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Factoring Quadratics: The Hard Case:
     The Modified "a-b-c" Method, or "Box"
(page 2 of 4)

Sections: The simple case, The hard case, The weird case

To factor a "hard" quadratic, we have to handle all three coefficients, not just the two we handled in the "easy" case, because the leading coefficient (the number on the x2 term) is not 1. The first step in factoring will be to multiply "a" and "c"; then we'll need to find factors of the product "ac" that add up to "b".

  • Factor 2x2 + x 6. 
  • Looking at this quadratic, I have a = 2, b = 1, and c = –6, so ac = (2)(–6) = –12. So I need to find factors of –12 that add up to +1. The pairs of factors for 12 are 1 and 12, 2 and 6, and 3 and 4. Since –12 is negative, I need one factor to be positive and the other to be negative (because positve times negative is negative). This means that I'll want to use the pair "3 and 4", and I'll want the 3 to be negative, because –3 + 4 = +1. Now that I have found my factors, I will use what my students refer to as "box": I will draw a two-by-two grid, putting the first term in the upper left-hand corner and the last term in the lower right-hand corner, like this:

      Inserting 2x^2 and -6x.

    Then I will take my factors –3 and 4 and put them, complete with their signs and variables, in the diagonal corners, like this:

      Inserting -3x and +4x.

    (It doesn't matter which way you do the diagonal entries; the answer will work out the same either way!)   Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

    Then I'll factor the rows and columns like this:

      from the top row from the bottom row
      From 2x^2 and -3x, factor out x. From +4x and -6, factor out +2.
      from the left column from the right column
      From 2x^2 and +4x, factor out 2x. From -3x and -6, factor out -3.

    (Note: The signs for the bottom-row entry and the right-column entry come from the closest term that you are factoring from. Do not forget your signs!)

    Now that I have factored the box, I can read off my answer from across the top and along the left-hand side:

      2x2 + x 6 = (2x 3)(x + 2).

If your text or teacher factors "by grouping", you'll find that it is very easy to make mistakes with the signs. Using that method, you'll still have to find the numbers that add to the coefficient in the middle (the "3 and 4" in the above example), but your steps would look like this:

    2x2 + x 6 = 2x2 + 4x – 3x – 6 = 2x(x + 2) – 3(x + 2) = (x + 2)(2x – 3)

You would get the same answer, but my students have always found "box" to be easier and more reliable, especially in cases like the one above where you're having to try to keep track of "minus" signs. In my experience, students using "by grouping" get in the habit of dropping the sign in the middle ("isn't it always a 'plus' sign?") and generally forget to factor the "minus" sign out of the second "group" correctly. (In this example, the student would either have gotten factors of "x + 2" and "x – 2", and been stuck, or else would have factored as "(x + 2)(2x + 3)". Either way, he would have gotten the wrong answer.) It was this continual confusion that led me to switch from factoring "by grouping" to using "box". My students just do better with "box", but you should use whatever works best for you.




  • Factor 4x2 – 19x + 12. 
  • The coefficients are a = 4, b = –19, and c = 12, so ac = 48. Since 48 is positive, I need two factors that are either both positive or else both negative (positive times positive is positive, and negative times negative is positive). Since –19 is negative, I need the factors both to be negative. The pairs of factors for 48 are 1 and 48, 2 and 24, 3 and 16, 4 and 12, and 6 and 8. Since –3 + (–16) = –19, I will use –3 and –16:

      4x^2 - 19x + 12 = (x - 4)(4x - 3)

      4x2 – 19x + 12 = (x – 4)(4x – 3).

  • Factor 5x2 – 10x + 6
  • I have a = 5, b = –10, and c = 6, so ac = +30. Since ac is positive and b is negative, I need to find two factors that are both negative and which add up to –10. But the pairs of factors for 30 are 1 and 30, 2 and 15, 3 and 10, and 5 and 6. None of these pairs adds to 10.

    Then this quadratic is said to be "unfactorable over the integers" (because I couldn't find integer factors that worked), or it might be called "prime". The specific terminology you should use will probably depend upon your text. If in doubt, ask your teacher what terminology you should use to refer to unfactorable quadratics.

The following example illustrates the fact that "box" works only if you have first removed all common factors.

  • Factor 2x2 – 4x – 16

If I don't first take out the common factor of "2", I'll find the factors of 2×(–16) = –32 that add to –4, which are –8 and +4. Doing "box", I then get:

    (2x - 8)(2x + 4)

In other words, I'll have factored as (2x – 8)(2x + 4). But when I multiply this back out, I'll get 4x2 for the leading term, instead of 2x2. By not taking that common factor out first, I'll have managed to get "extra" factors in "box"; in particular, I'll have gotten the wrong answer. To do things properly:

    First, I need to remove the common factor of 2 from each term, to get 2x2 – 4x – 16 = 2(x2 – 2x – 8). Then I need to factor the remaining quadratic: x2 – 2x – 8 = (x – 4)(x + 2). I need to be careful not to forget the factored-out "2" when I write down my final answer:

      2x2 – 4x – 16 = 2(x – 4)(x + 2).

Now YOU try one!

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Cite this article as:

Stapel, Elizabeth. "Factoring Quadratics: The Hard Case." Purplemath. Available from Accessed



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