How I factorise quadratics these days

It's nearly two years since I last tackled quadratics with a number in front. Recently, though, I stumbled on a slightly different method that's a bit less involved.

I won't say it's easier or better - different methods suit different people, after all - but I like it.

Let's factorise $5x^2 - 2x - 7$, the kind of thing that makes some students turn the page. The trick is to do a sneaky bit of variable-changing under the bonnet - details at the end of the post - which looks very much like moving the 5 to the end. What I mean by that is, you divide the $x^2$ by 5 and multiply the 7 by 5 to get a new expression (changing all the $x$s to $y$s for the sake of good form):

$y^2 -2y - 35$

This looks, and feels, completely non-kosher, but don't worry: it comes out fine. On the plus side, that's easy to factorise: $(y-7)(y+5)$.

The big minus: it doesn't multiply out to what we started with; we need to have a 5 with one of the $x$s. So, what we do is, we divide it out of the right-hand $5$ and multiply it back in the left-hand $x$ - and change the $y$s back into $x$s - making it $(5x -7)(x+1)$.

Now, that does multiply out to the right thing.

Similarly, to sort out $6x^2 + x - 2$, you would convert it into $y^2 + y - 12 = (y-3)(y+4)$, and move a three from left to right and a two from right to left (tricky!) to get $(2x -1)(3x + 2)$ - which multiplies out nicely.

But WHY, Colin? Why does it work?

That's a good question. Very well presented. Thank you for asking. I don't have a nice, simple explanation, or else I'd have asked the Mathematical Ninja to show you. I do have an observation, though:

$(ax + b)(cx + d) = ac x^2 + (ad + bc)x + bd$
$(y + bc)(y + ad) = y^2 + (ad + bc)y + abcd$

By shifting the $ac$ - the number in front of the $x^2$ - to the end of the expression, you end up with a factorisation in $y$ that permits switching back to the proper factorisation in $x$ - but I don't have a solid, 'here's why it happens, obviously' explanation. If you know why it works... do let me know!

* Thanks to @S3yM5n for pointing out a typo.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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16 comments on “How I factorise quadratics these days

  • Cav

    I use a similar method, although I don’t frame it like this. I simply examine factor pairs of ac (from the general quadratic formula) to see which cam be summed to find b. Then see how to achieve these factors using the necessary coefficients of x in brackets.

    The reason multiplying the constant term by the coefficient of x works is that the coefficient of x is multiplied by a factor of the constant term when expanding to give one of the x terms which you sum.

    • Colin

      I think that’s what my algebra said – I’ve just figured out why it works, though! The adjusted quadratic is stretched horizontally by a scale factor of $a$ – and in this context, $y = ax$. That gets you $\frac{y^2}a + \frac{by}a + c = 0$, or $y^2 + by + ac = 0$. Finally, that’s been bugging me for weeks!

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    • Sam

      I’m fascinated by your method for factorising quadratics and find it really fast, but after a string of correct answers I’ve suddenly hit a wall. I’m probably making a dumb mistake but I can’t get the right answer for: 3×2 – 28x + 20. I’ve got as far as (x + 2)(x – 30) and substituted the 3 back in to make (3x + 2) (x – 10).
      But this results in 3×2 – 28x – 20. What am I doing wrong?

      I also find the same problem with 3×2 + 10x – 8
      I can get it to work with factors 12 and 2, but not with factors 6 and 4. How do I know (without trial and error) which is the right pair of factors to use?
      Many thanks

      • Colin

        Hi, Sam, thanks for your comment!

        It looks like you’re being troubled by my old nemesis, the minus sign error. When you ‘move’ the 3, you get x^2 -28x + 60, so you’d need two numbers whose product is +60 and whose sum is -28, so they’d both need to be negative. Sadly, 3x^2 – 28x + 20 doesn’t factorise (you can check: b^2 – 4ac = 784 – 240 = 544, which isn’t a square.

        The second one is more fertile ground: you can move the 3 to make it x^2 + 10x -24, and you need the pair with a product of -24 and a sum of 10: they need to have different signs, so +12 and -2 are the pair that work.

        • Sam

          Thanks for your reply Colin. I hadn’t realised the products of the factors still had to make sense after adjusting the signs in the brackets. Your answer to the first question also makes complete sense to me now. Your method is totally brilliant!

          Can I just confirm one thing… Bearing in mind the lack of workings that your method requires (compared with using the ‘A – C’ method), is it still acceptable to use your method in public exams? What if examiners don’t understand the workings… could I lose marks even though the answer is correct?
          Thanks so much for all your help.

          • Colin

            Fantastic 🙂

            As far as I’m aware, factorising quadratics is (at least in any exam I can think of in the UK) an ‘any method goes’ sort of question – whether you do it by trial and error, by inspection, using the formula, or by receiving echoes of radio communications from the vicinity of Betelgeuse, you should always get the marks.

  • Steven Alexander

    Colin, you’ve multiplied the entire quadratic by “a” (5 in this case), to get 25xx –10x – 35 (I can’t enter superscripts). Then you’ve substituted y for 5x, to get yy – 2y – 35. You factor into (y – 7)(y + 5) and restore the x to get (5x – 7)(5x + 5). Then you divide by 5 to finish with (5x – 7)(x + 1). Well done, but only if the original quadratic equals 0.

    • Colin

      Yep, that’s it. You can set it up as $\frac15 (25x^2 – 10x – 35)$ etc, and bring the fifth back in at the end.

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    • Rayit@triggermaths

      Here goes my explanation
      $6x^2+x-2=6\left(x+\frac{x}{6}-\frac{2}{6}\right)$: factors of $ac=(6)(-2)$ that add up to give 1 are 4 and -3, therefore we can factorise by dividing each factor by 6 to give $6\left(x+\frac{4}{6}\right)\left(x-\frac{3}{6}\right)=6\left(x+\frac{2}{3}\right)\left(x-\frac{1}{2}\right)$
      $=6\left( \frac {3x+2}{3}\right)\left(\frac{2x-1}{2}\right)=6\frac{(3x+2)(2x-1)}{6}=(3x+2)(2x-1)$
      What do you think Colin?

      • Colin

        Nice!

  • Pat

    hey, colin, ive been using this trick for the past few years but i simply cannot solve this equation using it:

    2x^2-8x-24

    i get x-6 and x+2.

    any help?

    • Colin

      I think you’ve lost a factor of 2!

      Shuttling the 2 gives you $y^2 – 8x – 48$, which factorises as $(y-12)(y+4)$. Shuttling a 2 back gives either $(x-6)(2x+4)$ or $(2x-12)(x+2)$, both of which factorise to $2(x-6)(x+2)$.

      If you spot there’s a common factor in the original expression, you can divide it out before beginning and work on $2(x^2 – 4x – 12)$, which is less complicated.

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