NCERT Class 8 Maths Chapter 14 Factorisation Exercise 14.1

Chapter 14 Factorisation 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, 6x²y²z
Answer:

i)
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
2, 2, 3 are common.
= 2 × 2 × 3
= 12
Therefore, 12 is the common factor of the given terms.

ii)
2y = 2 × y
22xy = 2 × 11 × x × y
2, y are common.
= 2 × y
= 2y
Therefore, 2y is the common factor of the given terms.

iii)
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
2, 7, p, q are common.
= 2 × 7 × p × q
= 14pq
Therefore, 14pq is the common factor of the given terms.

iv)
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
Nothing is in common, so 1 is the common factor of the given terms.

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

vi)
16x³ = 2 × 2 × 2 × 2 × x × x × x
-4x² = 2 × 2 × -1 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
2, 2, x are common.
= 2 × 2 × x
= 4x
Therefore, 4x is the common factor of the given terms.

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

viii)
3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
x, x, y, y are common.
= x × x × y × y
= x²y²
Therefore, x²y² is common factor of the given terms.

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

i)
7x = 7 × x
-42 = 7 × -2 × 3
7 is common.
= 7(x - 6)

ii)
6p = 2 × 3 × p
-12q = -2 × 2 × 3 × q
2, 3 are common.
= 2 × 3 = 6
= 6(p - 2q)

iii)
7a² = 7 × a × a
14a = 7 × 2 × a
7, a are common.
= 7 × a = 7a
= 7a(a + 2)

iv)
-16z = -2 × 2 × 2 × 2 × z
20z³ = 2 × 2 × 5 × z × z × z
2, 2, z are common.
= 2 × 2 × z
= 4z
= 4z(-4 + 5z²)

v)
20l²m = 2 × 2 × 5 × l × l × m
30alm = 2 × 3 × 5 × a × l × m
2, 5, l, m are common.
= 2 × 5 × l × m = 10lm
= 10lm(2l + 3a)

vi)
5x²y = 5 × x × x × y
-15xy² = 5 × -3 × x × y × y
5, x, y are common.
= 5 × x × y = 5xy
= 5xy(x - 3y)

vii)
10a² = 2 × 5 × a × a
-15b² = -3 × 5 × b × b
20c² = 2 × 2 × 5 × c × c
5 is common.
= 5
= 5(2a² - 3b² + 4c²)

viii)
-4a² = 2 × 2 × -1 × a × a
4ab = 2 × 2 × a × b
-4ca = 2 × 2 × -1 × c × a
2, 2, a are common.
= 2 × 2 × a = 4a
= 4a(-a + b - c)

ix)
x²yz = x × x × y × z
xy²z = x × y × y × z
xyz² = x × y × z × z
x, y, z are common.
= x × y × z = xyz
= xyz(x + y + z)

x)
ax²y = a × x × x × y
bxy² = b × x × y × y
cxyz = c × x × y × z
x, y are common.
= x × y = xy
= 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
Answer:

i)
= x² + xy + 8x + 8y
= (x² + xy) + (8x + 8y)
= x(x + y) + 8(x + y)
= (x + y)(x + 8)

ii)
= 15xy - 6x + 5y - 2
= 15xy + 5y - 6x - 2
= (15xy + 5y) + (-6x - 2)
= 5y(3x + 1) - 2(3x + 1)
= (3x + 1)(5y - 2)

iii)
= ax + bx - ay - by
= (ax + bx) + (-ay - by)
= x(a + b) - y( a + b)
= (a + b)(x - y)

iv)
= 15pq + 15 + 9q + 25p
= (15pq + 9q) + (25p + 15)
= 3q(5p + 3) + 5(5p + 3)
= (5p + 3)(3q + 5)

v)
= z - 7 + 7xy - xyz
= (z - 7) + (7xy - xyz)
= 1(z - 7) + xy(7 - z)
= 1(z - 7) + xy(-z + 7)
= 1(z - 7) - xy(z - 7)
= (z - 7)(1 - xy)

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