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.


59 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?

    • Peter Griffan

      Thank you Colin you really helped with my math notes

  • Sakshi

    So what will be the factors of 2x^2+3x-90?

    • Colin

      Hi, Sakshi,

      What have you tried? I’d start by listing factors of -180 and seeing which pair has a difference of 3.


  • John

    this is a prime number what do I do then?

    • Colin

      Hi, John,

      In this context, there’s nothing special about prime numbers: you look at the possible factors of -3: either -1 and 3 (which sum to 2) or -3 and 1 (which sum to -2). There is no (integer) pair of factors that sums to -1, so this quadratic doesn’t factorise. Instead, you can solve it by completing the square or using the formula.

  • Rotimi

    How do I go about x³+8
    And x²-4

    • Colin

      The first of those is not a quadratic – you would probably need to spot a solution (there’s one obvious real number that works) and then use long division or similar to find the other factor.

      If you don’t spot the pattern for $x^2 – 4$, you can turn it into $x^2 + 0x – 4$ and use the usual factorising techniques.

  • Sakshi

    I got that sir
    It is 15 & 12

    • Colin

      OK, so now follow the method. You’ve got -12x and 15x in the middle, making it $2x^2 – 12x + 15x – 90$. What do you get if you factorise the two halves?

      • Sakshi

        After solving it i got (2x+15)(x-6)

  • Andrian

    I got this…

    • Colin

      For $6x^2 -7x + 2$?

      You would need a pair of numbers that added to -7 (which -1 and -6 do) but multiply to make $6\times 2=12$, which -1 and -6 don’t. See if you can make it work with another pair.

      • Unity

        How can i factorise this 5m²-45n²

        • Colin

          I would factor out the 5 and use the difference of two squares.

  • maeva

    Hello And for x²-2x-5 please cuz i’m stuck

    • Colin

      I’m afraid that one doesn’t factorise. If that’s equal to zero, you can solve it by completing the square or applying the formula – I get $x^2-2x-5 \equiv (x-1)^2 – 6$, so its zeros will be at $x=1\pm \sqrt{6}$.

  • Kate

    How do i solve this?

    • Colin

      Exactly the same way! You need to find two numbers that multiply to make -45, but add to make -4. I’d say 5 and -9 fit the bill nicely.

      Then $3a^2 – 4a – 15 = 3a^2 – 9a + 5a – 15 = 3a(a-3) + 5(a-3) = (3a+5)(a-3)$.

      That gives solutions of $a=-\frac{5}{3}$ and $a=3$.

  • Andrew Swift

    LEGEND!!!!!! I never understood factorisation so far this is the easiest method. Thank you so Much!!!!

  • Maina

    Very helpful, thanks

  • Farseena Aziz

    Thank you. It really helped me

    • Colin

      I’m happy to hear that! Thanks for letting me know :o)

  • emma

    do you have the answers for all the sample questions? thanks colin!

    • Colin

      I do, but you can check your work by expanding your factorised expressions (which is probably better practice). Alternatively, if you post your work, I’ll let you know whether they’re right.

  • Julia

    Hi! I love this website it has helped so much but could you please do an example with only 2 numbers, for example,
    how to factorise 16x^2 – 9 ?

    • Colin

      That works with the same pattern — as long as you write it as $16x^2 + 0x – 9$.

      (Typically, it’s simpler to use the ‘difference of two squares’ pattern to write it as $(4x + 3)(4x – 3)$ directly, though.)

  • scott robinson

    How do you expand this

    • Colin

      Expand what?

  • Courtney

    Just wondering what to do when the equation is like this? 3x^2-6x^4+3x

    • Colin

      That’s not a quadratic, but a quartic! I’d start by rearranging it into descending powers of $x$ to make it $-6x^4 + 3x^2 + 3x$.

      Then, there’s a common factor of $-3x$, which we can take out: $-3x \br{2x^3 – x – 1}$.

      There’s also a factor of $(x-1)$ in there, making it $-3x \br{x-1}\br{2x^2 + 2x + 1}$, with no other rational factors.

  • Zick Kolala

    Would like to be receiving Maths updates

  • Joseph

    Having problem with these

    • Colin

      What happened when you tried the method in the article?

  • Benz

    Thank very Much Mister Colin!
    Your method are super cool and easy to understand .I wish I had came across your website some few years earlier .But I’m still happy to have found it even now ;I now find great delight in solving those Equations .Thank you again

  • Aaris Amaan

    what to do with this I’m not getting…

    • Colin

      I would try letting $z = 3x+2y$ and going from there. (Don’t forget to replace z once you’ve factorised, though!)

  • Jarlath

    Hi Colin, nice site!

    On 3x^2-8x-3, I get as far as subbing in the factors so it becomes 3x^2+x-9x-3.

    It didn’t resolve properly when I factorised the halves as (3x^2+x) and (-9x-3), and noticed that it does resolve properly if I used a positive 9x.

    So is the operator supposed to be discounted if it is negative? Both halves always begin positive?

    Cheers 🙂

    • Colin

      It does resolve properly if you’re careful with your minus signs! $(3x^2 + x) + (-9x -3) = x(3x+1) – 3(3x+1) = (x-3)(3x+1)$.

      It does generally go easier if you can make the second lead term positive, though: $(3x^2 – 9x) +(x – 3) = 3x(x-3) + 1(x-3) = (3x+1)(x-3)$, the same thing with the factors reversed.

      (I’m not sure what you mean about ‘a positive 9x’ – I don’t think you can change the signs willy-nilly!)

      • Jarlath

        So obvious now!! I was not being very careful with my operators at all…

        Thank you very much :o)

  • Narayana

    Thank You Colin
    Your teaching helped me a lot

  • Feefee

    Very helpful! I was so confused and now after reading your instructions, it makes perfect sense.

  • brian

    I cannot solve this problem: factorise 2×3-5×2-4x+3

    • Colin

      Is that $2x^3 – 5x^2 – 4x + 3$? That’s not a quadratic, because the biggest power is bigger than 2.

      However, it does factorise – I think it’s $(2x-1)(x-3)(x+1)$. Some methods you could use are here.

  • Kam Atwal

    Hi, why does this method work?

    My students will want to know!

  • N.S

    How do we factorise 2x^2 + 13xy + 15y^2 ? I don’t understand because the method is different from the usual one (above.

    • Colin

      A very similar way, it turns out! We want two numbers that sum to $13y$ and multiply to make $30y^2$.

      $10y$ and $3y$ fit the bill.

      $2x^2 + 10xy + 3xy + 15y^2 = 2x(x+5y) + 3y(x+5y) = (2x + 3y)(x+5y)$

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I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

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