Class 6 Notes

I. Factor by Grouping
II. Factoring Quadratics
    A. Factoring When a = 1
    B. Factoring When a != 1
    C. The Discriminant
    D. Practice Problems

We've seen how to combine polynomials by multiplication, and how to
divide them by one another.

Now we're going to discuss a variant of division, called
factoring. The goal is to break polynomials down into their
constituent pieces.


I. Factor by Grouping

The simplest form of factoring is noticing that two or more monomials
have a factor in common.


As an example, consider

(on board)
2x^4 + 4x^3 + 6x^2

We can notice that all of the monomials in this polynomial contain an
x. We call x a common factor.

Thus we can re-write this as

(on board)
2x^4 + 4x^3 + 6x^2 = x (2x^3 + 4x^2 + 6x)

Just by looking at this expression, it is appearant that if I were to
multiply out the right-hand side I would get back to the original
expression.

The idea here is to find a factor that all the terms have in common,
and pull it out front.

In the expression we have now, we could in fact factor it further. Can
anyone suggest something else that these terms have in common? (let
class answer)

(on board)
2x^4 + 4x^3 + 6x^2 = x (2x^3 + 4x^2 + 6x)
     = (x)(x) (2x^2 + 4x + 6) = x^2 (2x^2 + 4x + 6)

We can pull out another factor of x. If we'd seen this originally, we
could just have pulled out a factor of x^2.

In fact, there's one more common factor we can pull out. Can anyone
see it? (let class answer)

(on board)
2x^4 + 4x^3 + 6x^2 = x (2x^3 + 4x^2 + 6x)
     = (x)(x) (2x^2 + 4x + 6) = x^2 (2x^2 + 4x + 6)
     = (x^2) (2) (x^2 + 2x + 3) = 2x^2 (x^2 + 2x + 3)

Once we've pull out the full 2x^2, there's no way to pull out more
common terms. There is no number or variable expression that goes into
each monomial term we have left.

Let's consider another example:

(on board)
x^2 y - 3x^2 - 2y + 6

In this case, there is nothing in common between all 4 terms, but we
can find pairs that have something in common.

Can anyone see a pair that of terms that could be factored? (let class
answer)

(on board)
x^2 y - 3x^2 - 2y + 6 = x^2 (y - 3) - 2y + 6

That's one pair. Can we see another pair? (let class answer)

(on board)
x^2 y - 3x^2 - 2y + 6 = x^2 (y - 3) - 2y + 6
    = x^2 (y - 3) + (-2) (y - 3)

Note that I chose to factor out a -2 rather than a 2. The reason for
this is that we can now see in this expression a way to factor even
further.

Can someone tell me how to factor this?

(on board)
x^2 y - 3x^2 - 2y + 6 = x^2 (y - 3) - 2y + 6
    = x^2 (y - 3) + (-2) (y - 3)
    = (y - 3) (x^2 - 2)

Now this has been factored as far as possible. This process is called
factoring by grouping.


II. Factoring Quadratics

Factoring by grouping is fine as far as it goes, but it's pretty
limited.

A much more useful type of factoring is breaking a quadratic
polynomical into pieces. We'll spend the rest of the class on this.

A quadratic is a polynomial of the form

(on board)
a x^2 + bx + c

where a, b, and c are numbers.

Thus,

(on board)
2x^2 + 3x + 4

is an example of a quadratic.

Recall that we got an expression of this form last week when we
multiplied certainbinomials using FOIL.


A. Factoring When a = 1

Let's consider an example. In this case, we'll start with the
binomials, then get the quadratic, then try to see how to go
backwards.

(on board)
(x + 3)(x - 2)

Let's multiply these using FOIL. Someone tell me how to do that. (let
class walk through it)

(on board)
(x + 3)(x - 2) = x^2 - 2x + 3x - 6= x^2 + x - 6

So, suppose we were presented with x^2 + x - 6 and asked to factor
it. How would we do that?

We know it's going to be something of the form

(on board)
(_x + _) (_x + _)

The challenge is to figure out what numbers to fill into the blanks.

The first thing to notice is the first part of foil. The quadratic we
have starts with 1x^2, not 2x^3, or 3x^2, or something like that.

Thus, we know that the coefficients in front of the x's in our fill-in
problem have to give us 1.

Since we're dealing with all integers here, and we want to keep it
that way, this means there's only one possibility: the fill-in slots
in front of x are both 1's.

This gives

(on board)
(x + _) (x + _)

Now we need to fill in the last two slots.

Now we can look at the "last" part of foil. The last terms have to
multiply to give -6, so we need to try pairs of numbers that, when
multiplied, give -6.

There's no way to do this other than by trial and error.

So let's make a list of integers we can multiply to get -6. Someone
tell me some. (let class make list)

(on board)
Possibilities for -6:
1,-6
-1,6
-2,3
2,-3

Note that we have to consider all possible permutations of where the -
sign could be.

