Objective: To factor general quadratic trinomials with integral coefficients.

If $ax^2 + bx + c$, $(a>1)$ can be factored, the factorization will have the pattern $$(px + r)(qx + s).$$ Factor $14x^2 - 17x + 5.$

Clue 1 Because the trinomial has a positive constant term and a negative linear term, both $r$ and $s$ will be negative.

Clue 2 List the factors of the quadradic term $14x^2$ and the negative factors of the constant term, $5$

Factors of $14x^2 $ : $(x, 14x)$ and $(2x, 7x)$

Factors of $5$: $(-1, -5)$ and $(-5, -1)$

Test the possibilities to see which produces the correct linear term $-17x$.

Since $(2x - 1)(7x - 5)$ gives the correct linear term,

$$14x^2 - 17x + 5 = (2x - 1)(7x - 5).$$

My question is if the negative factors of the constant term, 5 are listed as $(-1, -5)$ and $(-5, -1)$, then why aren't the factors of the quadratic term $14x^2$ listed as $(x, 14x)$ and $(14x, x)$ and $(2x, 7x)$ and $(7x, 2x)$? What is the general selection condition for the factors of the quadratic term, and the corresponding ordering of the factors of the constant term. Thank you for any contribution to this topic.

  • $\begingroup$ The point is that you want to make the factors of the coefficient of $x^2$ match up. If the quadratic factors properly, it will factor as a product of two degree one polynomials. The product of the coefficient of $x$ in each should result in the correct coefficient of $x^2$ in the product, ie $14.$ You could retain the $x$ as you did, but it doesn't really provide extra information. $\endgroup$ Mar 23, 2012 at 8:37

4 Answers 4


I was asked this same exact question in the chatroom, for the same exact polynomial example.

When trying to find $P,Q,R,S$ in the factorization $(Px+R)(Qx+S)$ it obviously matters which coefficients in front of an $x$ are paired with which constant terms - for example, if you switch the $R$ and the $S$ you get $(Px+S)(Qx+R)$, which is not the same polynomial. However, if you keep the pairs the same but switch their order, that is $P\leftrightarrow Q$ and $R\leftrightarrow S$, we get $(Qx+S)(Px+R)$, which is precisely the same polynomial as before, though written differently.

When you're looking for the unknown quantities then, you don't have to list every possible order, because some of them, as we've just seen, are equivalent, e.g. pairs $2x,7x$ and $-1,-5$ correspond exactly to the pairs $7x,2x$ and $-5,-1$. We can save time by not involving unnecessary, redundant configurations of the candidate factors. To this end, we fix the order of the factors in the first pair, and then let the factors in the second pair vary. Thus we arbitrarily decide to write $x,14x$ and $2x,7x$ and ignore the equivalent solutions that would come about if we also looked into $14x,x$ and $7x,2x$.

This saves time and energy. Alternatively, we could have done it the other way: fixed the order of the constant terms, and let the order of $x$-coefficient vary: we would fix $-1,-5$ and need to explore all of the possibilities listed in $(x,14x)$; $(14x,x)$; $(2x,7x)$; $(7x,2x)$.

$$\begin{array}{|c|c|} \hline (x-1)(14x-5) & (14x-5)(x-1) \\ (x-5)(14x-1) & (14x-1)(x-5) \\ (2x-1)(7x-5) & (7x-5)(2x-1) \\ (2x-5)(7x-1) & (7x-1)(2x-5) \\ \hline \end{array}$$

Above is a table of every possible ordering of both $(x,14x), (2x,7x)$ and $(-1,-5)$. The left side is created by fixing the order as $x,14x$ and $2x,7x$, and the right side is created by fixing the order as $14x,x$ and $7x,2x$ instead. As you can see, the ones on the left are equivalent to the ones on the right: we only need to search for the correct solution in one of the columns, not both. Alternatively:

$$\begin{array}{|c|c|}\hline (x-1)(14x-5) & (14x-5)(x-1) \\ (14x-1)(x-5) & (x-5)(14x-1) \\ (2x-1)(7x-5) & (7x-5)(2x-1) \\ (7x-1)(2x-5) & (2x-5)(7x-1) \\ \hline \end{array}$$

The above fixes the constant terms as $-1,-5$ on the left and $-5,-1$ on the right while letting the order of the $x$-coefficients vary. Just as before, the polynomials are equivalent, so we would only need to look at one of these columns for the true factorization, not both.


I'm going to have to take issue with the method in anon's answer, as it is tantamount to guessing and checking. Granted, he may have been restricting himself to the Clues provided, with which I also take issue.

Historical Note: I had been taught to guess and check in school, and never questioned this "method", until:

  • I started losing clients who couldn't guess as well as I could (or would falter with factoring, say, $12x^2 + 25x + 12 \ $ and $12x^2 + 24x + 12$ in rapid succession on tests).
  • One of my students - who had been kicked out of school for hacking and credit card fraud (!) - showed me the following ...

To factor $ax^2 + bx + c $, consider the factor pairs of $ac$.

"Clue" - when $ac <0$, one factor in each pair will be positive, and when $ac > 0$, both factors will have the same sign.

Now find two factors of $ac$, say $j \cdot k = ac$, such that $j + k = b$. Here $b$ is the coefficient of the linear term in $ax^2 + bx + c$.

