Main content

## Topic A: Lesson 1: Solutions to polynomial equations

Current time:0:00Total duration:8:02

# Factoring higher-degree polynomials

CCSS.Math:

## Video transcript

Plot the real zeroes of the
given polynomial on the graph below. And they give us p of x
is equal to 2 x to the 5th plus x to the 4th
minus 2x minus 1. And when they say
plot it, they give us this little widget here. Where if we click at any point
on this, we get our point. And we get as many
points as we would like. And we can drag
these points around. Or if we don't want
these points anymore, we can just dump them
in this little trash bin at the bottom right. So let's think about what
the zeroes of this polynomial actually are. And to do that I'll
take my scratch pad out. And this is a little
daunting at first. This is a 5th degree
polynomial here. Factoring 5th degree polynomials
is really something of an art. You're really going to have
to sit and look for patterns. If they're actually expecting
you to find the zeroes here without the help of a
computer, without the help of a calculator, then there
must be some type of pattern that you can pick out here. So let me just rewrite p of x. So p of x is equal to 2 x to
the 5th plus x to the 4th minus 2x minus 1. And one way that's
typically seen when you're trying to factor
this type of polynomial is to try to essentially undo
the distributive property a few times. And if you want to relate it
to techniques for factoring quadratics, it's essentially
factoring by grouping. So for example, you
see a 2x minus 1, or something that looks like
a 2x minus 1 right over here. And over here you have a 2x
to the 5th plus x to the 4th. So you have a 2x
of a higher degree term plus a 1 x of
a one degree lower. So there seems to be
some type of a pattern. 2 times x of a
higher degree-- this is a first degree
term-- minus 1 times-- you could do this as x to the
0-- of a lower degree term. And so let's think
about it a little bit. What happens if we essentially
try to group these two terms, and we group these two
terms right over here. And we try to
factor out anything to essentially clean
it up a little bit, to see if we can
make sense of it. Well these two terms, their
greatest common factor is x to the 4th. We could write this as x
to the 4th times 2x plus 1. And this should get us excited
because this looks pretty close to that,
especially if we were to factor out a negative 1 here. So we could factor
out a negative 1. And then this is
going to be 2x plus 1. And that's exciting because now
we can factor out a 2 x plus 1 from each of these terms. So you have a 2x plus 1. We're gonna factor
both of these out, to get 2x plus 1-- which
we just factored out. And if you factored it out
of this term right over here, you're left with x to the 4th. And if you factor
it out this term, you're left with
just the minus 1. Minus 1. And now this is exciting
because 2x plus 1, this is pretty
easy to figure out when does this thing equal 0. And we'll do that
in a little bit. And this is pretty
easy to factor. This is a difference of squares. This right over here
can be rewritten as x squared plus 1
times x squared minus 1. And of course we still have
this 2x plus 1 out front. 2x plus 1. And once again we have
another difference of squares. We have another difference
of squares right over here. That's the same thing as
x plus 1 times x minus 1. And let me just write all the
other parts of this expression. x squared plus 1. And you have 2x plus 1. 2x plus 1. And I think I factored
p of x about as much as could be reasonably expected. So p of x is equal to
all of this business. And remember, the whole reason
why I wanted to factor it is I wanted to figure out when
does this thing equal 0? So if p of x can be
expressed as the product of a bunch of these
expressions, it's going to be 0 whenever at
least one of these expressions is equal to 0. If any of these
is equal to 0 then that's just going to make this
whole expression equal to 0. So when does 2x plus 1 equal 0? So 2x plus 1 is equal to zero. Well you could probably
do this in your head, or we could do it
systematically as well. Subtract 1 from both sides,
you get 2x equals negative 1. Divide both sides by 2, you
get x is equal to negative 1/2. So when x equals negative 1/2--
or one way to think about it, p of negative 1/2 is 0. So p of negative 1/2 is 0. So this right over here
is a point on the graph, and it is one of
the real zeroes. Now we could try to solve this.
x squared plus 1 equals 0, and I'll just write it
down just to show you. If we try to isolate
the x term on the left-- subtract 1 from
both sides-- you get x squared is equal
to negative 1. Now if we were to start thinking
about imaginary numbers, we could think about
what x could be. But they want us to
find the real zeroes. The real zeroes. So there's no real
number where that number squared is equal to negative 1. So we're not going to
get any real zeroes by setting this
thing equal to 0. There's no real number
x where x squared plus 1 is going
to be equal to 0. Now let's think about when x
plus 1 could be equal to 0. We'll subtract 1
from both sides. You get x is equal
to negative 1. So p of negative 1
is going to be 0. So that's another one of
our zeroes right there. And then finally let's
think about when x minus 1 is equal to 0. We'll add one to both sides. X is equal to 1. So we have another real
0 right over there. And so we could plot them. So it's negative 1,
negative 1/2, and 1. So it's negative 1,
negative 1/2 and 1. And we can check our
answer, and we got it right. Now one thing that might be
bugging you, is like hey Sal, you just happened to group
this in exactly the right way. What if I try to group
it in a different way? What if I tried to-- Well
actually let's try to do that. That could be interesting. Just to show you that
this isn't voodoo. And actually there's
several ways to get there. There's several
ways to get there. So what if instead of writing
it like this, where you're writing it kind of
in the highest degree term and the next highest
degree and so on and so forth, you were to write it like
this: p of x is equal to 2 x to the 5th minus 2x plus
x to the 4th minus 1. Well actually, even
this way you could do a fairly
interesting grouping. If you grouped
these two together, you see that they have
the common factor 2 x. You factor 2x out, you get 2x
times x to the 4th minus 1. And I think you see
what's going on. And then this can be
rewritten as plus 1 times x to the 4th minus 1. Minus 1. And now you can factor out
an x to the 4th minus 1. And you're left with-- I'll
do this in a neutral color-- x to the fourth
minus 1 times 2 x plus 1, which is much
easier to factor now. A difference of squares. Exactly what we did
the last time around. So there are several
ways that you could have reasonably
grouped this, and reasonably undone the
distributive property. But I do admit it is
something of an art. You really just have
to play around in see, well let's group
the first two terms, let's see if there's
a common factor here. Let's group the
second two terms, let's see if there's
a common factor here. Hey, once we factor out
those common factors, it looks like both
of these two terms have this common
expression as a factor. And then you could start
to factor that out.