# 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)

**☛ Related Questions:**

- Factorize the following expressions. (i) a2 + 8a +16 (ii) p2 -10 p + 25 (iii) 25m2 + 30m + 9 (iv) 49 y2 + 84 yz + 36z2 (v) 4x2 - 8x + 4 (vi) 121b2 - 88bc +16c2 (vii) (l +m)2 - 4lm (Hint: Expand (l + m)2 first) (viii) a4+ 2a2b2 + b4.
- Factorize (i) 4 p2 - 9q2 (ii) 63a2 - 112b2 (iii) 49x2 - 36 (iv) 16x5 - 144x3 (v) (l + m)2 - (l - m)2 (vi) 9x2y2 - 16 (vii) (x2 - 2xy + y2) - z2 (viii) 25a2 - 4b2 + 28bc - 49c2
- Factorise the expressions. (i) ax2 + bx (ii) 7p2 + 21q2 (iii) 2x3 + 2xy2 + 2xz2 (iv) am2 + bm2 + bn2 + an2 (v) (lm + l) + m + 1 (vi) y(y + z) + 9( y + z) (vii) 5y2 - 20 y - 8z + 2yz (viii) 10ab + 4a + 5b + 2 (ix) 6xy - 4 y + 6 - 9x.
- Factorise (i) a4 - b4 (ii) p4 - 81 (iii) x4 - ( y + z)4 (iv) x4 - (x - z)4 (v) a4 - 2a2b2 + b4

visual curriculum