Possible roots are formed using the last term (12) of the given function and the 1st term (1) as your basis:
Since factors of 12 include plus or minus 1, 2, 3, 4, 6, 12,
possible rational roots include plus or minus 1/1, 2/1, 3/1, 4/1, 6,/1, 12/1 (and so on.) I will take a chance and try the possible root 6/1, or just plain 6.
Use synthetic division, with 6 as the divisor and all of the coefficients of x^5+3x^4-5x^3-15x^2+4x+12 as dividend:
______________________
6 / 1 3 -5 -15 4 12
6 54
--------------------------------
1 9 49 This is not going to work; 6 is not a root.
Try -3 as divisor:
___________________
-3 / 1 3 -5 -15 4 12
-3 0 15 0 -12
-------------------------- ------
1 0 -5 0 4 0
Since the remainder is zero, x = -3 is a root of the given polynomial.
Repeat this process, except skip x = -3 and x = 6 as divisors.
Use the coefficients 1 0 -5 0 4. Note that plus or minus 4 over 1 forms other possible rational roots: 4/1, -4/1, 2/1, -2/1, 1, -1
Let's check out
____________
1 / 1 0 -5 0 4
1 1 -4 -4
--------------------------
1 1 -4 -4 0
Thus, 1 is also a root of the polynomial, along with -3.
Repeat this process. As divisors try 4/1, -4/1, 2/1, -2/1, -1
Can you now finish this factoring?
The 2 roots found so far are {1, -3}.
