Keeping the examination point of view in mind the skmath.in team has prepared NCERT Solutions for Class-8-Maths-Factorisation Ex 12.1. The NCERT Solutions for chapter Factorisation solutions explains the easy and simple way to solve the problems. By understanding these ways in NCERT Solutions for Class 8, students will be confident while solving such problems found in Class-8 Maths- Factorisation Ex 12.1.
Question 1: Find the common factors of the terms
(i) 12x, 36
Solution:
\(12x = 2 × 2 × 3 × x\)
\(36 = 2 × 2 × 3 × 3\)
The common factors are 2, 2, 3.
And, \(2 × 2 × 3 = 12\)
\(36 = 2 × 2 × 3 × 3\)
The common factors are 2, 2, 3.
And, \(2 × 2 × 3 = 12\)
(ii) \(2y, 22xy\)
Solution:
\(2y = 2 × y\)
\(22xy = 2 × 11 × x × y\)
The common factors are \(2 × y = 2y\)
\(22xy = 2 × 11 × x × y\)
The common factors are \(2 × y = 2y\)
(iii) \(14pq, 28p^2q^2\)
Solution:
14pq = 2 × 7 × p × q
\(28p^2q^2 = 2 × 2 × 7 × p × p × q × q\)
The common factors are 2, 7, p, q=14pq
\(28p^2q^2 = 2 × 2 × 7 × p × p × q × q\)
The common factors are 2, 7, p, q=14pq
(iv) \(2x, 3x^2, 4\)
Solution:
2x = 2 × x
\(3x^2 = 3 × x × x\)
4 = 2 × 2
The common factor is 1.
\(3x^2 = 3 × x × x\)
4 = 2 × 2
The common factor is 1.
(v) \(6abc, 24ab^2, 12a^2b\)
Solution:
6abc = 2 × 3 × a × b × c
\(24ab^2\) = 2 × 2 × 2 × 3 × a × b × b
\(12a^2b\) = 2 × 2 × 3 × a × a × b
The common factors are 2 × 3 × a × b = 6ab
\(24ab^2\) = 2 × 2 × 2 × 3 × a × b × b
\(12a^2b\) = 2 × 2 × 3 × a × a × b
The common factors are 2 × 3 × a × b = 6ab
(vi) \(16x^3, −4x^2, 32x\)
Solution:
\(16x^3 = 2 × 2 × 2 × 2 × x × x × x\)
\(−4x^2 = −1 × 2 × 2 × x × x\)
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors are 2 × 2 × x = 4x
\(−4x^2 = −1 × 2 × 2 × x × x\)
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors are 2 × 2 × x = 4x
(vii) 10pq, 20qr, 30rp
Solution:
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
The common factors are 2 × 5 = 10
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
The common factors are 2 × 5 = 10
(viii) \(3x^2y^3, 10x^3y^2, 6x^2y^2z\)
Solution:
\(3x^2y^3 = 3 × x × x × y × y × y\)
\(10x^3y^2 = 2 × 5 × x × x × x × y × y\)
\(6x^2y^2z = 2 × 3 × x × x × y × y × z\)
The common factors are \(x × x × y × y = x^2y^2\)
\(10x^3y^2 = 2 × 5 × x × x × x × y × y\)
\(6x^2y^2z = 2 × 3 × x × x × y × y × z\)
The common factors are \(x × x × y × y = x^2y^2\)
Question 2: Factorise the following expressions
(i) 7x − 42
Solution:
7x = 7 × x
42 = 2 × 3 × 7
The common factor is 7.
∴ 7x − 42 = (7 × x) − (2 × 3 × 7) = 7 (x − 6)
42 = 2 × 3 × 7
The common factor is 7.
∴ 7x − 42 = (7 × x) − (2 × 3 × 7) = 7 (x − 6)
(ii) 6p − 12q
Solution:
Solution:
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
The common factors are 2 and 3.
∴ 6p − 12q = (2 × 3 × p) − (2 × 2 × 3 × q)
= 2 × 3 [p − (2 × q)]
= 6 (p − 2q)
(iii) \(7a^2 + 14a\)
Solution:
\(7a^2 = 7 × a × a\)
14a = 2 × 7 × a
The common factors are 7 and a.
\(∴ 7a^2 + 14a = (7 × a × a) + (2 × 7 × a)\)
= 7 × a [a + 2] = 7a (a + 2)
(iv) \(−16z + 20z^3\)
Solution:
16z = 2 × 2 × 2 × 2 × z
\(20z^3 = 2 × 2 × 5 × z × z × z\)
The common factors are 2, 2, and z.
\(∴ −16z + 20z^3 = − (2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)\)
= (2 × 2 × z) [− (2 × 2) + (5 × z × z)]
= 4z (− 4 + 5z2)
(v) \(20l^2m + 30 alm\)
Solution:
\(20l^2m = 2 × 2 × 5 × l × l × m\)
\(30alm = 2 × 3 × 5 × a × l × m\)
The common factors are 2, 5, l, and m.
\(∴ 20l^2m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)\)
= (2 × 5 × l × m) [(2 × l) + (3 × a)]
= 10lm (2l + 3a)
(vi) \(5x^2y − 15xy^2\)
Solution:
\(5x^2y = 5 × x × x × y\)
\(15xy^2 = 3 × 5 × x × y × y\)
The common factors are 5, x, and y.
