# How to factor $\,9x^2-80x-9?\,$ (AC-method) [closed]

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 $$1$$?

• There's nothing that multiplies to be 9 but at the same time adds up to 80. I'm so lost. Jan 22, 2014 at 3:48
• How would that look in between paranthesees? I have this figured out so far: (9x )(x ) Jan 22, 2014 at 3:53

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...

• This is the AC method - see my answer. Feb 5, 2019 at 17:46

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

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.

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 if 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 $$c$$ be, 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 for $$b$$ 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.

$$9x^2-80x-9=9x^2-81x+x-9=9x(x-9)+x-9=(x-9)\cdot(9x+1).$$

I recommend 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 $$k=c-\frac{b^2}{4a}$$