Then write $$\dfrac{1}{a}(ax + j)(ax + k)$$ The product of the GCDs of the terms $(ax + j)$ and $(ax + k)$ will be $a$, which will cancel with the $\dfrac{1}{a}$ in front.

example: Factor $6x^2 - x - 15$. Here $ac = -90; b = -1$. Keeping in mind that since $ac$ is negative, one factor from each pair will be negative, we list the pairs of positive factors:

$$(1, 90) \ (2, 45) \ (3, 30) \ (5, 18) \ (6, 15) \ (9,10)$$

And look for the pair with a difference of $1$ (since the sum of a positive and a negative can be thought of as the difference of two positives). We find that for $j = 9; k = -10$, we have both of what we want: $$jk = ac; j + k = b$$

So we write: $$\dfrac{1}{a}(ax +j)(ax +k) = \dfrac{1}{6}(6x + 9)(6x - 10) = \dfrac{1}{6}(3)(2x + 3)(2)(3x - 5) = (2x +3)(3x - 5)$$

This works because when we know that we have $jk = ac; j + k = b$, we can multiply $ax^2 + bx + c$ by $a$ and make substitutions to get $$a^2x^2 + (aj + ak)x + jk = a^2x^2 + ajx + akx + jk = ax(ax + j) + k(ax + j) = (ax + k)(ax + j)$$

Notice that we have multiplied our original expression by $a$, whence the need to put the $\dfrac{1}{a}$ outside to compensate.

  • 2
    $\begingroup$ My experience has been almost the same: guess and check means try, fail, and despair. But then I was tutoring in a local highschool, and learned this method from one of the teachers. It makes factoring a nonmonic polynomial every bit as easy as factoring a monic: simple, direct, and foolproof. $\endgroup$
    – Lubin
    Mar 29, 2012 at 20:10
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    $\begingroup$ Alternatively, rewrite $-x$ as $-10x + 9x$, thus: $6x^2 -10x +9x -15$. This leads to a natural factoring in pairs: $2x(3x-5) + 3(3x-5)$, so the polynomial is $(2x+3)(3x-5)$. $\endgroup$
    – Théophile
    Mar 29, 2012 at 20:40
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    $\begingroup$ @Théophile: that works, too. One key observation is that in both methods (yours and mine, if you will), it is necessary to determine which factor pair of $ac$ has sum $b$. $\endgroup$ Mar 29, 2012 at 20:46
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    $\begingroup$ @Lubin See also my answer here where I explain why this AC Method works, and how it generalizes. $\endgroup$ May 16, 2012 at 19:48
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    $\begingroup$ That is quite the post script! $\endgroup$ Dec 10, 2015 at 18:02

Well, since $(ax+b)(cx+d)=(cx+d)(ax+b)$, if we switch both $a$ with $c$ and $b$ with $d$, then we get the same thing. We only need to swap one pair. So, it is arbitrary which pair of: $a$ with $c$ or $b$ with $d$, we swap. But, once that choice has been made: the other pair of term's factors must be combined in every possible way. Once you pick which pair (quadratic or constant terms) to fix the order of, with the other pair you'll have to check all of the different orders.

If you fix the order of the quadratic terms (e.g. $x,14x$ and $2x,7x$ as opposed to $14x,x$ and $7x,2x$), then you let the other pair vary (you have to check both $-1,-5$ and $-5,-1$). If, however, you fix the order of the constant terms (e.g. $-1,-5$ as opposed to $-5,-1$), then you have to check all of the orderings of quadratic terms: $14x,1x$ and $1x,14x$ and $2x,7x$ and $7x,2x$.

So for setting up the first side of the table order does not matter, but for the second side of the table it does. Both tables include all possible orderings and demonstrates that one half of the table is equivalent to the other half. Each column involves either fixing the quadratic terms one way (first table) or the constant terms one way (second table).

A Useful Analogy

Suppose that there are two taxis parked outside. They are going to provide transport home from the party, to two men and their wives. The men come out first, and it does not matter which man gets into which taxi. A minute later, the women come out. It's really important which woman gets into which taxi; otherwise, they both end up with the wrong man. But if the women came out first, it wouldn't matter which taxi each woman got into, but it would be important which taxi each man got into.

Therefore, choose one pair to fix, the other pair must be permuted to give all possible match ups. It doesn't matter which pair you fix, the other pair needs to go through all the orders and choices.

The reason the other pair needs to go through all the orders and choices is because the linear term is found by multiplying the first term of each binomial by the last term of the other and adding these products.

Special thanks to anon, robjohn, and David W. for their insightful comments.

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    $\begingroup$ I have beautified this post. It might be good to point out what those first and second tables are. (They appear in the brackets.) $\endgroup$
    – user21436
    Mar 29, 2012 at 19:03
  • $\begingroup$ That is an insightful analogy, but you are still guessing! There is only one "married couple" that matters: the two factors of $ac$ whose sum is $b$. $\endgroup$ Mar 29, 2012 at 20:49

Remember; $$ \begin{eqnarray} (ax+b)(cx+d) &=& acx^2 + adx + bcx + bd \notag \\ &=& acx^2 + (ad+bc)x + bd \end{eqnarray}$$

It is a matter for you to choose those values of $a$, $b$, $c$, & $d$ so that the coefficients of your polynomial match to these coefficients here, (i.e, $ac=14, ad+bc = -17$ etc) .

  • $\begingroup$ This doesn't answer the question: why fix the order $a,c$ but not $b,d$ when searching through candidate solutions? $\endgroup$
    – anon
    Mar 23, 2012 at 8:57

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