PSEB 8th Class Maths Solutions Chapter 14 Factorization InText Questions

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 14 Factorization InText Questions

Try These : [Textbook Page No. 219]

1. Factorise:

Question (i)
12x + 36
Solution:
12x = 2 × 2 × 3 × x and
36 = 2 × 2 × 3 × 3
Common factors = 2 × 2 × 3
∴ 12x + 36 = (2 × 2 × 3 × x) + (2 × 2 × 3 × 3)
= (2 × 2 × 3) (x + 3)
= 12 (x + 3)

Question (ii)
22y – 33z
Solution:
22y = 2 × 11 × y and 33z = 3 × 11 × z
Common factor =11
∴ 22y – 33z = (2 × 11 × y) – (3 × 11 × z)
= (11) × (2 × y – 3 × z)
= 11 (2y – 3z)

Question (iii)
14pq + 35pqr
Solution:
14pq = 2 × 7 × p × q and
35pqr = 7 × 5 × p × q × r
Common factor = 7pq
∴ 14pq + 35pqr = (2 × 7 × p × q) + (7 × 5 × p × q × r)
= 7 × p × q (2 + 5 × r)
= 7pq (2 + 5r)

Try These : [Textbook Page No. 225]

1. Divide:

Question (i)
24xy²z³ by 6yz²
Solution:
= \(\frac{24 x y^{2} z^{3}}{6 y z^{2}}\)
= \(\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\)
= \(\frac{2 \times 2 \times x \times y \times z}{1}\) = 4xyz
∴ 24xy²z³ ÷ 6yz²
= 4xyz

Question (ii)
63a²b⁴c⁶ by 7a²b²c³
Solution:
= \(\frac{63 a^{2} b^{4} c^{6}}{7 a^{2} b^{2} c^{3}}\)
= \(\frac{3 \times 3 \times 7 \times a^{2} \times b^{4} \times c^{6}}{7 \times a^{2} \times b^{2} \times c^{3}}\)
= 3 × 3 × \(\frac{a^{2}}{a^{2}} \times \frac{b^{4}}{b^{2}} \times \frac{c^{6}}{c^{3}}\)
= 9 × a^2-2 × b^4-2 × c^6-3
= 9 × a⁰ × b² × c³
= 9 × 1 × b² × c³
= 9b²c³
∴ 63a²b⁴c⁶ ÷ 7a²b²c³
= 9b²c³