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# Factorise the following expressions

(i) p^{2} + 6 p + 8 (ii) q^{2} - 10q + 21

(iii) p^{2} + 6 p - 16

**Solution:**

In general, for factorizing an algebraic expression of the type x^{2} + px + q, we find two numbers a and b such that ab = q and a + b = p.

(i) p^{2} + 6 p + 8

It can be observed that, 8 = 4 × 2 and 4 + 2 = 6

Therefore,

p^{2} + 6 p + 8 = p^{2} + 2p + 4p + 8

= p(p + 2) + 4(p + 2)

= (p + 2)(p + 4)

(ii) q^{2} - 10q + 21

It can be observed that, 21 = (-7) × (-3) and (-7) + (-3) = -10

Therefore, q^{2} - 10q + 21 = q^{2} - 7q - 3q + 21

= q(q - 7) - 3(q - 7)

= (q - 7)(q - 3)

(iii) p^{2} + 6 p -16

It can be observed that, -16 = (-2) × 8 and 8 + (-2) = 6

p^{2} + 6 p -16 = p^{2} + 8p - 2p -16

= p(p + 8) - 2(p + 8)

= (p + 8)(p - 2)

**Video Solution:**

## Factorise the following expressions (i) p² + 6 p + 8 (ii) q² - 10q + 21 (iii) p² + 6 p - 16

Class 8 Maths NCERT Solutions Chapter 14 Exercise 14.2 Question 5

**Summary:**

The following expressions (i) p^{2} + 6 p + 8 (ii) q^{2} - 10q + 21 (iii) p^{2} + 6 p - 16 are factorised as (i) (p + 2)(p + 4) (ii) (q - 7)(q - 3) (iii) (p + 8)(p - 2)

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