You may have noticed that I started the exercises on the previous page by first checking to see if there is any factor common to all three terms. Taking something common and factoring it out of everything is the simplest factoring, but is often forgotten once the student reaches quadratics. But this is still a consideration. In fact, "box" won't work if a common factor is left inside the quadratic.
Here's what that error looks like:
There's a factor of 2 that's common to all three terms. But let's suppose that I'm being lazy and haven't checked this first:
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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. In other words, I'll need factors of opposite sign (so I'll be subtracting them) which are four units apart.
The factor pairs of 32, and their differences, are:
1, 32: 32 – 1 = 31
2, 16: 16 – 2 = 14
4, 8: 8 – 4 = 4
Okay; so 8 and 4 are four units apart. Since I'm adding to a "minus", I'll slap the "minus" sign on the larger of the two numbers. I'll be using –8 and +4 to split the middle term's coefficient of –4. My "box" looks like this:
This is wrong!
My factorization is (2x – 8)(2x + 4), which I can check by multiplying this back together. But right at the start of multiplying this back out, I see that I'm getting a leading term of 4x^{2}, which is not what I'd started with. So clearly this is wrong!
By not taking that common factor out first, I have managed to create extra factors in "box"; in particular, by not pulling the 2 out front first, I've created an extraneous factor of 2.
To do things properly, I start by pulling out that common factor of 2. This gives me:
2(x^{2} – 2x – 8)
Now I need to factor the remaining quadratic:
x^{2} – 2x – 8
Oh, hey! After pulling out the common factor, this turned into one of the simple-case factorizations! All I need are factors of –8 that sum to –2. In other words, I need factors of 8 that are two units apart, where the larger of the factors (other than sign) gets the "minus" sign. That's easy; I'll use –4 and +2. The quadratic factors as:
(x – 4)(x + 2)
Then my answer (remembering to include the 2 that I'd factored out at the start) is:
2(x – 4)(x + 2)
The one special case that often causes students some trouble is when the leading coeffiecient is a "minus" value. A good first step is to factor that value out of the entire quadratic (or, at least factor the "minus" out of the whole thing).
There are no (non-trivial) common factors, so there's nothing "interesting" (like a 2) that I can pull out of all three terms. The leading coefficient is not 1, so I will need to use "box". Because the leading coefficient is also "minus", my first step will be to pull a –1 from all three terms. This gives me:
–1(6x^{2} + x – 2)
(I need to remember that every sign changes when I multiply or divide through by a "minus". 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!)
I can see that I'll need factors of ac = (6)(–2) = –12 (so one "plus" and one "minus") that add to the middle term's coefficient of 1 (so the factors will be one unit apart, with the larger one getting the "plus" sign). This one is simple enough that I don't need to list factor pairs; I already know that 3 and 4 are one unit apart, so I'll split the 1 by using +4 and –3.
Plugging the contents of the parentheses into "box" gives me:
Remembering the –1 that I factored out front in my first step, my answer then is:
–1(2x – 1)(3x + 2)
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By putting these two techniques together (that is, by factoring out anything common to all three terms, and by taking out a leading "minus" sign), we can now handle factoring messier quadratics, such as:
The leading coefficient is not 1, so I'll be needing to use "box" to factor, as things stand now. However, I can see that there is a factor of 3 that's common to all three terms, and also that the leading coefficient is a "minus"; this tells me that a good first step will be to pull a –3 out front, applying "box" to whatever is left. So my first step gives me:
–3(2x^{2} – 5x – 12)
Now I need to factor the remaining quadratic. I need to find factors of ac = (2)(–12) = –24 (so one will be "plus" and the other will be "minus") that add to –5 (so they'll be five units apart, and the larger factor will get the "minus" sign). Assuming I didn't notice the required factor pair right away, here are the pairs, and their differences:
1, 24: 24 –1 = 23
2, 12: 12 –2 = 10
3, 8: 8 – 3 = 5
There is another factor pair, but I can stop at this point, because I've got the pair I need; namely, 3 and 8. Because I need them to have opposite signs and to add to a "minus", I will put the "minus" sign on the 8. Then "box" looks like this:
When I write down my answer, I need to remember to include the –3 that I factored out front in my first step:
–3(x – 4)(2x + 3)
