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# NCERT Solutions(Part- 1)- Factorisation Class 8 Notes | EduRev

## Mathematics (Maths) Class 8

Created by: Full Circle

## Class 8 : NCERT Solutions(Part- 1)- Factorisation Class 8 Notes | EduRev

The document NCERT Solutions(Part- 1)- Factorisation Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.
All you need of Class 8 at this link: Class 8

Question: Factorise (i) 12x + 36           (ii) 22y – 33z          (iii) 14pq + 35pqr
Solution:
(i) We have 12x = 3 * 2 * 2 * x = (2 * 2 * 3) * x   EXERCISE 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, 6x2y2xz

Solution: (ii) ∵                         2y = 2 X y = (2 X y)
and                        22y = 2 X 11 X y = (2 X y) X 11
∴ the common factor   = 2 X y
= 2y 3x2 = 1 * 3 * x * x = 1 * 3 * x
and                                  4 = 1 * 2 * 2 = 1 * 2 * 2
∴ the common factor = 1
Note: 1 is a factor of every term.

(v) ∵            6abc = 2 * 3 * a * b * c = (2 * 3 * a * b) * c
24 ab2 = 2 * 2 * 2 * 3 * a * b * b = (2 * 3 * a * b) * 2 * 2 * b
12 a2b = 2 * 2 * 3 * a * a * b = (2 * 3 * a * b) * 2 * a
∴ the common factor = 2 * 3 * a * b
= 6ab

(vi) ∵                   16x3 = 2 * 2 * 2 * 2 * x * x * x = (2 * x * x
–4x2 = –1 * 2 * 2 * x * x = (2 * 2 * x) * (–1) * x
32x = 2 * 2 * 2 * 2 * 2 * x = (2 * 2 * x) * 2 * 2 * 2
∴ the common factor = 2 * 2 * x
= 4x

(vii) ∵                10pq  = 2 * 5 * p * q = (2 * 5) * p * q
20qr = 2 * 2 * 5 * q * r = (2 * 5) * 2 * q * r
30rp = 2 * 3 * 5 * r * p = (2 * 5) * 3 * r * p
∴ the common factor  = 2 * 5
= 10

(viii) ∵                3x2y3 = 3 * x * x * y * y * y
= (x * x * y * y) * 3 * y
10x3y2 = 2 * 5 * x * x * x * y * y
= (x * x * y * y) * 2 * 5 * x
6x2y2z = 2 * 3 * x * x * y * y * z
= (x * x * y * y) * 2 * 3 * z
∴ the common factor   = (x * x * y * y)
= x2y2

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

Solution:
(i) ∵           7x = 7 * x = (7) * x
42 = 2 * 7 * 3 = (7) * 2 * 3
∴                        7x – 42 = 7[(x) – (2 * 3)]
= 7[x – 6]

(ii) ∵          6p = 2 * 3 * p = (2) * (3) * p = (2 * 3) * p
12q = 2 * 2 * 3 * q = (2) * 2 * (3) * q = (2 * 3) * 2 * q
∴     6p – 12q = (2 * 3) [(p) – (2 * q)]
= 6[p – 2q]

(iii) ∵           7a2 = 7 * a * a = (7 * a) * a
14a = 2 * 7 * a = (7 * a) * 2
∴       7a2 – 14a = (7 * a)[a + 2]
= 7a(a + 2)

(iv) ∵       –16z = (–1) * 2 * 2 * 2 * 2 * z = (2 * 2 * z) * (–1) * 2
20z3 = 2 * 2 * 5 * z * z * z = (2 * 2 * z) * 5 * z * z
∴  –16z + 20z3 = (2 * 2 * z) [(–1) * 4 + 5 * z * z]
= 4z[–4 + 5z2]

(v) ∵         20l2m = 2 * 2 * 5 * l * l * m = (2 * 5 * l * m) * 2 * l
30alm = 2 * 3 * 5 * a * l * m = (2 * 5 * l * m) * 3 * a
∴ 20l2m + 30alm = (2 * 5 * l * m)[ 2 * l + 3 * a]
= 10lm[2l + 3a]

(vi) ∵           5x2y = 5 * x * x * y = (5 * x * y)[x]
15xy2 = 5 * 3 * x * y * y = (5 * x * y)[3 * y]
∴    5x2y – 15xy2 = (5 * x * y)[x – 3 * y]
= 5xy(x – 3y)

(vii) ∵                10a2 = 2 * 5 * a * a = (5)[2 * a * a]
15b2 = 3 * 5 * b * b = (5)[3 * b * b]
20c2 = 2 * 2 * 5 c * c = (5)[2 * 2 * c * c]
∴ 10a2 – 15b2 + 20c2 = (5)[2 * a * a + 3b * b + 2 * 2 * c * c]
= 5[2a2 + 3b2 + 4c2]

(viii) ∵              –4a2 = (–1) * 2 * 2 * a * a = (2 * 2 * a)[(–1) * a]
4ab = 2 * 2 * a * b = (2 * 2 * a)[b]
–4ca = (–1) * 2 * 2 * c * a = (2 * 2 * a)[(–1) * c]
∴ –4a2 + 4ab – 4ca = (2 * 2 * a)[(–1) * a + b – c]
= 4a[–a + b –c]

(ix) ∵                  x2yz = x * x * y * z = (xyz)[x]
xy2z = x * y * y * z = (xyz)[y]
xyz2 = x * y * z * z = (xyz)[z]
∴ x2yz + xy2z + xyz2 = (xyz)[x + y + z]
= xyz(x + y + z)

(x) ∵                  ax2y = a * x * x * y = (x * y)[a * x]
bxy2 = b * x * y * y = (x * y)[b * y]
cxyz = c * x * y * z = (x * y)[c * z]
∴ ax2y + bxy2 + cxyz = (x * y)[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) x2 + xy + 8x + 8y = x[x + y] + 8[x + y]
= (x + y)(x + 8)

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

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

(iv) Regrouping the terms, we have
15pq + 15 + 9q + 25p = 15pq + 9q + 25 p + 15
= 3q[5p + 3] + 5[5p + 3]
= (5p + 3)[3q + 5]

(v) Regrouping the terms, we have
z – 7 + 7xy – xyz = z – 7 – xyz + 7xy
= 1[z – 7] – xy[z – 7]
= (z – 7)(1 – xy)

FACTORISATION USING IDENTITIES
We know that
(a + b)= a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a – b) = a2 – b2
Using these identities, we can say that
a2 + 2ab + b= (a + b)2
= (a + b)(a + b)
a2 – 2ab + b2 = (a – b)2
= (a – b)(a – b)
a2 – b2 = (a + b)(a – b)
Note: For factorising an expression of the type x2 + px + q [where p = (a + b) and q = (ab)], we have
x2 + px + q = x2 + (a + b)x + ab
= x2 + ax + bx + ab
= x(x + a) + b(x + a)
= (x + a)(x + b)

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