Factoring Quadratics: The Hard Case:
Examples of How to Use "Box"
(page 3 of 4)

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

The one special case that often causes students some trouble is when the leading coeffiecient is a negative one. A good first step is to factor out the –1.

• Factor –6x2x + 2
• I will first take out the minus one to get 6x2 x + 2 = 1(6x2 + x 2). (I need to remember that every sign changes when I multiply or divide by a negative. I mustn't fall into the trap of taking the 1 out of only the first term; I must take it out of all three!) Factoring the contents of the parentheses then gives me:

Then:   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved

6x2 x + 2 = 1(6x2 + x 2) = –1(2x – 1)(3x + 2)

Putting these two techniques together (factoring out anything common, and taking out a leading negative sign), you can handle such problems as:

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• Factor –6x2 + 15x + 36
• First, I will first remove the common factor of 3, taking the leading negative sign with it: –6x2 + 15x + 36 =
–3(2x2 – 5x – 12
). Then I'll factor the remaining quadratic: 2x2 – 5x – 12
= (x – 4)(2x + 3)
:

When I write down my answer, I need to remember to include the –3 factor:

–6x2 + 15x + 36 = –3(x – 4)(2x + 3).

A disguised version of this factoring-out-the-negative case is when they give you a backwards quadratic where the squared term is subtracted. For example, if they give you something like 6 + 5x + x2, you would just reverse the quadratic to put it back in the "normal" order, and then factor: 6 + 5x + x2 = x2 + 5x + 6 = (x + 2)(x + 3). You can do this because order doesn't matter in addition. In subtraction, however, order does matter, and you need to be careful with signs.

• Factor 6 + xx2
• First, I will want to reverse the quadratic, but I'll need to take care with the signs: 6 + xx2
= –x2 + x + 6
. Then I'll factor out the –1, and factor the remaining quadratic as usual:

x2 + x + 6 = –1(x2x – 6) = –1(x + 2)(x – 3)

Sometimes the numbers in a factorization are large enough that the factor pair you need is hard to find. But if you list all the factor pairs, in order, you will eventually find the pair you need.

• Factor 20x2 – 17x – 63
• Multiplying a and c, I get (20)(–63) = –1260. Off the top of my head, I have no idea what factors I'll need to use. All I know so far is that those factors will have opposite signs, and that they'll be seventeen units apart. So I'll make a list of factor pairs, and see where that leads. (My calculator can be very helpful for this!)

 factor pairs the differences 1, 1260 2, 630 3, 420 4, 315 5, 252 6, 210 7, 180 9, 140 10, 126 12, 105 14, 90 15, 84 18, 70 20, 63 21, 60 28, 45 30, 42 35, 36 As you can see (to the left), I get  a very long list of factor pairs.   Now that I have my list of factor pairs, I can subtract the pairs to find the differences. If there is a pair of factors with a difference of 17, then I can factor the quadratic. If not, then I will know that the quadratic is prime.       As you can see (to the right), there is one pair of factors that suits my needs; namely, 45 and 28. 1260 – 1 = 1259  630 – 2 = 628 420 – 3 = 417 315 – 4 = 311 252 – 5 = 247 210 – 6 = 204 180 – 7 = 173 140 – 9 = 131 126 – 10 = 116 105 – 12 = 93 90 – 14 = 76 84 – 15 = 69 70 – 18 = 52 63 – 20 = 43 60 – 21 = 39 45 – 28 = 17 42 – 30 = 12 36 – 35 = 1

Now that I have my factor pair (with the larger number having the "minus" sign), I can factor the quadratic:

20x2 – 17x – 63 = (4x – 9)(5x + 7)

By the way, you should expect an exercise as long as this on the next test. Don't waste a lot of time trying to "eyeball" the solution; when you have numbers this big, it is actually faster to write down the list of factor pairs.

There is one other type of quadratic that looks kind of different, but the factoring works in exactly the same way:

• Factor 6x2 + xy – 12y2.

This may look bad, what with the y2 at the end, but it factors just like all the ones above. Remember, from the "simple" case, that we knew that the factors had to be of the form:

(x + something)(x + something else)

...because we knew we'd multiplied factors that looked like this in order to get the quadratic in the first place. This was how we knew that we needed x's in the fronts of our parentheses. In the same way, we know that we must have multiplied factors of the form:

(an x term + a y term)(another x term + another y term)

...to get that y2 term at the end of the quadratic. So we'll need to put y's at the ends of our parentheses. But other than this, the process will work as usual.

First, I need to find factors of (6)(–12) = –72 that add to +1; I'll use +9 and –8.  Then "box" gives me:

So 6x2 + xy – 12y2 factors as (2x + 3y)(3x – 4y).

Quadratic factoring can pop up in even more exotic forms than the last example above....

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 Cite this article as: Stapel, Elizabeth. "Factoring Quadratics: Examples Using the 'Box'; Method." Purplemath. Available from http://www.purplemath.com/modules/factquad3.htm. Accessed [Date] [Month] 2016

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