S.chand publication New Learning Composite mathematics solution of class 8 Chapter 6 Factorisation of Algebraic Expressions Exercise 6A

 Exercise 6A

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Q1 | Ex-6A | Class 8 | SChand | New Learning Composite maths | Factorisation of Algebraic Expressions | myhelper

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Question 1

Complete the factorisation.

(a) 4x + 8y = 4(___)

(b) 9m – 3 = 3(____)

(c) 5b² – 20b = 5b (____)

(d) -3ax + 6ay = -3a (___)

(e) 3x²y² – xy² = xy²(___)

(f) 14a³b – 7a²b² = 7a²b(___)

(g) –xy – 3x = -x (___)

(h) 8x³ – 28x²– 4x = 4x(___)

Sol :

(a) 4x + 8y = 4(x+4y)

(b) 9m – 3 = 3(3m-1)

(c) 5b² – 20b = 5b (b-4)

(d) -3ax + 6ay = -3a (x-2y)

(e) 3x²y² – xy² = xy²(3x-1)

(f) 14a³b – 7a²b² = 7a²b(2a-b)

(g) –xy – 3x = -x (y+3)

(h) 8x³ – 28x²– 4x = 4x(2x²-7x-1)

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Q2 | Ex-6A | Class 8 | SChand | New Learning Composite maths | Factorisation of Algebraic Expressions | myhelper

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Question 2

One factor is given for each polynomial. Write the other factor.

(a) 7x + 28 = (___) (x + 4)

(b) 15y – 3 = (____) (5y – 1)

(c) 600x – x² = (___) (600 –x)

(d) – 3x – 27 = (___) (x + 9)

(e) 50x⁵ – 25x⁴ + 100x³ = 25x³ (____)

(f) -3y² – 9y³ = (___) (1 + 3y)

Sol :

(a) 7x + 28 = (7) (x + 4)

(b) 15y – 3 = (3) (5y – 1)

(c) 600x – x² = (x) (600 –x)

(d) – 3x – 27 = (-3) (x + 9)

(e) 50x⁵ – 25x⁴ + 100x³ = 25x³ (2x²-x+4)

(f) -3y² – 9y³ = (-3y²) (1 + 3y)

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Q3 | Ex-6A | Class 8 | SChand | New Learning Composite maths | Factorisation of Algebraic Expressions | myhelper

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Question 3

Factorise the following completely.

(a) ax²+ay

(b) 4a²–2a²c

(c) 28p³–42p²q

(d) -15p²–20pr

(e) –m²n–mn²

(f) 48y⁵z³ – 64y⁴z⁴

(g) 21y³ – 14y² – 7y

(h) 16x³y² – 20x²y³ – 28x²y²z

Sol :

(a) ax² + ay=a(x²+y)

(b) 4a² – 2a²c=2a²(2-c)

(c) 28p³ – 42p²q=14p²(2p-3q)

(d) -15p² – 20pr=-5p(3p+4r)

(e) –m²n – mn²=-mn(m+n)

(f) 48y⁵z³ – 64y⁴z⁴=16y⁴z³(3y-4z)

(g) 21y³ – 14y²– 7y=7y(3y²-2y-1)

(h) 16x³y² – 20x²y³ – 28x²y²z=4x²y²(4x-5y-7z)