\(∴ 5x^2y − 15xy^2 = (5 × x × x × y) − (3 × 5 × x × y × y)\)
= 5 × x × y [x − (3 × y)]
= 5xy (x − 3y)
(vii) \(10a^2 − 15b^2 + 20c^2\)
Solution:
\(10a^2 = 2 × 5 × a × a\)
\(15b^2 = 3 × 5 × b × b\)
\(20c^2 = 2 × 2 × 5 × c × c\)
The common factor is 5.
\(10a^2 − 15b^2 + 20c^2 = (2 × 5 × a × a) − (3 × 5 × b × b) + (2 × 2 × 5 × c × c)\)
\(= 5 [(2 × a × a) − (3 × b × b) + (2 × 2 × c × c)]\)
\(= 5 (2a^2 − 3b^2 + 4c^2)\)
(viii) \(−4a^2 + 4ab − 4 ca\)
Solution:
\(4a2 = 2 × 2 × a × a\)
\(4ab = 2 × 2 × a × b\)
\(4ca = 2 × 2 × c × a\)
The common factors are 2, 2, and a.
\(∴ −4a^2 + 4ab − 4ca = − (2 × 2 × a × a) + (2 × 2 × a × b) − (2 × 2 × c × a)\)
= 2 × 2 × a [− (a) + b − c]
= 4a (−a + b − c)
(ix) \(x^2yz + xy^2z + xyz^2\)
Solution:
\(15xy^2 = 3 × 5 × x × y × y\)
The common factors are 5, x, and y.
\(∴ 5x^2y − 15xy^2 = (5 × x × x × y) − (3 × 5 × x × y × y)\)
= 5 × x × y [x − (3 × y)]
= 5xy (x − 3y)
(vii) \(10a^2 − 15b^2 + 20c^2\)
Solution:
\(10a^2 = 2 × 5 × a × a\)
\(15b^2 = 3 × 5 × b × b\)
\(20c^2 = 2 × 2 × 5 × c × c\)
The common factor is 5.
\(10a^2 − 15b^2 + 20c^2 = (2 × 5 × a × a) − (3 × 5 × b × b) + (2 × 2 × 5 × c × c)\)
\(= 5 [(2 × a × a) − (3 × b × b) + (2 × 2 × c × c)]\)
\(= 5 (2a^2 − 3b^2 + 4c^2)\)
(viii) \(−4a^2 + 4ab − 4 ca\)
Solution:
\(4a2 = 2 × 2 × a × a\)
\(4ab = 2 × 2 × a × b\)
\(4ca = 2 × 2 × c × a\)
The common factors are 2, 2, and a.
\(∴ −4a^2 + 4ab − 4ca = − (2 × 2 × a × a) + (2 × 2 × a × b) − (2 × 2 × c × a)\)
= 2 × 2 × a [− (a) + b − c]
= 4a (−a + b − c)
(ix) \(x^2yz + xy^2z + xyz^2\)
Solution:
\(x^2yz = x × x × y × z\)
\(xy^2z = x × y × y × z\)
\(xyz^2 = x × y × z × z\)
The common factors are x, y, and z.
∴ \(x^2yz + xy^2z + xyz^2 = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)\)
\(xy^2z = x × y × y × z\)
\(xyz^2 = x × y × z × z\)
The common factors are x, y, and z.
∴ \(x^2yz + xy^2z + xyz^2 = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)\)
= x × y × z [x + y + z]
= xyz (x + y + z)
(x) \(ax^2y + bxy^2 + cxyz\)
Solution:
\(ax^2y = a × x × x × y\)
\(bxy^2 = b × x × y × y\)
\(cxyz = c × x × y × z\)
The common factors are x and y.
\(ax^2y + bxy^2 + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)\)
\(= (x × y) [(a × x) + (b × y) + (c × z)]\)
\(= xy (ax + by + cz)\)
\(bxy^2 = b × x × y × y\)
\(cxyz = c × x × y × z\)
The common factors are x and y.
\(ax^2y + bxy^2 + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)\)
\(= (x × y) [(a × x) + (b × y) + (c × z)]\)
\(= xy (ax + by + cz)\)
Question 3: Factorise
(i) \(x^2 + xy + 8x + 8y\)
Solution:
\(x^2 + xy + 8x + 8y = x × x + x × y + 8 × x + 8 × y\)
\(= x (x + y) + 8 (x + y)\)
\(= (x + y) (x + 8)\)
(ii) \(15xy − 6x + 5y − 2\)
Solution:
\(15xy − 6x + 5y − 2 = 3 × 5 × x × y − 3 × 2 × x + 5 × y − 2\)
\(= 3x (5y − 2) + 1 (5y − 2)\)
\(= (5y − 2) (3x + 1)\)
(iii) \(ax + bx − ay − by\)
Solution:
\(ax + bx − ay − by = a × x + b × x − a × y − b × y\)
\(= x (a + b) − y (a + b)\)
\(= (a + b) (x − y)\)
(iv) \(15pq + 15 + 9q + 25p\)
Solution:
\(15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15\)
\(= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5\)
\(= 3q (5p + 3) + 5 (5p + 3)\)
\(= (5p + 3) (3q + 5)\)
(v) \(z − 7 + 7xy − xyz\)
Solution:
\(z − 7 + 7xy − xyz = z − x × y × z − 7 + 7 × x × y\)
\(= z (1 − xy) − 7 (1 − xy)\)
\(= (1 − xy) (z − 7)\)
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