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

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)

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

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)

MP Board Class 8th Maths Solutions

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