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!
$9(a-1)^2 +3(a-1) - 2=(3(a-1)-1)(3(a-1)+2)$
$=(3a-4)(3a-1)$
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
$$(3(a-1)-1)(3(a-1)+2)=(3a-4)(3a-1)$$
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.
$x=a-1\\ 9x^2+3x-2=0\\ \Delta=9+72=81\\ \sqrt{\Delta}=9\\ x=\frac{-3 \pm 9}{18} = \pm \frac{1}{2}-\frac{1}{6}\\ a=x+1=\pm \frac{1}{2}-\frac{1}{6}+1 = \pm \frac{1}{2}+\frac{5}{6}$
$$9(a-1)^2 +3(a-1)-2=9(a^2-2a+1)+3a-3-2=$$ $$=9a^2-18a+9+3a-5=9a^2-15a+4=$$ $$(9a^2-1)-(15a-5)=(3a+1)(3a-1)-5(3a-1)=$$ $$(3a-1)(3a+1-5)=(3a-1)(3a-4)$$
$\begin{align}{\bf Hint}\,\ {\rm Let}\,\ x = 3(a\!-\!1).\ {\rm Then}\qquad &9(a\!-\!1)^2 +3(a\!-\!1)-2\\ =\ &x^2 + x - 2\\ =\ &(x+2)(x-1)\end{align}$
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.
