Right now we are going to work on an important
technique that is used often for finding zeros
of polynomials or factoring. And that is of
course the process of long division using
polynomials.
So the big idea is that you have one polynomial
and you are dividing it by another one.
And through the process of long division you'll
be able to figure out what other smaller polynomials
or even what remainder is left over.
Now it can look a complicated process but
one thing you want to keep in mind is that
it basically mirrors the process of long division
with just normal numbers. So watch for those
good parallels as we go through some examples.
The very first thing you want to do, when
doing long division with polynomials, is make
sure you have the correct set up.
So in my problem that I had on the first board,
I have the polynomial 5x^3 - 6x^2 - 28x - 2.
All of that is being divided by x + 2.
So you'll notice I've set it up into a normal
division bar.
So this first polynomial is what I'm dividing, so
it goes underneath the bar, and I'm dividing
it by the x+2 so it goes on the outside.
Now what we are going to do is basically take
a look at the very first two terms of each
of these polynomials.
I want to think to myself, "What do I have
to multiply x by in order to get a 5x^3 ?"
Now if you think about that for a moment you'll
see that you'd have to multiply x by a 5x^2.
Let's write that on top. 5x^2
Now that will be essentially the first term
of my new polynomial.
So let's double check this, 5x^2 times x indeed
does give us a 5x^3.
I'm going to write that right underneath this
one.
Now we also want to take this and multiply
it by the next term. So multiply by the 2.
So 5x^2 multiplied by 2, +10x^2.
So just like a normal long division problem
you figure out how many times it goes in there,
you write it one top.
Then you multiply through, write it underneath,
and of course the most important step, you
want to subtract that away.
I always like to put a big set of parenthesis
and a negative sign when I'm subtracting so
I can keep track of all of my signs.
Lets see how this works.
So I have 5x^3 minus 5x^3, that will be a
zero.
If you do this process correctly these first
terms will always drop away, always should
give you zero.
If it doesn't double check this term that
you've picked the right one.
Now on to the next term.
A -6x^2 minus 10x^2
So -10 minus 6, looks like I'm at -16x^2.
Again, the reason why I'm putting on these
parenthesis is so you can subtract away all
of the terms. You don't want to miss any signs
in there.
Alright, so looks like we have a -16x^2 and
we want to continue doing this division process
so I'll bring down the next term, and we'll
repeat this one more time.
So again, I want to look at my first two terms
and figure out, "What would I have to multiply
x by in order to get a -16x^2 ?"
So I'd have to multiply by -16x.
Now we'll take this and multiply through by
both of the terms.
So -16x times x, -16x^2
-16x times 2, a -32x.
Alright, looking good.
Of course, now let's subtract that away.
Be very careful with your signs. Notice how
it does work out pretty good for our first
terms.
So a -16x^2 minus a -16x^2, that essentially
like adding a 16x^2.
They'll cancel each other out and give me
a zero.
-28x minus a -32x. So -28x plus 32x.
4x.
Awesome.
So we've done our subtraction, we'll bring
down our next term and of course repeat this.
So what do I have to multiply x by in order
to get a 4x?
Well, I'll have to multiply by 4.
So 4 times x, 4x.
4 times a 2, I'll get an 8.
And then of course we want to subtract this
away.
Alright so 4x minus 4x, gone, like it should
be.
-2 minus an 8, giving me a -10.
So since this now lower in power than what
I'm dividing it by, so it's only like x^0.
This is considered my remainder.
So what we have found is that when take our
original polynomial and we divide by x+2,
we get the answer 5x^2 - 16x + 4 with a remainder
of -10.
Now there are a few things you want to remember
when doing the long division process, that
will help you out for some polynomials.
Let's see those examples.
One of the biggest things you want to remember
is that you should always put in place holders
for missing powers of x.
So this problem looks like a fairly small
one. I'm taking an x^3 - 1 divided by an x-1.
And you know, it doesn't look like a whole
lot is involved.
But as I go to set this up, watch how many
placeholders I have to put in here. So this
is what's being divided.
x^3.
And I have to put in all of those missing
powers of x.
So since I don't have an x^2, I'll put in
a 0x^2.
Since I don't have an x, I'll put in a 0x.
Minus 1.
Now you may be wondering, "Doesn't that mess
up the problem?"
But when we are multiplying by zero here,
essentially that term is zero, and its not
really there anyway. We are just keeping them
in there as a placeholder.
Alright, dividing by x-1.
So if you are missing any powers of x, make
sure you put it those placeholders for the
missing powers.
Alright, now let's continue through the problem.
What do I have to multiply x by in order to
get an x^3?
How about an x^2.
Multiply through, get an x^3.
Multiply by the next term, -x^2.
Once you have your terms, go ahead and subtract
those away.
First two terms are canceling each other out,
awesome.
Zero minus a -x^2, so zero plus a x^2, x^2.
Bring down the next term.
Continue again.
x times what would give me an x^2?
Well I think x is the thing that will do it.
So x times x, x^2
x times a -1, -1x.
And of course we want to subtract that away.
So x^2 minus x^2 is gone.
0x minus -1x, that like plus 1.
So 1x, and we'll bring down our -1.
Alright, looks good.
Let's do this one more time.
x times what would give me a 1x?
A 1.
So 1 times x, 1x
1 times a -1, -1.
And now we can subtract away.
And since these are exactly the same, it looks
like we'll just get zero.
In other words, we have no remainder.
It looks like it went in evenly, awesome.
So remember that when doing the long division process
it looks a lot like normal division with numbers.
Make sure that you write your polynomial that
is being divided in descending order, and
you put in any placeholders for those missing
powers of x.
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