The easy way to factorise nasty quadratics

Until about three years ago, I had literally no idea how to factorise nasty quadratics. I would turn straight to the quadratic formula, go bing bang boom and say ‘there, job done.’ This was a very effective short-cut – I got a long way with my ignorance – but I’m faintly embarrassed about it now. Especially since I was taught how to do it properly by two of my students (Kat and Cos – to whom I’m still grateful.)

Let’s start with a simple-ish quadratic expression, such as $x^2 -9x + 18$. The way to factorise is to find two numbers that multiply together to make 18 but add to make -9. Eighteen doesn’t have all that many factor pairs – (1, 18), (2, 9), (3,6) and their negative counterparts. The one we’re after is (-3, -6), which just drop into brackets with the $x$s to make $(x-3)(x-6)$. If you multiply that out, you get back to the original equation. So far so simple.

But when you have a number in front of the $x^2$, it’s a bit trickier. How about $2x^2 + 5x – 3$?

The method’s quite similar – except that the number you’re interested in is the first number times the last one ($2 \times -3 = -6$). The possible factors are (1, -6), (2, -3) and the negatives – so to get 5, I want (-1, 6). Here comes the clever bit!

You’re going to change the middle number (with the $x$) into the numbers you just found – so you rewrite the expression as $2x^2 – x + 6x – 3$.

Factorise the two halves of the equation to get $x(2x -1) + 3(2x -1)$, and glomp them together to get $(x + 3)(2x – 1)$. Finished!

One more example: $6x^2 + 5x – 6$.

You’re interested in $6 \times -6 = -36$, which has plenty of factors: (1,-36), (2,-18), (3,-12),(4,-9) and (6,-6) – and their negative friends. So (-4, 9) will give me 5 in the middle.

That leaves me with $6x^2 – 4x + 9x – 6$, which factorises as $2x(3x – 2) + 3(3x – 2)$, or $(2x + 3)(3x – 2)$.

Try it with these!

  • $3x^2 – 8x – 3$
  • $5x^2 – 4x – 1$
  • $8x^2 – 10x + 3$
  • $4x^2 + 7x + 3$

(If you get stuck, try using the formula and see where it’s gone pear-shaped!)


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.


12 comments on “The easy way to factorise nasty quadratics

  • elvaton elkayz

    This place is just great!….i can now manage to solve quadratic equations using factorisation method….thanks Colin.

    • Colin


  • Algy

    Hmmmm; where are I going wrong here!?

    8x^2 -10x +3

    8*3 = 24
    -12 and +2 factors

    8x^2 -12x +2x +3
    => 4x(2x-3) = 8x^2 – 12x

    ??(2x-3) = 2x+3

    • Colin

      Be careful: -12 × 2 isn’t 24. Try -4 × -6.

  • Adrian

    When you say ‘Glomp them together’ how do you get x(2x−1)+3(2x−1) to become (x+3)(2x–1)? I am stuck on that bit.
    BTW thanks great site – I’m enjoying reading about completing the square!

    • Colin

      Hi, Adrian, great question — and thanks for your kind words, glad you’re finding it useful!

      Right, so you’re happy that something like $x\text{ Apples} + 3\text{ Apples} = (x+3)\text{ Apples}$? Similarly, $xz + 3z = (x+3)z$. In general, if you’ve got $x$ of something and $3$ of the same thing, you end up with $(x+3)$ of the thing. Here, the ‘thing’ is $(2x-1)$ — you have $x$ of it, added to $3$ of it, making $(x+3)$ of it!

      Hope that helps 🙂

  • Andy

    Loved this method.

  • Jonathan

    3x^2 -8x-3
    Result : 3x^2 -9x+x-3

    • Colin


  • trinankur saha

    It is helpful


    That really helps but please help me with the following equation:

    6m2-2mn-3mp+np (The 2 after 6m is supposed to be squared.)

    • Colin

      Hi, Emkay!

      My approach to that would be pretty similar, actually: treat $m$ as your $x$, and everything else as numbers.

      You’ve then got $6m^2 + (-2n – 3p)m + np$, and you want a pair of factors that multiply together to make $6np$ but add together to make $-2n-3p$. Can you take it from there?

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