RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 14 Factorization Ex 14.1

Question 1.
Find the common factors of the given terms :
(i) 12x, 36
Answer:
The numerical coefficients in the given monomials are 12 and 36.
The highest common factor of 12 and 36 is 12. But there is no common literal appearing in the given monomials 12 and 36.
∴ The highest common factor =12

(ii) 2y, 22xy
Answer:
The numerical coefficients of the given monomials are 2 and 22.
The highest common factor of 2 and 22 is 2. The common literal appearing in the given monomial is y.
The smallest power of y in the two monomials = 1
The monomial of common literals with smallest powers = y
∴ The highest common factor = 2y

(iii) 14pq, 28p²q²
Answer:
The numerical coefficients of the given monomials are 14 and 18.
The highest common factor of 14 and 28 is 14.
The common literals appearing in the given monomials are p and q.
The smallest power of p and q in the two monomials = 1
The monomials of common literals with smallest powers = pq
∴ The highest common factor = 14pq

(iv) 2x, 3x², 4
Answer:
The numerical coefficients of the given monomials are 2, 3 and 4.
The highest common factor of 2, 3 and 4 is 1.
There is no common literal appearing in the three monomials.
∴ The highest common factor = 1.

(v) 6abc, 24ab², 12a²b
Answer:
The numerical coefficients of the given monomials are 6, 24 and 12.
The highest common factor of 6, 24 and 12 is 6.
The common literals appearing in the three monomials are a and b.
The smallest power of a and b in the three monomials =1
The monomial of common literals with smallest powers = ab
∴ The highest common factor = 6ab

(vi) 16x³, - 4x², 32x
Answer:
The numerical coefficients of the given monomials are 16, 4 and 32.
The highest common factor of 16, 4 and 32 is 4.
The common literal appearing in the three monomials is x.
The smallest power of x in the three monomials = 1
The mondmial of common literal with smallest power = x
∴ Hence, the highest common factor = 4x

(vii) 10pq, 20qr, 30rp
Answer:
The numerical coefficients of the given monomials are 10, 20 and 30.
The highest common factor of 10, 20 and 30 is 10.
There is no common literal appearing in die three monomials.
Hence, the highest common factor = 10

(viii) 3x²y³, 10x³y², 6x²y²z
Answer:
The numerical coefficients of the given monomials are 3, 10 and 6.
The highest common factor of 3, 10 and 6 is 1.
The common literals appearing in the three monomials are x and y.
The smallest power of x and y in the three monomials = 2
The monomials of common literals with smallest powers = x²y²
The highest common factor = x²y²

Question 2.
Factorise the following expressions.
(i) 7x - 42
Answer:
We have 7x = 7 × x
and 42 = 2 × 3 × 7
common factor of two terms is 7.
∴ 7x - 42 = (7 × x) - 2 × 3 × 7
= 7 × (x - 2 × 3)
= 7(x - 6)

(ii) 6p - 12q
Answer:
We have, 6p = 2 × 3 × p
and 12q = 2 × 2 × 3 × q
The two terms have 2 and 3 as a common factor
.'. 6p - 12q = (2 × 3 × p) - (2 × 2 × 3 × q) = 2 × 3 × (p - 2 × q)
= 6(p - 2q)

(iii) 7a² + 14a
Answer:
We have, 7a² = 7 × a × a
and 14a = 2 × 7 × a
The two terms have 7 and 9 as a common factor
7a² + 14a = (7 × a × a) + (2 × 7 × a) = 7 × a × (a + 2)
= 7a(a + 2)

(iv) - 16z + 20z³
Answer:
We have 16z = 2 × 2 × 2 × 2 × z
and 20z³ = 2 × 2 × 5 × z × z × z
The two terms have 2, 2 and z as common factors
- 16z + 20z³ = (-2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)
= 2 × 2 × z(-2 × 2 + 5 × z × z) = 4z(- 4 + 5z²)

(v) 20l²m + 30alm
Answer:
We have, 20l²m = 2 × 2 × 5 × l × l × m
and 30 alm = 3 × 2 × 5 × a × l × m
The two terms have 2, 5, l and m as a common factors
20l2m + 30alm = (2 × 2 × 5 × l × l × m) + (3 × 2 × 5 × a × l × m)
= 2 ×5 × l × m × (2 × l + 3 × a)
= 10lm(2l + 3a)

(vi) 5x²y - 15xy²
Answer:
We have, 5x2y = 5 × x × x × y
and 15xy2 = 3 × 5 × x × y × y
The two terms have 5, x and y as a common factors
5x²y - 15xy² = (5 × x × x × y) - (3 × 5 × x × y)
= 5 × x × y × (x - 3xy)
= 5xy(x - 3y)

(vii) 10a² - 15b² + 20c²
Answer:
We have, 10a² = 2 × 5 × a × a,
15b² = 3 × 5 × b × b
and 20c² = 2 × 2 × 5 × c × c
The three terms have 5 as a common factor 10a² -15b² + 20c² =(2 × 5 × a × a) - (3 × 5 × b × b × b) + (2 × 2 × 5 × c × c)
= 5 × (2 × a × a - 3 × b × b + 4 × c × c)
= 5(2a² - 3b² + 4c²)

(viii) - 4a² + 4ab - 4ca
Answer:
We have, 4a2 = 2 × 2 × a × a
4ab = 2 × 2 × a × b
4ca= 2 × 2 × c × a

The three terms have 2, 2 and a as common factors
∴ - 4a² + 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) x²yz + xy²z + xyz²
Answer:
We have, x²yz = x × x × y × z
xy²z = x × y × y × z
and xyz² = x × y × z × z

The three terms have x, y and z as common factors
x²yz + x²yz + xyz² = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)
= x × y × z × (x + y + z)
= xyz (x + y + z)

(x) ax²y + bxy² + cxyz
Answer:
We have, ax²y = a × x × x × y
bxy² = b × x × y × y
and cxyz² = c × x × y × z
The three terms have x and y as common factors
ax²y + bxy² + 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) 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 - 6x) + (5y - 2)
. = 3x(5y - 2) + 1(5y - 2)
= (5y - 2) (3x + 1)

(iii) ax + bx - ay - by = (ax + bx) - (ay + by)
[combining the terms]
= (a + b) x - (a + b)y
= (a + b) (x - y)

(iv) 15pq +15 + 9q + 25p
= 15pq + 9q + 25p + 15
[Rearranging the terms]
= 3 q(5p + 3) + 5(5p+ 3)
= (5p + 3) (3q + 5)

(v) z - 7 + 7xy - xyz = z - 7 - xyz + 7xy
[Rearranging the terms]
= 1(z - 7) - xy(z - 7)
= (z - 7) (1 -xy)