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 –6x^{2} – x + 2

I will first take out
the minus one to get –6x^{2} – x + 2 = –1(6x^{2} + 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:

–6x^{2} – x + 2 = –1(6x^{2} + 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:

ADVERTISEMENT

Factor –6x^{2} + 15x + 36

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

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

–6x^{2} + 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 + x^{2},
you would just reverse the quadratic to put it back in the "normal"
order, and then factor: 6
+ 5x + x^{2} = x^{2} + 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
+ x – x^{2}

First, I will want to
reverse the quadratic, but I'll need to take care with the signs: 6
+ x – x^{2}
= –x^{2} + x + 6. Then I'll factor
out the –1,
and factor the remaining quadratic as usual:

–x^{2} + x + 6 = –1(x^{2} – x – 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 20x^{2} – 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!)

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.

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

20x^{2} – 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 6x^{2} + xy – 12y^{2}.

This may look bad, what
with the y^{2} 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 + ay term)(another x term + another y term)

...to get that y^{2} 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 6x^{2} + xy – 12y^{2} factors as (2x + 3y)(3x – 4y).

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

Stapel, Elizabeth.
"Factoring Quadratics: Examples Using the 'Box'; Method." Purplemath. Available from http://www.purplemath.com/modules/factquad3.htm.
Accessed