Factorise the following expressions: (i) a² + 8a + 16 (ii) p² - 12p + 25 (iii) 25m² + 30m + 9 (iv) 49y² + 84yz + 36z² (v) 4x² - 8x + 4 (vi) 121b² - 22bc + 16c² (vii) (l + m)² - 8lm (viii) a⁴ + 2a²b² + b⁴

(i) Given a² + 8a + 16
Here, the middle term 8a split into two terms 4a and 4a such that its product 4a × 4a = 16a² is same as product of last term and first term of given equation.
= a² + 4a + 4a + 16
= a(a + 4) + 4(a + 4)
= (a + 4)(a + 4)

Similar way,
(ii) p² - 12p + 25
Here the middle term -12p = -6p - 6p
= p² - 6p - 6p + 25
= p(p - 6) - 5(p - 5)
This is not a perfect square trinomial. It cannot be factored using integers.
However, if the question intended p² - 12p + 36, then:
= p² - 6p - 6p + 36
= p(p-6) - 6(p-6)
= (p-6)(p-6)

(iii) 25m² + 30m + 9
Here, middle term 30m = 15m + 15m
= 25m² + 15m + 15m + 9
= 5m(5m + 3) + 3(5m + 3)
= (5m + 3)(5m + 3)

(iv) 49y² + 84yz + 36z²
Middle term 84yz = 42yz + 42yz
= 49y² + 42yz + 42yz + 36z²
= 7y(7y + 6z) + 6z(7y + 6z)
= (7y + 6z)(7y + 6z)

(v) 4x² - 8x + 4
Middle term -8x = -4x - 4x
= 4x² - 4x - 4x + 4
= 4x(x - 1) - 4(x - 1)
= (4x - 4)(x - 1)
= 4(x - 1)(x - 1)

(vi) 121b² - 22bc + 16c²
Here, middle term -22bc = -11bc -11bc
=121b² - 11bc - 11bc + 16c²
= 11b(11b - c) - 16c(11b -c)
This expression cannot be factored using integers. If the question intended 121b² - 22bc + c², then:
= 121b² - 11bc - 11bc + c²
= 11b(11b - c) - c(11b - c)
= (11b - c)(11b - c)

(vii) (l + m)² - 8lm = l² + m² + 2lm - 8lm (Since (a + b)² = a² + 2ab + b²)
= l² + m² - 6lm
This expression cannot be factored easily.

(viii) a⁴ + 2a²b² + b⁴ = (a²)² + 2a²b² + (b²)²
It is in the form of x² + 2xy + y² = (x + y)²
Replace x by a² and y by b², we get
(a²)² + 2a²b² + (b²)² = [a² + b²]² = (a² + b²)(a² + b²)