Factor $9(a-1)^2 +3(a-1) - 2$ I got the equation $9(a-1)^2 +3(a-1) - 2$ on my homework sheet. I tried to factor it by making $(a-1)=a$ and then factoring as a messy trinomial. But even so, I couldn't  seem to get the correct answer; they all seemed incorrect. 
Any help would be greatly appreciated.
Thank you so much in advance! 
 A: $9(a-1)^2 +3(a-1) - 2=(3(a-1)-1)(3(a-1)+2)$
$=(3a-4)(3a-1)$
A: If you write $x=a-1$ then you get $9x^2+3x-2$. The last expression has roots $-2/3$ and $1/3$ so we can write it like 
$$9x^2+3x-2=9(x-1/3)(x+2/3)=(3x-1)(3x+2)$$
Now use again $x=a-1$ and get
$$(3(a-1)-1)(3(a-1)+2)=(3a-4)(3a-1)$$
A: Making "a-1" be "a" doesn't make sense, but using some other name for it does; you could write
$$
9t^2 + 3t - 2
$$
for instance. The roots of that quadratic are 
\begin{align}
\frac{-3 \pm \sqrt{9 + 4 \cdot 2 \cdot 9}}{18} 
&= \frac{-3 \pm 3\sqrt{1 + 4 \cdot 2 }}{18} \\
& = \frac{-1 \pm 1\sqrt{9}}{6}\\
& = \frac{-1 \pm 3}{6}\\
& = \frac{1}{3}, \frac{-2}{3}\\
\end{align}
So it factors as something proportional to $$(t-1/3)(t + 2/3)$$. Given the leading coefficient is $9$, you know that it must be
$$
(3t-1)(3t+2)
$$
And then remembering that $t = a-1$, you can get a final answer. 
A: $x=a-1\\
9x^2+3x-2=0\\
\Delta=9+72=81\\
\sqrt{\Delta}=9\\
x=\frac{-3 \pm 9}{18} = \pm \frac{1}{2}-\frac{1}{6}\\
a=x+1=\pm \frac{1}{2}-\frac{1}{6}+1 = \pm \frac{1}{2}+\frac{5}{6}$
A: $$9(a-1)^2 +3(a-1)-2=9(a^2-2a+1)+3a-3-2=$$
$$=9a^2-18a+9+3a-5=9a^2-15a+4=$$
$$(9a^2-1)-(15a-5)=(3a+1)(3a-1)-5(3a-1)=$$
$$(3a-1)(3a+1-5)=(3a-1)(3a-4)$$
A: $\begin{align}{\bf Hint}\,\ {\rm  Let}\,\ x = 3(a\!-\!1).\ {\rm Then}\qquad &9(a\!-\!1)^2 +3(a\!-\!1)-2\\ =\ &x^2 + x - 2\\ =\ &(x+2)(x-1)\end{align}$
Remark $ $ Above is a special case of the AC-method, which gives a general way to change variables to transform polynomials to have leading coefficient $=1.\,$ This general method is well-worth learning since it often proves useful.
