Answer
$3y^2+24y+45 = (3)~(y+5)~(y+3)$
Work Step by Step
Let's consider a trinomial in this form: $~x^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$
Then we can factor the trinomial as follows:
$~x^2+bx+c = (x+r)~(x+s)$
Let's consider this trinomial: $~3y^2+24y+45$
First we can factor by using the GCF:
$3y^2+24y+45 = (3)~(y^2+8y+15)$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = 8$ and $r\times s = 15$. We can see that $3+5 = 8~$ and $3\times 5 = 15$
We can factor the trinomial as follows:
$3y^2+24y+45 = (3)~(y^2+8y+15)$
$3y^2+24y+45 = (3)~(y+5)~(y+3)$
