WEBVTT

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>> Welcome to the Cypress
College Math Review

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for Dividing Polynomials.

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Objective 1, Long Division
Without Missing Terms.

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So, for our first
example we want

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to divide the polynomial
2x squared plus 3x minus 5

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by the polynomial x plus 1.

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So, the first thing that you'll
want to notice is that both

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of our polynomials are written
in descending order which means

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that the exponents of x go from
largest to smallest and both

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of these polynomials do
not have any missing terms.

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Later on, we'll look
at a couple examples

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where we do have missing terms

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and what we'll have
to do in that case.

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So, when they present you

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with the problem the first
polynomial, the polynomial

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that you're dividing, that's
going to be called the dividend

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and the second polynomial
that you're dividing

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by is called the divisor.

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So, when you're setting up
your problem the first thing

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that you'll do is write your
long division symbol and inside

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of the long division symbol
should be your dividend.

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Then outside of the long
division symbol will be

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your divisor.

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So, for our example,

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the dividend 2x square plus
3x minus 5 will be written

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underneath the long
division symbol.

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And then the divisor of x plus
1 will be written outside.

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So, to start dividing you want
to focus on the first terms

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of your two polynomials.

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So, we want to focus on
the x and the 2x squared.

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You want to ask yourself
what you need to multiple x

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by to get to 2x squared.

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Well, you would have to multiply
x by 2x to get to 2x squared.

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So, we're going to
write that on top.

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And then from here, we're going
to multiple 2x by the x plus 1.

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So, the 2x times x will
give us 2x squared.

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And then 2x times a positive
1 will give us plus 2x.

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From here, we should
subtract, but to help us

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with our subtraction it
makes it a little bit easier

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if we change the signs.

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So, then the positive 2x squared
will become a negative 2x

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squared and the plus 2x
will become a minus 2x.

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So, once you've changed your
signs you can now just add

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straight down.

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So 2x squared minus 2x
squared will just give us zero.

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So, I won't write
anything there.

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And the 3x minus 2x will
leave us with just an x. Then

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from here you're going to
bring down your next term.

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Bring down the minus 5
and then we're just going

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to repeat the process.

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So, what do you have to
multiply x by to get to x?

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Well, you'd have to multiply
x by a positive 1 to get to x

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and then from here, again,

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we multiply the positive
1 by the x plus 1.

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So positive 1 times
x gives us x.

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And then positive 1 times
a positive 1 will leave us

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with a plus 1.

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And then from here
we should subtract.

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But again, to help us subtract
we'll change our signs.

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So that'll become a
negative x minus 1,

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and then we can just
add straight down.

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So, x minus x will give us zero.

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So again, I won't
write anything there.

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And then negative 5 minus 1
will give us a negative 6.

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So, writing our final answer,
we want to write it of the form,

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quotient plus the
remainder over the divisor.

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So, your quotient is
what we have on top

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of the long division
symbol, the 2x plus 1.

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Our remainder is the negative 6.

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And our divisor is x plus 1.

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So, then our final answer
should be written in this form,

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2x plus 1, our quotient,
plus our remainder negative 6

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over the divisor of x plus 1.

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To make it a little
bit neater though,

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because you have the
plus and the negative 6,

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we can rewrite that
as just minus.

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So, then we can rewrite this

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as 2x plus 1 minus
a 6 over x plus 1.

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For our next example we want

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to divide the polynomial 12x
cubed minus 11x squared plus 9x

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plus 18 by the polynomial
4x plus 3.

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So again, notice that
both our polynomials are

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in descending order and both

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of our polynomials do not
have any missing terms.

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So, to start the division, our
first polynomial, the dividend,

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will go inside of the
long division symbol.

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And then, our second polynomial,
the divisor, will go outside.

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To start the division, you
will focus on the first terms

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of the two polynomials,
so we'll focus

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on the 4x and the 12x cubed.

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And you want to ask yourself
what you need to multiply 4x

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by to get to 12x cubed.

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Well, you would have
to multiply 4x

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by 3x squared to
get to 12x cubed.

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And then, now we multiply the
3x squared by the 4x plus 3.

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So 3x squared times 4x
will give us 12x cubed,

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and then 3x squared times a
positive 3 will give us a plus

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9x squared.

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And then, from here
we should subtract,

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but to make the subtraction
easier we'll change the signs.

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So, the 12x cubed will
become a negative 12x cubed

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and the plus 9x squared will
become a minus 9x squared.

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And then from here, we
can add straight down.

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So 12x cubed minus 12x
cubed just gives us zero.

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And negative 11x squared minus
9x squared will give us a

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negative 20x squared.

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And then, from here
we will bring

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down our next term
of the plus 9x.

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And then we start the
process over again.

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So now we'll focus on the 4x
and the negative 20x squared.

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So, you need to ask yourself
what you need to multiply the 4x

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by to get to negative
20x squared.

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Well, you would have to multiply
4x by a negative 5x to get

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to negative 20x squared.

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And then, from here we multiply
the negative 5x times the 4x

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plus 3.

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So negative 5x times 4x will
give us a negative 20x squared

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and the negative 5x times
a positive 3 will give us a

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negative 15x.

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And then from here
we should subtract,

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but before we subtract,
we will change the signs.

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So, the negative 20x squared
will become a plus 20 x squared

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and the minus 15x will
become a plus 15x.

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And then from here, we
will add straight down.

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So negative 20x squared plus 20
x squared just gives us zero.

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And then 9x plus 15x
will give us 24x.

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And then from here we'll bring

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down our next term
of the plus 18.

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And then we restart
the process again.

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So, looking at the 4x and the
24x you want to ask yourself,

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what do you need to
multiply 4x by to get to 24x.

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Well, you would have
to multiply 4x

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by a positive 6 to get to 24x.

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And then from here we will
multiply the positive 6

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by the 4x plus 3.

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So positive 6 times
4x will give us 24x.

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And then 6 times a positive
3 will give us a positive 18.

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And then from here
we should subtract,

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but before we subtract,
we'll change our signs.

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So, then this becomes a
negative 24x minus 18,

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and then we can add
just straight

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down 24x minus 24x
just gives us zero.

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And then 18 minus 18
also gives us zero.

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So, in this case we got
a remainder of zero.

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So, to write our final
answer in the form

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of quotient plus
remainder over divisor,

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well our quotient is
what we got on top.

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The 3x squared minus 5x plus 6.

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And since we got a remainder

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of zero this will
be our final answer.

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Pause the video and
try these problems.

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With the previous two
examples we worked on,

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we always tracked two things
about the polynomials.

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We checked that they are
written in descending order

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and that there were
no missing terms.

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So now what we're going to look
at will be what we need to do

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if a polynomial does
have missing terms.

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So, let's consider
this polynomial.

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Notice that its written
in descending order

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since the exponents
of x are written

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from largest to smallest.

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And notice that we do not
have any missing terms

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since we have an x to the fourth
term, an x to the third term,

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an x squared term, and x
term, and a constant term.

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So now let's consider
this next polynomial.

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Notice that this polynomial does
have a couple of missing terms.

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So, we start off with
an x to the fourth term,

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and then we have our
x to the third term,

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but we are missing
our x squared term.

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And notice that we're also
missing our constant term.

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So, when a polynomial has
missing terms such as this one,

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we need to rewrite
the polynomial

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so that it has placeholders.

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So, rewriting this polynomial
to include placeholders

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for the missing x squared term

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and the missing constant
term we would write 7x

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to the fourth minus x

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to the third plus 0x
squared plus 5x plus zero.

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Objective 2, Long Division
With Missing Terms.

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So, for our next example we
want to divide the polynomial 3x

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to the fourth minus 12x plus
5 by the polynomial x plus 1.

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So, if you notice that
with our first polynomial,

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we actually have
two missing terms.

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We're missing our
x to the third term

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and also our x squared term.

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So, when we set up our problem,

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we're going to include
placeholders

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to replace those
two missing terms.

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If you look at our second
polynomial, the x plus 1,

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you don't have any missing
terms there so we're good to go

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on the second polynomial.

00:11:47.146 --> 00:11:48.486 A:middle
So, writing our problem,

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underneath the long division
symbol we'll write our first

00:11:52.726 --> 00:11:54.106 A:middle
polynomial, the dividend.

00:11:54.316 --> 00:11:55.916 A:middle
Again, including
the placeholders

00:11:55.916 --> 00:11:57.156 A:middle
for our two missing terms.

