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

Solution 1:

Hint $\ \ $ Reduce to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $=1)$ as follows:

$$\quad\ \ \begin{eqnarray} f &\,=\,& \ \ 9\ x^2-\ 80\ x\ -\,\ 9\\ \Rightarrow\ 9f &\,=\,& (9x)^2\! -80(9x)-81\\ &\,=\,& \ \ \ \ \color{#c00}{X^2\!- 80\ X\ -\,\ 81},\,\ \ X\, =\, 9x\\ &\,=\,& \ \ \ \,(X-81)\ (X+\,1)\\ &\,=\,& \ \ \ (9x-81)\,(9x+1)\\ \Rightarrow\ f\,=\, 9^{-1}(9f) &\,=\,& \ \ \ \ \ (x\ -\ 9)\,(9x+1)\\ \end{eqnarray}$$

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. This is sometimes called the AC method. It works for higher degree polynomials too. As above, we can reduce the problem of factoring a non-monic polynomial to that of factoring a monic polynomial by scaling by a $ $ power of the lead coefficient $\rm\:a\:$ then changing variables: $\rm\ X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\rm\: X^2 + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\smash[t]{\overbrace{ac}^{\rm\qquad\ \ \ \ \ {\bf AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ 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 $\ $ Those 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.

For completeness we do the general case as above

$$ \begin{eqnarray} f &\,=\,&\ \ \ a\, x^2 + b\ x\ +\,\ c\\ \Rightarrow\ af &\,=\,&\ (ax)^2\! +b (ax)+ ac\\ &\,=\,&\ \ \ \ \ {X^2\! + b\ X\ +\,\ ac},\,\ \ X\, =\, ax\\ &\,=\,&\ \ \ \ \,(X-k_1)\ \ (X\,-\,k_2)\\ &\,=\,&\ \ \ \ (ax-k_1)\ \ (ax-k_2)\\ &\,=\,& a_1(a_2x-j_1)\,(a_1 x-j_2)a_2\\ \Rightarrow\ f\, =\, a^{-1}(af)&\,=\,& \ \ \ \ (a_2 x-j_1)\,(a_1 x-j_2)\\ \end{eqnarray}$$

where, by Primal Law, $\,a\mid \underbrace{k_1 k_2}_{\large ac}\Rightarrow\, a = a_1 a_2,\ \begin{align}&k_1 = a_1 j_1\\ &k_2 = a_2 j_2\end{align},\,\ j_i\in\Bbb Z$

Solution 2:

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?

Solution 3:

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...

Solution 4:

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 $b = -80$.

Solution 5:

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