How to factor $2x^3 + 21x^2 +27x$ I am having trouble with factoring $2x^3 + 21x^2 +27x$. The answer is $x(x+9)(2x+3)$ but not sure how that was done. Obviously I factored out the $x$ to get $x(2x^2+21x+27)$ then from there I am lost. I tried the AC method and grouping. Can someone show the steps? Thanks! 

1) 
$2x^3 + 21x^2 +27x = x(2x^2 + 21x + 27) = x(2x(x + 9) + 3x + 27) =x(2x(x+9) + 3(x + 9)) = x(2x + 3)(x+9)$.
2) 
$2x^3 + 21 x^2 + 27x = x(2x^2 + 21x + 27) = x*2*(x - a)(x+b)$ where $a,b$ are solutions to $2x^2 + 21x + 27=0$.  i.e. $x = \frac {-21 \pm {21^2 - 8*27}}{4} = \frac {-21 \pm \sqrt {225}}4 = \frac {-21 \pm 15}4 = -9, -\frac 32$ so 
$2x^3 + 21 x^2 + 27x = x*2*(x + \frac 32)(x + 9) = x(2x +3)(x+9)$.
3)
$2x^3 + 21x^2 + 27x = (ax + b)(cx + d)(ex + f)$ and 
$ace = 2$.  Wolog, $a = 2$ and $c, e = 1$.
So $2x^3 + 21x^2 + 27x = (2x + b)(x + d)(x + f)$
$2f + 2d + b = 21$
$2df + bf + bd = 27$
$bdf = 0$
$b$ is odd so $b\ne 0$.  Wolog $d= 0$.
$2f + b = 21$ 
$bf = 27$
So $b=3; f= 9$
So $x(2x + 3)(x + 9)$.
