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

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

The prime factorization of 84 is 2 x 2 x 3 x 7 x 1 = 22 x 3 x 7 x 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.

Factors of 70 are any integer that can be multiplied by another integer to make exactly 70.

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)

Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. There are 8 integers that are factors of 24.

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)

Factors of 144 are any whole number that can be multiplied by another whole number to make exactly 144.