Now we just have to try these possibilities one by one until one of
them works.

Let's do that:

(on board)
(x + 1)(x - 6) = x^2 - 6x + x - 6 = x^2 - 5x - 6 -- doesn't work
(x - 1)(x + 6) = x^2 + 6x - x - 6 = x^2 + 5x - 6 -- doesn't work
(x - 2)(x + 3) = x^2 + 3x - 2x - 6 = x^2 + x - 6 -- this work

Once we find the right pair, we're done. We've factored the quadratic:

x^2 + x - 6 = (x - 2)(x + 3)

Let's try another example, this time without knowing the answer in
advance.

(on board)
Factor x^2 - 8x + 16.

How do we do this? (let class walk through it)

(on board)
(x + _)(x + _)

(on board)
Possibilities for 16:
1, 16
-1, -16
2, 8
-2, -8
4, 4
-4, -4

(on board)
Trying each:
(x + 1)(x + 16) = x^2 + 17x + 16 -- doesn't work
(x - 1)(x - 16) = x^2 - 17x + 16 -- doesn't work
(x + 2)(x + 8) = x^2 + 10x + 16 -- doesn't work
(x - 2)(x - 8) = x^2 - 10x + 16 -- doesn't work
(x + 4)(x + 4) = x^2 + 8x + 16 -- doesn't work
(x - 4)(x - 4) = x^2 - 8x + 16 -- this works

(on board)
x^2 - 8x + 16 = (x - 4)(x - 4)


B. Factoring When a != 1

Things are only slightly more complicated when the thing in front of
the x^2 isn't a 1.

Let's consider:

(on board)
Factor 2x^2 + 11x + 12

We can go back to our template for what the answer has to look like:

(on board)
(_x + _)(_x + _)

Looking at the first term, we notice that we have 2x^2, so we're going
to have to multiply two things to get 2.

We can make a possibilities list here:

(on board)
Possibilities for 2:
1,2
-1,-2

We make the same list for 12:

(on board_)
Possibilities for 12:
1,12
-1,-12
2,6
-2,-6
3,4
-3,-4

To check these systematically, unfortunately, we need to consider
every possible was of combining the pairs for 2 and for 12.

For the pairs (1,2) and (1,12), note that there are two possible ways
of combining them.

(on board)
(1x + 1)(2x + 12)
(2x + 1)(x + 12)

For the pairs (1,2) and (2,6) we have:

(on board)
(1x + 2)(2x + 6)
(2x + 2)(x + 6)

We need to check each possible combination, and two orders for each
combination.

As you can see, this quickly becomes extremely long.

There is a way of shortcutting this process, which we'll learn next
week.

However, for now there's nothing to do but try to use some intuition
to see which pair can work.

It helps to be able to do the multiplications in your head, since
that's quicker than writing them down.

Does anyone have a guess for the combination they'd like to try here?
(let class answer)

The on that works turns out to be:

(on board)
(2x + 3)(x + 4) = 2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12

Really the only way to be able to do these quickly is to practice a
lot of them.

However, I can point out some tips that may help shorten your
search. These are rules of thumb to keep in mind.

(on board)
1. If all the numbers in the quadratic are positive, don't bother
checking negative possibilities.
2. You get larger numbers in the middle by taking possibilities for
first and last that are far apart, and smaller numbers in the middle
by taking first and last possibilities that are close to each other.


C. The Discriminant

Notice that not everything is factorable.

For example:

x^2 + 10 x + 1

You can immediately see that this can't be factored: the only things
you could multiply to get 1 are 1 . 1 and (-1) . (-1), and neither of
this will add up to give you 10.

However, it's not always this obvious, so here's a little trick.

There's quantity called the discriminant. It's defined as

b^2 - 4 a c.

If you want to see if something can be factored, just compute this
quantity. If the result is a positive perfect square, you can factor
it. If not, you can't.

Example: for x^2 + 10x + 1, a = 1, b = 10, c = 1, so

b^2 - 4ac = 10^2 - 4(1)(1) = 100 - 4 = 96

The number 96 is not a perfect square, so this quadratic can't be
factored.

On the other hand, let's try:

x^2 - 6x + 8

Here a = 1, b = -6, c = 8, so

b^2 - 4ac = (-6)^2 - 4(1)(8) = 36 - 32 = 4

4 is a perfect square, since 2^2 = 4, so this can be factored:

x^2 - 6x + 8 = (x-2)(x-4).

We'll hear more about the discriminant in a few weeks.


D. Practice Problems

Some practice problems, to be done as time permits:

Factor: ax + bx - ay - by
Factor: b^2+2b-35
Factor: x^2 + 5xy + 6y^2
Factor: 4y^2 - 15y + 9