Factorisation of quadratics [duplicate]

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 $1$?

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, $\,a\mid \underbrace{k_1 k_2}_{\large \color{#c00}{ac}}\Rightarrow\, a = a_1 a_2,\ \begin{align}&k_1 = a_1 j_1\\ &k_2 = a_2 j_2\end{align},\,\ j_i\in\Bbb Z$

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?

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...

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 $b = -80$.

Use the quadratic formula. $ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$