Let us factorise the following polynomials:

4a³ – 9a² + 3a + 2

Given, f(a) = 4a³ – 9a² + 3a + 2

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

f(1)=4.(1)³ −9.(1)² +3.(1)+2=0

We observe that f(1) = 0

From factor theorem, we can say, (a−1) is a factor of f(a)

4a³–9a²+3a+2= 4a³−4a²−5a²+5a−2a+ 2

= 4a²(a−1)−5a(a−1)−2(a−1)

= (a−1)(4a²−5a−2)