca
Factors of – 4a² = (-1) × 2 × 2 × a × a
4ab = 2 × 2 × a × b and
-4ca = -1 × 2 × 2 × c × a
∴ -4a² + 4ab – 4ca = -1 × 2 × 2 × a × a + 2 × 2 × a × b – 1 × 2 × 2 × c × a
= 2 × 2 × a (-1 × a + b – 1 × c)
= 4a(-a + b – c)

(ix) The expression is x²yz + xy²z + xyz²
Factors of x²yz = x × x × y × z and
xy²z = x × y × y × z and xyz² = x × y × z × z
∴ x² yz + 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) The expression is ax²y + bxy² + cxyz
Factors of ax²y = a × x × x × y,
bxy² = b × x × y × y and cxyz = c × x × y × z
∴ ax²y + bxy² + cxyz = a × x × x × y + b × x × y × y + c × x × y × z
= xy (a × x + b × y + c × z) = xy (ax + by + cz)

Question 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) The given expression is
x² + xy + 8x + 8y
= x × x + x × y + 8 × x + 8xy
= x(x + y) + 8(x + y) = (x + 8) (x + y).

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

(iii) The given expression is ax + bx – ay – by
= x (a + b) – y (a + b) = (x – y) (a + b)

(iv) The given expression is
15 pq + 15 + 9 q + 25 p
= 15pq + 25p + 15 + 9q = 5p(3q + 5) + 3(5 + 3q)
= (5p + 3) (3q + 5)

(v) The given expression is z – 7 + 7xy – xyz
= z – 7 + 7 × x × y – x × y × z
= z – 7 + xy (7 – z) = 1 (z – 7) – xy (z – 7)
= (z – 7)(1 – xy)