How to factor $\,9x^2-80x-9?\,$ (AC-method) Is there a general method to factor a quadratic like $9x^2-80x-9?\,$ I'm having a lot of difficulty due to the leading coefficient being unequal to $1$?
 A: You can do it like this:
First, write the polynomial like this:
$$\frac{9(9x^2-80x-9)}{9}$$
Then expand the numerator as 
$$81x^2-720x-81$$
which can be written in the form
$$(9x)^2-80(9x)-81$$
If we let $y=9x$, then the polynomial becomes
$$\frac{y^2-80y-81}{9}$$
Can you continue from here?
A: Use the fact that: $$-80 = 1 - 81 = 1 -9\times9$$
In general, if you want to find $A,B$ such that $(ax+A)(x+B) = ax^2+bx+c$, you need them to satisfy:
$$aB + A= b,\ AB = c$$
If you assume integer factors, you can see $A,B$ must be either $3,-3$ or $\pm 9,\mp 1$. Only of of these three options gives $b = -80$.
A: To factor a trinomial:$$ax^2+bx+c$$
First multiply $a\times c$;  pay attention to the signs of $a$ and  $c$
Now find two numbers that multiply to give this product and add to the middle coefficient, $b$.
All this work to split the middle term into two, so you can factor by grouping.
For example:$$24x^2+31x-15$$
The product is $-360$: eventually you'll find $40$ and $-9$  
So now you factor $$24x^2+40x-9x-15$$ by grouping the first two terms, taking a common factor, and the same for the second pair...
A: Use the quadratic formula.  $ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$
A: Why not. You know the quadratic formula,
$$   \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$
For your example, $B^2 - 4 AC = 6724$ and $\sqrt {B^2-4AC}=82.$
TRICK: one of the $x$ coefficients is
$$ \gcd \left(A,  \frac{-B + \sqrt{B^2 - 4AC}}{2} \right) =   \gcd \left(9,  \frac{80 + 82}{2} \right) = \gcd(9,81)=9.   $$ See how you write the quadratic formula, but pull the $A$ out from the denominator and find the $\gcd$ with what's left.
I prove an expanded (and very slightly different) version of this, with entirely elementary methods, at How to factor the quadratic polynomial $2x^2-5xy-y^2$? 
EDIT: I had not wanted to muddy the waters...what happens if we switch the $\pm$ sign in the "trick" above? One of the $x$ coefficients is
$$ \gcd \left(A,  \frac{-B - \sqrt{B^2 - 4AC}}{2} \right) =   \gcd \left(9,  \frac{80 - 82}{2} \right) = \gcd(9,-1)=1.   $$ You get the other $x$ coefficient, that's all.
A: Hint $  $ Reduce to factoring a polynomial that is $\,\rm\color{#0a0}{monic}\,$ (lead coeff $=1)$ as follows:
$$\quad\ \ \begin{eqnarray}
f &\,=\,& \ \  9\ x^2-\ 80\ x\ -\,\ 9\\
\Rightarrow\ 9f &\,=\,& (9x)^2\! -80(9x)-81\\
         &\,=\,& \ \ \ \  \color{#0a0}{X^2\!- 80\  X\ -\,\ 81},\,\ \ X\, =\, 9x\\
         &\,=\,&  \ \ \ \,(X-81)\ (X+\,1)\\
         &\,=\,& \ \ \ (9x-81)\,(9x+1)\\
\Rightarrow\  f\,=\, 9^{-1}(9f)  &\,=\,&   \ \ \ \ \  (x\ -\ 9)\,(9x+1)\\
\end{eqnarray}$$
Below we show that the above method works not only for quadratic $f\,$ but also for higher  degree polynomials (see the Note below for the above computation done for an arbitrary quadratic).
If we denote our factoring algorithm by $\,\cal F\,$ then the above transformation is simply
$$\cal F f\,  = a^{-1}\cal F\, a\,f\quad\,$$
Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials.
In elementary treatments (e.g. high school level) the quadratic case is sometimes called the $\rm\color{#c00}{AC}$ method. It also works for higher degree polynomials, i.e. as above, we can reduce the problem of factoring a non-monic polynomial to that of factoring a monic polynomial by scaling the polynomial by a  $ $ power of the lead coefficient $\rm\:a\:$ then changing variables: $\rm\ X = a\:x,\, $ as below
$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c)  &\,=\,&\rm\: X^2 + b\:X + \color{#c00}{ac} =\, g(X),\ \ \ X = a\:x \\
\\
\rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c
\end{eqnarray}$$
After factoring the monic $\rm\,g(X)\, =\, a^{n-1}f(x),\,$ we are guaranteed that the transformation reverses to yield a factorization of $\rm\:f,\ $ since $\rm\ a^{n-1}$ must divide into the factors of $\rm\ g\ $ by Gauss' Lemma, i.e. primes  $\,p\in\rm\mathbb Z\,$ remain prime in $\rm\,\mathbb Z[X],\,$ so $\rm\ p\ |\ g_1(x)\:g_2(x)\,$ $\Rightarrow$ $\,\rm\:p\:|\:g_1(x)\:$ or $\rm\:p\:|\:g_2(x).$
This method also works for multivariate polynomial factorization, e.g. it applies to  this question.
