10a³ + 9a² + 2a

Factor out "a"

a(10a² + 9a + 2)

Now we'll factor  10a² + 9a + 2 

Multiply the 10 by the 2, get 20

Write down all the ways to have two positive integers
which have product 20, starting with 20*1

20*1
10*2
 5*4

Since the last sign in 10a² + 9a + 2 is +, ADD them,
and place the SUM out beside that:


 20*1   20+1=21
 10*2   10+2=12
  5*4    5+4=9

Now, again ignoring signs, we find in that list of
sums the coefficient of the middle term in 10a² + 9a + 2

So we replace the number 9 by 5+4

10a² + 9a + 2
10a² + (5+4)a + 2

Then we distribute to remove the parentheses:

10a² + 5a + 4a + 2

Factor the first two terms 10a² + 5a by taking out the
greatest common factor, getting 5a(2a + 1)

Factor the last two terms +4a + 2 by taking out the
greatest common factor, +2, getting +2(2a + 1)

So we have

5a(2a + 1) + 2(2a + 1) 

Notice that there is a common factor, (2a + 1)

5a(2a + 1)+ 2(2a + 1)

which we can factor out leaving the 5a and the +2 to put 
in parentheses:

(2a + 1)(5a + 2)

Now remember there was an "a" factor in front that
we took out first

a(10a² + 9a + 2)

So the answer is:

a(2a + 1)(5a + 2)

Edwin