Factoring Quadratics: More "Hard" Examples

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:

To do things properly, I start by pulling out that common factor of 2. This gives me:

Now I need to factor the remaining quadratic:

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:

Then my answer (remembering to include the 2 that I'd factored out front at the start) is:

There are no (non-trivial) common factors — that is, the only common factor is 1 — so there's nothing interesting (like, say, a factor of 2) that I can pull out of all three terms. The leading coefficient is not 1, so I will need to use Box to factor. Because the leading coefficient is also "minus", my first step will be to pull a −1 from all three terms. This gives me:

(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 terms.)

I can see that I'll need factors of ac = (6)(−2) = −12 — so I'll need one "plus" factor and one "minus" factor — 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 complete answer then is:

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:

Now I need to factor the remaining quadratic. I need to find factors of ac = (2)(−12) = −24 — so one of the factors will be a "plus" and the other will be a "minus" — that add to −5 — so the two factors will 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:

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:

6 + 5x + x²

x² + 5x + 6

(x + 2)(x + 3)

The leading term (that is, the term with the largest degree, being in this case the squared-variable 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 associated terms. This gives me:

Now I have a quadratic with a "minus" leading coefficient, so I'll factor a −1 out of all three terms:

This leaves me with a simple quadratic to factor. I need factors of −6 — which is negative, so one factor will be a "plus" and the other will be a "minus" — that add up to −1 — so the factors will be one unit apart, and the larger factor will get the "minus" sign. Clearly, I should use +2 and −3. Including the −1 that I pulled out front, this gives me:

For this quadratic, ac = (20)(−63) = −1260. (Holy fudge!)

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.

My calculator tells me that 8 isn't a factor of 1,260 (how? by returning a value with decimal places when I divided 1260 by 8), but 9 is a factor (and I can plainly see that 10 is), so:

Skipping the remaining numbers that my calculator tells me aren't factors, I get the rest of my list:

Finally! Yes, I could keep going and find the two more factor pairs for 1,260, but I've finally found a pair that works (and there will be only one pair that will work for a given quadratic), 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 Box gives me this:

Then my factorization is:

You should expect at least one 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 classmate's paper).

This may look bad, what with the y² 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:

For this slightly-different quadratic, they must have multiplied factors in this slightly-different form:

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:

See? Easy! And my answer is: