vimeo.com/.../755665546

College Algebra students,
Dillon here.

We're going to do a little quick
wrap up before Section 3.3.

Here we go.

We're studying polynomials, and we’ve
learned there's a connection between

the x-intercepts (which we also call zeros)
and factors of the polynomial.

An important theorem—

We've learned to do
synthetic division also.

That's what we're going to apply here
that we're learning in Section 3.3.

We have this factor theorem
that says “c is a zero“—

or x-intercept of the polynomial—
“P if and only if x minus c is a factor.”

So, if c is a zero, this
polynomial has to be a factor.

If this is a factor,
then it has to be a zero.

That's what “if and only if” means:
it works in both directions.

Let's see an application of this.

Let P of x, be this
following polynomial.

We're going to show, first of all
that P of 1 equals zero.

Now, according to this factor,
if we could show it does equal zero,

then x minus 1 would
have to be a factor.

Then we can use that to
actually factor the polynomial.

Let me show you what
I'm talking about.

First of all, let's just check
is P of 1 equal to zero.

We're going to have 1 cubed
minus 7 times 1 plus 6.

We have 1 plus 6 is 7,
7 minus 7, is zero, check.

Okay, so if it's zero,
then what has to be true?

Then x minus 1
must be a factor.

We can factor it by using our new skill
that we learned: synthetic division.

Remember, the factor—

if the binomial that we're trying
to divide by is x minus 1,

we put a little 1
here in the corner.

We fill in the coefficients including
a zero if we’re missing any terms

(and we are missing
the squared term).

The coefficient on
x to the third is 1.

And then we'll have a zero,
minus 7, and 6.

And draw the box
below the 6.

And then we apply
synthetic division.

Bring down the 1,
1 times 1 is 1.

Add here, we get 1.

1 times 1 is 1.

Add here we get negative 6, 1 times
negative 6 is negative 6 and we get zero.

Now first of all, we knew that
was going to happen because look,

if you evaluate 1 in the
polynomial, we get zero.

We knew that the remainder
here was going to be zero.

The important part is right here.

Okay, so our polynomial
P of x can now be written

as x minus 1 times—
Read this off.

Remember, this is a
third-degree polynomial.

This quotient must be second degree,
so x squared plus x minus 6.

Now we have a quadratic
that we can try and factor.

I suggest you pause the
video and give it a shot.

It does factor;
I can see that it factors.

I hope you tried
[and] paused the video.

If x minus 1 times x
plus 3 times x minus 2.

Now we can use the skills
that we learned earlier.

We could plot the zeros,
we know the end behavior,

and we could sketch
the polynomial.

Okay, one more problem
to finish up Section 3.3.

We're going to find a polynomial
(fourth-degree) that has these zeros

for which the coefficient
x to the third is negative 6.

Okay, we’ll get
to this part first.

This first part is not that
difficult because each

zero (each x-intercept)
corresponds to a factor.

Let's build this polynomial;
we'll just call a P of x.

As x minus negative 3,
x minus zero, x minus 1, x minus 5.

Each one of those factors
corresponds to one of these zeros,

one of these x-intercepts.

We can clean it up a little.

We are going to have just x here,
so let's put that up front.

And then we're going to have
x plus 3 and x minus 1 and x minus 5.

Now this is a little tougher.

They want the coefficient of the
x to the third term to be negative 6.

Okay? I hope you can see
that if we multiply this out,

it's going to be a
fourth-degree polynomial.

Let's do that.

Let's multiply these two,

then these two.

And then we'll multiply the
two resulting polynomials.

We get x squared plus 3x.

I just multiply these two terms.

Then here we're
going to have x squared.

There's negative 5x minus 1x
is minus 6x plus 5.

Let's multiply that out.

Now I'm going to distribute
the two terms here

to all three terms in
the second polynomial.

Keep an eye on me.

X to the fourth, there's minus
6x to the third, plus 5x squared.

I distributed this term
to all three terms here.

Now let's take the 3x
and distribute it.

We're going to have
plus 3x to the third

minus 18x squared plus 15x.

Let's clean up here;
this is a 15x.

So, we have
x to the fourth.

How many x to the thirds are there?
Well, here's negative 6,

here's positive 3,
so minus 3x to the third.

Here's 5x squared minus
18x squared is minus 13x squared.

And then plus 15x.

Now they wanted this—
Let's go back to the problem.

First of all, we found a polynomial
of degree four that has these zeros.

This polynomial for
sure has these zeros.

Okay, now there's a catch.

They want the coefficient on
x to the third to be negative 6.

The way that we can
force this to be negative 6

is actually to take our polynomial P of x,
multiply everything by positive 2,

so 2x to the fourth, minus 6x
to the third, minus 26x squared—

I'm just multiplying
everything by two— plus 30x.

Okay, so this polynomial—
and I would dare you to graph this—

it's still going to have these
four zeros guaranteed.

But it also has this condition
that the x to the third term

in fact has a coefficient
of negative 6 as desired.

Okay, so there we go.

That's the end of section 3.3.

Finally, now we move on to 3.4.

I think I said that already in the
previous video, but here we go.

Bye.