Class 8 NCERT Solutions - Chapter 14 Factorization - Exercise 14.1

In Class 8, Chapter 14 of the NCERT Mathematics textbook focuses on Factorization an essential algebraic concept. Factorization involves expressing a polynomial as a product of its factors. This chapter is critical as it lays the foundation for understanding algebraic expressions and equations. Exercise 14.1 specifically deals with the basic factorization techniques offering the students practice in simplifying and solving the polynomial expressions. Mastery of these concepts is crucial for tackling more advanced algebraic problems in the higher classes.

Factorization

Factorization is a process of breaking down complex algebraic expressions into simpler components, called factors that when multiplied together yield the original expression. In the context of Class 8, factorization primarily deals with the polynomials and their factors. It simplifies expressions making it easier to solve equations and inequalities. The importance of factorization extends to the various areas in mathematics including solving quadratic equations and simplifying rational expressions.

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)12x, 36

Factors of 12x and 36 are
⇒ 12x = 2 × 2 × 2 × 3 × x
⇒ 36 = 2 × 2 × 3 × 3
So, common factors are
⇒ 2 × 2 × 3 × 3 = 12

(ii) 2y, 22xy

Factors of 2y, 22xy
⇒ 2y = 2 × y
⇒ 22xy = 2 × 11 × x × y
So, common factors are
⇒ 2 × y = 2y

(iii) 14pq, 28p²q²

Factors of 14pq, 28p²q²
⇒ 14pq = 2 × 7 × p × q
⇒ 28p²q² = 2 × 2 × 7 × p × p × q × q
So, common factors are
⇒ 2 × 7 × p × q = 14pq

(iv) 2x, 3x², 4

Factors of 2x, 3x², 4
⇒ 2x = 2 × x
⇒ 3x² = 3 × x × x
⇒ 4 = 2 × 2
So, common factor is 1 (∵ 1 is a factor of every number)

(v) 6abc, 24ab², 12a²b

Factors of 6abc, 24ab², 12a²b
⇒ 6abc = 2 × 3 × a × b × c
⇒ 24ab² = 2 × 2 × 2 × 3 × a × b × b
⇒ 12a²b = 2 × 2 × 3 × a × a × b
So, common factors are
⇒ 2 × 3 × a × b = 6ab

(vi) 16x³, -4x², 32x

Factors of 16x³, -4x², 32x
⇒ 16x³ = 2 × 2 × 2 × 2 × x × x × x
⇒ -4x² = -1 × 2 × 2 × x × x
⇒ 32x = 2 × 2 × 2 × 2 × 2
So, common factors are
⇒ 2 × 2 × x = 4x

(vii) 10pq, 20qr, 30rp

Factors of 10pq, 20qr, 30rp
⇒ 10pq = 2 × 5 × p × q + 
⇒ 20qr = 2 × 2 × 5 × q × r
⇒ 30rp = 2 × 3 × 5 × r × p
So, common factors are
⇒ 2 × 5 = 10

(viii) 3x²y³, 10x³y², 6x²y²z

Factors of 3x²y³, 10x³y², 6x²y²z
⇒ 3x²y³ = 3 × x × x × y × y × y
⇒ 10x³y² = 2 × 5 × x × x × x × y × y
⇒ 6x²y²z = 2 × 3 × x × x × y × y
So, common factors are
⇒ x × x × y × y = x²y²

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

Solution: 

(i) 7x − 42

⇒ 7x = 7 × x
⇒ 42 = 2× 3 × 7
So, common factor is 7
Therefore, 7x − 42 = 7(x − 6)

(ii) 6p − 12q

⇒ 6p = 2 × 3 × p
⇒ 12q = 2 × 2 × 3 × q
So, common factors are 2 × 3
Therefore, 6p − 12q = 2 × 3[p − (2 × q)]
⇒ 6(p − 2q)

(iii) 7a² + 14a

⇒ 7a² = 7 × a × a
⇒ 14a = 2 × 7 × a
So, common factors are 7 × a
Therefore, 7a² + 14a = 7 × a(a + 2)
⇒ 7a(a + 2)

(iv) −16z + 20z³ 

⇒ 16z = 2 × 2 × 2 × 2 × z
⇒ 20z² = 2 × 2 × 5 × z × z × z
So, common factors are 2 × 2 × z
Therefore, −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

⇒ 20l²m = 2 × 2 × 5 × l × l × m
⇒ 30alm = 2 × 3 × 5 × a × l × m
So, common factors are 2 × 5 × l × m
Therefore, 20l²m + 30alm = 2 × 5 × l × m[(2 × l) + (3 × a)]
⇒ 10lm(2l + 3a)

(vi) 5x²y − 15xy²

⇒ 5x²y = 5×x×x×y
⇒ 15xy² = 3×5×x×y×y
So, common factors are 5×x×y
Therefore, 5x²y − 15xy² = 5×x×y[(x) − (3×y)]
⇒ 5xy(x − 3y)

(vii)  10a² − 15b² + 20c²

⇒ 10a² = 2×5×a×a
⇒ 15b² = 3×5×b×b
⇒ 20c² = 2×2×5×c×c
So, common factor is 5
Therefore, 10a² − 15b² +20c² = 5[(2×a×a) − (3×b×b) + (2×2×c×c)]
⇒ 5(2a² − 3b² + 4c²)

(viii) −4a² + 4ab − 4ca

⇒ 4a² = 2×2×a×a
⇒ 4ab = 2×2×a×b
⇒ 4ca = 2×2×c×a
So, common factors are 2×2×a = 4a
Therefore, −4a² + 4ab − 4ca = 4a(−a + b − c)

(ix) x²yz + xy²z + xyz²

⇒ x²yz = x×x×y×z
⇒ xy²z = x×y×y×z
⇒ xyz² = x×y×z×z
So, common factors are x×y×z = xyz
Therefore, x²yz + xy²z + xyz² = xyz(x +y + z)

(x) ax²y + bxy² + cxyz

⇒ ax²y = a×x×x×y
⇒ bxy² = b×x×y×y
⇒ cxyz = c×x×y×z
So, common factors are x×y = xy
Therefore, ax²y + bxy² + cxyz = 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 + x×y + 8×x + 8×y
Assembling the terms,
⇒ x(x + y) + 8(x + y)
Therefore, the factors are
⇒ (x + y)(x + 8)

(ii) 15xy − 6x + 5y − 2

⇒ 3×5×x×y − 2×3×x + 5×y − 2
Assembling the terms
⇒ 3x(5y − 2) + 1(5y − 2)
Therefore, the factors are
⇒ (5y − 2)(3x + 1)

(iii) ax + bx − ay − by

⇒ a×x + b×x − a×y − b×y
Assembling the terms
⇒ x(a + b) − y(a + b)
Therefore, the factors are
⇒ (a + b)(x − y)

(iv) 15pq + 15 + 9q + 25p

⇒ 3×5×p×q + 3×5 + 3×3×q + 5×5×p
Assembling the terms
⇒ 3q(5p + 3) + 5(5p + 3)
Therefore, the factors are
⇒ (5p + 3)(3q + 5)

(v) z − 7 + 7xy − xyz

⇒ z − 7 + 7×x×y − x×y×z
Assembling the terms
⇒ z(1 − xy) − 7(1 − xy)
Therefore, the factors are
⇒ (1 − xy)(z − 7) 

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