Question 131119
Factor:
1) {{{12(p+3)^2+13(p+3)+3}}} 
To make things a little easier, you can temporarily substitute another variable, say x,for (p+3), to give you:
{{{12x^2+13x+3}}} Factoring this trinomial, you'll get:
{{{(3x+1)(4x+3)}}} Now you can re-substitute the (p+3) for x, to get:
{{{(3(p+3)+1)(4(p+3)+3)}}} Now you simplify this expression.
{{{(3p+9+1)(4p+12+3)}}}
{{{(3p+10)(4p+15)}}} ...and there you have it!
---------------------------------------------
Check:
Let's first multiply the factors using FOIL:
{{{(3p+10)(4p+15) = 12p^2+45p+40p+150}}} = {{{12p^2+85p+150}}}
Now we'll expand the given trinomial that you were asked to factor:
{{{12(p+3)^2+13(p+3)+3 = 12(p^2+6p+9)+13(p+3)+3}}} = {{{(12p^2+72p+108)+(13p+39)+3 = 12p^2+85p+150}}}

2) Factor:
{{{18z^8-15z^5-18z^2}}}  First factor {{{3z^2}}}
{{{3z^2(6z^6-5z^3-18)}}} Now, to simplify the calculations a bit, temporarily substitute x for {{{z^3}}}, but only inside the parentheses!
{{{3z^2(6x^2-5x-6)}}} Now factor the trinomial in x. 
{{{3z^2(3x+2)(2x-3)}}} Finally, you can replace the x with {{{z^3}}}
{{{3z^2(3z^3+2)(2z^3-3)}}} ...so you were close!
------------------------------
Check:
{{{3z^2(3z^3+2)(2z^3-3) = 3z^2(6z^6-9z^3+4z^3-6)}}} = {{{3z^2(6z^6-5z^3-6) = 18z^8-15z^5-18z^2}}}...which is exactly what you started with!