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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 above. In this case, we first need to multiply "a" and "c", and then find factors of the product "ac" that add up to "b". For instance:

Factor 2x² + x – 6. 

Looking at this quadratic, we have a = 2, b = 1, and c = –6, so ac = (2)(–6) = –12. Then we 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, we need one factor to be positive and one to be negative (positve times negative is negative). Then we want to use the pair "3 and 4", and we want the 3 to be negative, because –3 + 4 = +1. Now that we have found our factors, we use what my students call "box": we draw a two-by-two grid, and put the first term in the upper left-hand corner, and the last term in the lower right-hand corner, like this:

Then we take our factors –3 and 4, and put them, complete with signs and variables, in the diagonal corners, like this:

(It doesn't matter which way you do the diagonal entries; the answer will work out the same anyway!)

Then factor out like this:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

from the top row	from the bottom row
from the left column	from the right column

(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 we have factored the box, we can read off our answer:

2x² + x – 6 = (2x – 3)(x + 2).

If your text or teacher has you factoring "by grouping", you'll find that it is very easy to make mistakes with the signs. You'll still have to find the numbers that add to the coefficient in the middle, 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 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 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 case, 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". Let's try another one.

Factor 4x² – 19x + 12. 

Then a = 4, b = –19, and c = 12, so ac = 48.  Since 48 is positive, we 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, we 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, we will use –3 and –16:




4x² – 19x + 12 = (x – 4)(4x – 3).

Factor 5x² – 10x + 6

We have a = 5, b = –10, and c = 6, so ac = +30. Since ac is positive and b is negative, we 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. In this case, the quadratic is said to be "unfactorable over the integers" (because we couldn't find integers 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.)

Now YOU try one!

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