Answer
$4a^2+12ab+9b^2 = (2a+3b)~(2a+3b)$
Work Step by Step
Let's consider a trinomial in this form: $~ax^2+bxy+cy^2$
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:
$~x^2+bxy+cy^2 = ax^2+rxy+sxy+cy^2$
Let's consider this trinomial: $~4a^2+12ab+9b^2$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 12$ and $r\times s = 36$. We can see that $6+6 = 12~$ and $6\times 6 = 36$
We can factor the trinomial as follows:
$4a^2+12ab+9b^2 = 4a^2+6ab+6ab+9b^2$
$4a^2+12ab+9b^2 = (4a^2+6ab)+(6ab+9b^2)$
$4a^2+12ab+9b^2 = (2a)(2a+3b)+(3b)(2a+3b)$
$4a^2+12ab+9b^2 = (2a+3b)~(2a+3b)$