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So, we'll write 3x
to the fourth.

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And since we need a
placeholder for our missing x

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to the third term, we'll
write plus 0x to the third

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and then we also
need a placeholder

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for our x squared term, so
we'll write plus 0x squared

00:12:16.796 --> 00:12:21.276 A:middle
and then we have
minus 12x plus 5.

00:12:22.146 --> 00:12:24.356 A:middle
And then, writing our
second polynomial,

00:12:24.426 --> 00:12:26.566 A:middle
our divisor, on the outside.

00:12:28.886 --> 00:12:31.036 A:middle
And then from here we just
follow the same process

00:12:31.036 --> 00:12:32.686 A:middle
as our previous two examples.

00:12:32.906 --> 00:12:36.766 A:middle
So, to start our division
we'll focus on the first terms

00:12:36.766 --> 00:12:40.136 A:middle
of our two polynomials,
so we'll focus on the x

00:12:40.336 --> 00:12:41.806 A:middle
and the 3x to the fourth.

00:12:42.196 --> 00:12:44.846 A:middle
So, you need to ask yourself
what you need to multiply x

00:12:44.976 --> 00:12:47.416 A:middle
by to get to 3x to the fourth.

00:12:47.896 --> 00:12:52.466 A:middle
Well, we would have to
multiply x by 3x to the third

00:12:52.846 --> 00:12:54.776 A:middle
to get to 3x to the fourth.

00:12:56.106 --> 00:12:58.936 A:middle
And then from here,
we'll multiply the 3x

00:12:59.126 --> 00:13:01.726 A:middle
to the third times the x plus 1.

00:13:02.516 --> 00:13:08.046 A:middle
So 3x to the third times x
will give us 3x to the fourth

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and then 3x to the third times a
positive 1 will give us plus 3x

00:13:14.536 --> 00:13:15.856 A:middle
to the third.

00:13:16.356 --> 00:13:18.956 A:middle
And then from here,
we should subtract,

00:13:19.136 --> 00:13:21.696 A:middle
but before we subtract,
we'll change our signs.

00:13:22.126 --> 00:13:27.676 A:middle
So, then this becomes a minus
3x to the fourth minus 3x cubed.

00:13:28.546 --> 00:13:30.806 A:middle
And then from here we can
add just straight down,

00:13:31.246 --> 00:13:35.136 A:middle
so 3x to the fourth minus 3x
to the fourth gives us zero.

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And then zero x cubed minus
3x cubed will leave us

00:13:40.516 --> 00:13:43.626 A:middle
with a negative 3x cubed.

00:13:44.096 --> 00:13:49.696 A:middle
And then we bring down our
next term of plus 0x squared.

00:13:50.776 --> 00:13:54.226 A:middle
And then we repeat the process.

00:13:54.906 --> 00:13:58.986 A:middle
So now we'll focus on the x
and the negative 3x cubed.

00:13:59.026 --> 00:14:02.036 A:middle
So, you should ask yourself
what you need to multiply x

00:14:02.176 --> 00:14:04.956 A:middle
by to get to negative 3x cubed.

00:14:05.356 --> 00:14:11.496 A:middle
Well, we would have to multiply
by a negative 3x squared to get

00:14:11.496 --> 00:14:13.226 A:middle
to the negative 3x cubed.

00:14:14.156 --> 00:14:18.326 A:middle
And then from here we multiply
negative 3x squared times the x

00:14:18.396 --> 00:14:18.986 A:middle
plus 1.

00:14:19.636 --> 00:14:25.446 A:middle
So negative 3x squared times
x gives us negative 3x cubed

00:14:26.016 --> 00:14:30.076 A:middle
and then negative 3x squared
times a positive 1 will give us

00:14:30.076 --> 00:14:32.976 A:middle
a negative 3x squared.

00:14:35.076 --> 00:14:38.036 A:middle
From here we should subtract,
but before we subtract,

00:14:38.036 --> 00:14:39.566 A:middle
we will change our signs.

00:14:39.896 --> 00:14:45.106 A:middle
So, then this becomes a positive
3x cubed plus 3x squared.

00:14:46.536 --> 00:14:48.386 A:middle
And then we'll add down.

