Answer
$12a^2+31a+20 = (3a+4)~(4a+5)$
Work Step by Step
Let's consider a trinomial in this form: $~ax^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = a\times c$
Then the next step is to write the trinomial as follows:
$~x^2+bx+c = ax^2+rx+sx+c$
Let's consider this trinomial: $~12a^2+31a+20$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 31$ and $r\times s = 240$. We can see that $15+16 = 31~$ and $15\times 16 = 240$
We can factor the trinomial as follows:
$12a^2+31a+20 = 12a^2+15a+16a+20$
$12a^2+31a+20 = (12a^2+15a)+(16a+20)$
$12a^2+31a+20 = (3a)(4a+5)+(4)(4a+5)$
$12a^2+31a+20 = (3a+4)~(4a+5)$
