I'm teaching my students about factoring quadratics. We've done GCF, difference of two squares, squared binomials, and grouping. One of my colleagues then found this asterisk method on line.

It's basically the grouping method but presented in a little different light. Has anyone ever seen this or used it? My bosses would prefer learning the guess and check method, or the grouping method, but if you factor out a CGF before, this method works for all of them. I would like my students to be able to recognize the special cases and not have to resort to any work if they see $a^2x^2-c^2$ or if they see $(ax)^2+2abx+b^2$, but if they are struggling with those and can grasp this method, would anyone here use this method?


2 Answers 2


It's a pretty slick method, which will solve most of the same problems that factoring by grouping will solve. Using the example in the video, we can instead proceed as follows: $$\begin{align}12x^2-5x-2 &= 12x^2+(3-8)x-2\\ &= 12x^2+3x-8x-2\\ &= 3x(4x+1)-2(4x+1)\\ &= (4x+1)(3x-2).\end{align}$$

I tend to prefer completing the square for its general utility, but for "nicely factorable" trinomials, the asterisk method works just fine.

Added: As you point out, the method will fail for differences of squares with a common factor, unless you pull out the GCF first. That is a drawback to this method, as opposed to factoring by grouping, which does not require us to pull out the GCF first.

  • 1
    $\begingroup$ It's indeed pretty much the grouping method. It looks nice, but I don't think it adds anything other than it may help the so called "visual" learner? $\endgroup$
    – imranfat
    Oct 11, 2013 at 20:08
  • $\begingroup$ Yeah I showed it to them because there is a "slickness" to it and my kids definitely grasped it and saw its utility. Indeed, they saw grouping and shook their head. Some of them who just couldn't wrap their head around the special binomials or the difference of two squares got this immediately, so long as the GCF is removed first (try a difference of two squares where there is a common factor and it doesn't work!) $\endgroup$ Oct 11, 2013 at 20:11
  • $\begingroup$ @Christopher: Why wouldn't that work? For example, $$9x^2-36=(3x^2)-6^2=(3x+6)(3x-6)=3(x+2)(3x-6)=9(x+2)(x-2).$$ $\endgroup$ Oct 11, 2013 at 20:22
  • $\begingroup$ As for special trinomials and differences of squares, I've always preferred a visual proof, myself. Likewise for completing the square. $\endgroup$ Oct 11, 2013 at 20:23
  • $\begingroup$ It works, but in the video it tells you to "reduce the fractions" and then form the linear product. If you factor out the GCF first it works but if you don't take the GCF out you then reduce and you're left with no GCF. I watched it happen with my kids a lot. $\endgroup$ Oct 11, 2013 at 20:34

At How to factor the quadratic polynomial $2x^2-5xy-y^2$?

I prove that, for $a x^2 + b x + c,$ with discriminant $\Delta = b^2 - 4 a c,$ the given quadratic factors if and only if $\Delta$ is a (non-negative) square. If so, we are guaranteed to be able to factor as $$ a x^2 + b x + c = (a_1 x + c_1)(a_2 x + c_2) $$ by beginning with $$ a_1 = \gcd \left( \; a, \; \frac{b + \sqrt \Delta}{2} \right), $$ next taking $a_2 = a / a_1.$

Maybe you can do something with that.


$$ a x^2 + b x + c = \; \left(a_1x+ \left( \frac{b - \sqrt \Delta}{2a_2} \right) \right) \; \; \left(a_2x+ \left( \frac{b + \sqrt \Delta}{2a_1} \right) \right) \; $$ in integers.

$$ 24929 x^2 + 15966 x - 10403. $$

$$ \Delta = 15966^2 - 4 \cdot 24929 \cdot(-10403) = 1292258704 $$ $$ \sqrt \Delta = 35948 $$ $$ \frac{b + \sqrt \Delta}{2} = 25957 $$ $$ \gcd (24929, 25957) = \gcd ( 24929, 1028) = \gcd ( 1028, 257) = 257. $$ $$ a_1 = 257.$$ $$ a_2 = 24929/257 = 97.$$ $$ c_1 = \frac{b - \sqrt \Delta}{2a_2} = \frac{15966 - 35948}{2 \cdot 97} = -103 $$ $$ c_2 = c / c_1 = 101. $$ OR $$ c_2 = \frac{b + \sqrt \Delta}{2 a_1} = \frac{25957}{257} = 101, \; \; \; c_1 = c / c_2 = -10403 / 101 = -103. $$ $$ 24929 x^2 + 15966 x - 10403 = (257 x - 103)(97 x + 101). $$

  • $\begingroup$ I appreciate the method. We're moving slowly through quadratics. I teach them complex numbers next week and completing the square, then I do quadratic equation the following week and I can hit them with that. $\endgroup$ Oct 11, 2013 at 21:28
  • $\begingroup$ @ChristopherErnst, edited in, it is a bit messy to write out as a pure recipe but it works, no choices ever. Let me put in an example with big numbers. $\endgroup$
    – Will Jagy
    Oct 11, 2013 at 21:39

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