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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?

Cameron Buie
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Eleven-Eleven
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2 Answers2

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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.

Cameron Buie
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    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? – imranfat Oct 11 '13 at 20:08
  • 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!) – Eleven-Eleven Oct 11 '13 at 20:11
  • @Christopher: Why wouldn't that work? For example, $$9x^2-36=(3x^2)-6^2=(3x+6)(3x-6)=3(x+2)(3x-6)=9(x+2)(x-2).$$ – Cameron Buie Oct 11 '13 at 20:22
  • As for special trinomials and differences of squares, I've always preferred a visual proof, myself. Likewise for completing the square. – Cameron Buie Oct 11 '13 at 20:23
  • 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. – Eleven-Eleven Oct 11 '13 at 20:34
  • @Christopher: Ah! I see what you meant now. Hmm...that is a drawback to this method, as opposed to grouping. – Cameron Buie Oct 11 '13 at 20:36
  • I've been teaching that step 1 is always GCF anyway, but the kids are stubborn... and they don't listen. :) It's because you're putting the $a$ term in both linear factors by placing them at the upper part of the asterisk. You are guaranteed that at least one will reduce...but with DOTS, and not GCF taken out first, they both disappear. – Eleven-Eleven Oct 11 '13 at 20:41
  • @ChristopherErnst, I give an alternative as an answer. Whether stubborn kids can use it is another matter, but it is a pure algorithm, with no choices in the first part. – Will Jagy Oct 11 '13 at 21:15
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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.

EEDDDDIIIITTTT:

$$ a x^2 + b x + c = \; \left(a_1x+ \left( \frac{b - \sqrt \Delta}{2a_2} \right) \right) \; \; \left(a_2x+ \left( \frac{b + \sqrt \Delta}{2a_1} \right) \right) \; $$ 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 $$ c_2 = \frac{b + \sqrt \Delta}{2 a_1} = \frac{25957}{257} = 101, \; \; \; c_1 = c / c_2 = -10403 / 101 = -103. $$ $$ 24929 x^2 + 15966 x - 10403 = (257 x - 103)(97 x + 101). $$

Community
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Will Jagy
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  • 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. – Eleven-Eleven Oct 11 '13 at 21:28
  • @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. – Will Jagy Oct 11 '13 at 21:39