Question 1097337
You need a pair of factors of 18 that add up to 11.
The pairs of factors for 18 are        ;
1 and 18 (adding to 19),
2 and 9 (adding to 11),
3 and 6 (adding to 9).
So, your pair of factors is 2 and 9.
 
Next, you could re-write the polynomial uswing 2 and 9 as coefficients for a split middle term as
6(p-5)^2+2(p-5)+9(p-5)+3 ,
and then factor by parts
6(p-5)^2+11(p-5)+3 = 6(p-5)^2+2(p-5)+9(p-5)+3 = [6(p-5)^2+2(p-5)]+[9(p-5)+3]
= 2(p-5)[3(p-5)+1] + 3[3(p-5)+1] = [2(p-5)+3][3(p-5)+1] 
= (2p-10+3)(3p-15+1) = (2p-7)(3p-14)
 
It looks complicated, and looks like it would be easy to make some mistake.
An easier option would be to start by making a cahnge of variable,
with y = p-5 , and factor 6y^2+11y+3.
6y^2+11y+3 = 6y^2+2y+9y+3 = (6y^2+2y) + (9y+3) = 2y(3y+1) + 3(3y+1)
= (2y+3)(3y+1) 
At that point, you can change back to p as a variable, rplacing p-5 for y:
[2(p-5)+3] [3(p-5)+1] = (2p-10+3)(3p-15+1) = (2p-7)(3p-14)