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A disguised version of this factoring-out-the-"minus" case is when they give us a backwards quadratic where the squared term is subtracted, like this:
6 + 5x + x^{2}
To do the factorization, the first step would be to reverse the quadratic to put it back in the "normal" order
x^{2} + 5x + 6
Then we'd factor in the usual way:
(x + 2)(x + 3)
We can do this "swapping the terms around" in the above case because order doesn't matter in addition. In subtraction, however, order does matter; when flipping around a quadratic with "minus" signs, we need to be careful with those signs. For instance:
The "leading" term isn't up front, so I'll want to start by flipping things around. However, I'll need to take care with the signs, making sure that I carry them along with their (following) terms. This gives me:
–x^{2} + x + 6
This gives me a quadratic with a "minus" leading coefficient, so I'll factor a –1 out of all three terms:
–1(x^{2} – x – 6)
This leaves me with a simple quadratic to factor. The factors of –6 (so one is "plus" and the other is "minus") add up to –1 (so the factors are one unit apart, and the larger one gets the "minus" sign); clearly, I should use +2 and –3. Including the –1 that I pulled out front, this gives me:
–1(x + 2)(x – 3)
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Usually, they'll give us quadratics for which the product ab doesn't have too terribly many factor pairs. However, sometimes ab is large enough that it's painful to find the necessary factor pair. In such a case, we can always create a list (using our calculator to divide progressively larger values into ab to find those pairs), quitting when we find the pair we need. That process looks like this:
For this quadratic, ac = (20)(–63) = –1260. Off the top of my head, I have no frikkin' 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, with the larger of the two getting the "minus" sign. So I'll start listing pairs of factors of 1,260, and see if I can find a pair that works.
1, 1260: 1260 – 1 = 1259
2, 630: 630 – 2 = 628
3, 420: 420 – 3 = 417
4, 315: 315 – 4 = 311
5, 252: 252 – 5 = 247
6, 210: 210 – 6 = 204
7, 180: 180 – 7 = 173
My calculator tells me that 8 isn't a factor of 1,260, but 9 is (and I can plainly see that 10 is), so:
9, 140: 140 – 9 = 131
10, 126: 126 – 10 = 116
Skipping the remaining numbers that my calculator tells me aren't factors, I come up with:
12, 105: 105 – 12 = 93
14, 90: 90 – 14 = 76
15, 84: 84 – 15 = 69
18, 70: 70 – 18 = 52
20, 63: 63 – 20 = 43
21, 60: 60 – 21 = 39
28, 45: 45 – 28 = 17
Finally! Yes, there are two more factor pairs for 1,260, but I've finally found a pair that works, so I can stop here. Since I'm adding to a "minus", I'll be putting the "minus" sign on the larger of the two factors, and my "box" looks like this:
Then my factorization is:
(4x – 9)(5x + 7)
You should expect an exercise as long as this on the next test. When you get a value for ac that's ridiculously large, don't waste a lot of time trying to "eyeball" the solution. When you have numbers this big, it can actually be faster to write down the list of factor pairs. In such a case, write your pairs in your hand-in work, as this will be important evidence that you found the factorization, rather than your calculator or your phone (or "your neighbor's paper").
There is one other type of quadratic that looks kind of different, but the factoring itself works in exactly the same way; it's the case of a quadratic with two squared variables, one at each end of the quadratic:
This may look bad, what with the y^{2} at the end, but it factors just like all the previous quadratics. I recall, from simple factoring, that I know that the factors of a simple-case quadratic had to be in the following form:
(x + something)(x + something else)
For this slightly-different quadratic, they must have multiplied factors in this slightly-different form:
(x-term + y-term)(x-term + y-term)
So I'll have y's at the ends of each of my parentheticals. But other than this, the process will work as usual.
First, I need to find factors of ac = (6)(–12) = –72 (so one "plus" and one "minus") that add to +1 (so they'll be one unit apart, with the larger one getting the "plus" sign). This factorization is easy; I'll use +9 and –8 to split the middle term. So "box" looks like this:
Easy! And my answer is:
(2x + 3y)(3x – 4y)
You can use the Mathway widget below to practice factoring quadratics (or, as the widget calls them, "trinomials"). Try the entered exercise, or type in your own exercise. (Or skip the widget and continue on the next page.) Then click the button to compare your answer to Mathway's.
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They also have an option for checking if a polynomial (such as a quadratic) is "prime"; enter the quadratic, click the button, and select "Determine if Prime".)
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URL: https://www.purplemath.com/modules/factquad3.htm
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