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How do I factor a trinomial like this? I'm having a lot of difficulty. How do I deal with the $9x^2$?

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closed as off-topic by user296602, user91500, Claude Leibovici, JMP, user1551 Jul 27 '16 at 10:29

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  • $\begingroup$ There's nothing that multiplies to be 9 but at the same time adds up to 80. I'm so lost. $\endgroup$ – user3131263 Jan 22 '14 at 3:48
  • $\begingroup$ How would that look in between paranthesees? I have this figured out so far: (9x )(x ) $\endgroup$ – user3131263 Jan 22 '14 at 3:53
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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.

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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?

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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...

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  • $\begingroup$ This is the AC method - see my answer. $\endgroup$ – Bill Dubuque Feb 5 at 17:46
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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 $b = -80$.

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

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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.

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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.

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

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

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