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College Algebra students, 
Dillon here.

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

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Here we go.

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

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

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An important theorem—

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We've learned to do 
synthetic division also.

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

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

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

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

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

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

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Let's see an application of this.

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Let P of x, be this
following polynomial.

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

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

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then x minus 1 would 
have to be a factor.

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Then we can use that to 
actually factor the polynomial.

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Let me show you what
I'm talking about.

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

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

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

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

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Then x minus 1 
must be a factor.

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

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Remember, the factor—

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

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we put a little 1 
here in the corner.

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

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(and we are missing
the squared term).

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The coefficient on 
x to the third is 1.

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

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And draw the box
below the 6.

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And then we apply 
synthetic division.

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Bring down the 1, 
1 times 1 is 1.

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Add here, we get 1.

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1 times 1 is 1.

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

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

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

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

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The important part is right here.

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

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as x minus 1 times— 
Read this off.

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Remember, this is a 
third-degree polynomial.

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

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

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

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It does factor; 
I can see that it factors.

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I hope you tried 
[and] paused the video.

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

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Now we can use the skills
that we learned earlier.

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

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and we could sketch 
the polynomial.

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Okay, one more problem
to finish up Section 3.3.

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

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

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Okay, we’ll get 
to this part first.

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This first part is not that 
difficult because each

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

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

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

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

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one of these x-intercepts.

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We can clean it up a little.

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

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

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Now this is a little tougher.

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

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

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

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Let's do that.

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Let's multiply these two,

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then these two.

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And then we'll multiply the 
two resulting polynomials.

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We get x squared plus 3x.

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I just multiply these two terms.

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Then here we're
going to have x squared.

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

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Let's multiply that out.

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Now I'm going to distribute 
the two terms here

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to all three terms in 
the second polynomial.

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Keep an eye on me.

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

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I distributed this term
to all three terms here.

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

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

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minus 18x squared plus 15x.

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

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So, we have 
x to the fourth.

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

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

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

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And then plus 15x.

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

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

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This polynomial for 
sure has these zeros.

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Okay, now there's a catch.

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

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The way that we can 
force this to be negative 6

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

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

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

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

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

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

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in fact has a coefficient
of negative 6 as desired.

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Okay, so there we go.

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That's the end of section 3.3.

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Finally, now we move on to 3.4.

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

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Bye.