Let us factorise the following polynomials:

a³ – 12a – 16

Given, f(a) = a³ – 12a – 16

In f(a) putting a=±1, ±2, ±3, we see for which value of a, f(a)=0

f(−2)=(−2)³−12.(−2)−16=0

We observe that f(−2) = 0

From factor theorem, we can say, (a+2) is a factor of f(a)

a³ – 12a – 16 = a³ – 12a – 16

= a³+2a²–2a²– 4a – 8a − 16

= a²(a+2)−2a(a+2)−8(a+2)

= (a+2)(a²−2a−8)

= (a+2)(a²−4a+2a−8)

= (a+2)( a(a−4) + 2(a−4) )

= (a+2)(a−4)(a+2)

= (a+2)²(a−4)