Remark $ $ Readers who know university algebra might be interested to know that this works not only for UFDs and GCD domains but also for integrally-closed domains satisfying
$\qquad\qquad$ Primal Divisor Property $\rm\ \ c\ |\ AB\ \ \Rightarrow\ \ c = ab,\ \ a\ |\: A,\ \ b\ |\ B$
Elements $c$ satisfying this are called primal. One easily checks that atoms are primal $\!\iff\!$ prime. Also products of primes are also primal. So "primal" may be viewed as a generalization of the notion "prime" from atoms (irreducibles) to composites.
Integrally closed domains whose elements are all primal are called $ $ Schreier rings by Paul Cohn (or Riesz domains, because they satisfy a divisibility form of the Riesz interpolation property). In Cohn's Bezout rings and their subrings
he proved that if $\rm\:D\:$ is Schreier then so too is $\rm\,D[x],\:$ by using a primal analogue of Nagata's Lemma: an atomic domain $\rm\:D\:$ is a UFD if some localization $\rm\:D_S\:$ is a UFD, for some monoid $\rm\:S\:$ generated by primes. These primal and Riesz interpolation viewpoints come to the fore in a refinement view of unique factorization, which proves especially fruitful in  noncommutative rings (e.g. see Cohn's 1973 Monthly survey Unique factorization domains).
In fact Schreier domains can be characterized equivalently by a suitably formulated version of the above "factoring by conjugation" property. This connection between this elementary AC method and Schreier domains appears to have gone unnoticed in the literature.

Note $ $ For completeness we do the general case as above
$$ \begin{eqnarray}
f &\,=\,&\ \ \  a\, x^2 + b\ x\ +\,\ c\\
\Rightarrow\ af &\,=\,&\ (ax)^2\! +b (ax)+ \color{#0a0}{ac}\\
         &\,=\,&\ \ \ \ \  {X^2\! + b\  X\ +\,\ ac},\,\ \ X\, =\, ax\\
         &\,=\,&\  \ \ \ \,(X-k_1)\ \ (X\,-\,k_2)\\
         &\,=\,&\ \ \ \ (ax-k_1)\ \ (ax-k_2)\\
         &\,=\,&  a_1(a_2x-j_1)\,(a_1 x-j_2)a_2\\
\Rightarrow\  f\, =\, a^{-1}(af)&\,=\,& \ \ \ \   (a_2 x-j_1)\,(a_1 x-j_2)\\
\end{eqnarray}$$
where,  by Primal Law, $\,a\mid \underbrace{k_1 k_2}_{\large \color{#c00}{ac}}\Rightarrow\, a = a_1 a_2,\ \begin{align}&k_1 = a_1 j_1\\ &k_2 = a_2 j_2\end{align},\,\ j_i\in\Bbb Z$
A: From the looks of the problem, it looks like they want to you to make educated guesses. If the factors are $(a x+ b)$ and $(cx+d)$ then you need
$$
(a x + b) (c x + d) = 9x^2 -80 x + 9$$
If you look at the $x^2$ term, on the left you have $a\,c$ and on the right you have $9$. So you want
$$
a\,c = 9 \tag 1$$
Similarly of you look at the constant term, on the left you have $b\,d$ and on the right you have $9$. So
$$
b\, d = 9 \tag 2$$
Finally the $x$ term on both sides gives
$$ a d + bc = -80 \tag 3$$
Since you want $-80$ on the right, all the four can't be positive. So if $a$ is negative, so should $c$ since $a \,c = 9$. 
At this stage you need a big leap of faith. This will usually be true at an introductory course. The leap of faith is that all the numbers have to be whole numbers. This means $a$ can only be $-1$ or $-3$ or $-9$ since $a$ divides $9$. Same is true or $b$ and $d$ and they have to be positive. So $b$ can only be $1$, $3$ or $9$. Now try all possibilities for $a$ and $b$. For each $a$ and $b$ you try, you can get $c$ and $d$ from equations (1) and (2). Once you have all the four, check (3). With experience you can skip a few steps.
A: $$9x^2-80x-9=9x^2-81x+x-9=9x(x-9)+x-9=(x-9)\cdot(9x+1).$$
A: I recomend using the method called "completing the square" (the quadratic formula can be derived from it)
A quadratic of the form $$a^2x+bx+c=0$$
can be written in the form $$a(x-h)^2+k=0$$
where $$h=\frac{-b}{2a}$$
and $$ k=c-\frac{b^2}{4a}$$
