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.