Let us factorise the following polynomials:

5a³ + 11a² + 4a –2

Given, f(a)= 5a³ + 11a² + 4a –2

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

f(−1)=5.(−1)³+11.(−1)²+4.(−1)−2=0

We observe that f(−1) = 0

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

5a³ + 11a² + 4a –2 = 5a³ + 11a² + 4a –2

= 5a³+5a²+6a²+6a−2a−2

= 5a²(a+1)+6a(a+1)−2(a+1)

= (a+1)(5a²+6a−2)