00:14:48.826 --> 00:14:52.966 A:middle
So negative 3x cubed plus
3x cubed gives us zero.

00:14:53.866 --> 00:15:00.356 A:middle
And then 0x squared plus 3x
squared gives us 3x squared.

00:15:01.486 --> 00:15:04.006 A:middle
And then from here we will
bring down our next term.

00:15:04.506 --> 00:15:06.816 A:middle
So, we'll bring down
the negative 12x

00:15:06.816 --> 00:15:09.586 A:middle
and then we repeat the process.

00:15:10.336 --> 00:15:14.286 A:middle
So, focusing on the x and
the 3x squared you want

00:15:14.366 --> 00:15:16.336 A:middle
to ask yourself what
you need to multiply x

00:15:16.486 --> 00:15:18.506 A:middle
by to get to 3x squared.

00:15:19.006 --> 00:15:20.416 A:middle
Well, I'll have to multiply x

00:15:21.436 --> 00:15:25.736 A:middle
by a positive 3x to
get to 3x squared.

00:15:26.676 --> 00:15:29.256 A:middle
So 3x times x plus 1.

00:15:29.766 --> 00:15:33.666 A:middle
So 3x times x will
give us 3x squared

00:15:34.276 --> 00:15:38.916 A:middle
and then 3x times a positive
1 will give us plus 3x.

00:15:40.216 --> 00:15:42.216 A:middle
And from here we should
subtract, but again,

00:15:42.216 --> 00:15:44.576 A:middle
before we subtract,
we'll change our signs.

00:15:45.146 --> 00:15:49.816 A:middle
So, then this becomes a
negative 3x squared minus 3x.

00:15:49.816 --> 00:15:52.696 A:middle
And then from here,
we can add straight

00:15:52.696 --> 00:15:56.846 A:middle
down so 3x squared minus
3x squared gives us zero.

00:15:57.306 --> 00:16:03.306 A:middle
And then negative 12x minus 3x
will give us a negative 15x.

00:16:04.766 --> 00:16:08.926 A:middle
And then from here we will drop
down our next term of plus 5.

00:16:09.726 --> 00:16:12.206 A:middle
So, we'll focus on the
x and the negative 15x.

00:16:12.206 --> 00:16:15.636 A:middle
So, you want to ask yourself
what you have to multiply x

00:16:15.786 --> 00:16:17.706 A:middle
by to get to negative 15x.

00:16:18.306 --> 00:16:21.476 A:middle
Well, you'd have to
multiply x by negative 15

00:16:21.966 --> 00:16:23.386 A:middle
to get to negative 15x.

00:16:23.386 --> 00:16:28.066 A:middle
And then we'll multiply the
negative 15 by the x plus 1.

00:16:28.506 --> 00:16:32.806 A:middle
So negative 15 times x
gives us negative 15x

00:16:33.506 --> 00:16:37.066 A:middle
and then negative 15 times
a positive 1 gives us a

00:16:37.066 --> 00:16:38.156 A:middle
negative 15.

00:16:40.206 --> 00:16:41.356 A:middle
From here we will subtract,

00:16:41.476 --> 00:16:44.196 A:middle
but before subtracting
we will change our signs

00:16:44.576 --> 00:16:48.546 A:middle
so that it becomes a
positive 15x plus 15.

00:16:50.586 --> 00:16:52.226 A:middle
From here we will
add straight down,

00:16:52.556 --> 00:16:56.426 A:middle
so negative 15x plus
15x gives us zero.

00:16:56.966 --> 00:17:01.666 A:middle
And then a positive 5 plus 15
will give us a remainder of 20.

00:17:02.866 --> 00:17:05.486 A:middle
And now to write our final
answer, we should write it

00:17:05.486 --> 00:17:09.366 A:middle
of the form quotient plus
remainder over divisor.

00:17:09.796 --> 00:17:12.396 A:middle
So, our quotient is
what we got on top,

00:17:13.186 --> 00:17:21.946 A:middle
the 3x to the third minus
3x squared plus 3x minus 15,

00:17:23.146 --> 00:17:30.176 A:middle
plus our remainder which was 20
over the divisor of x plus 1.

00:17:35.576 --> 00:17:37.646 A:middle
Pause the video and
try these problems.

00:17:47.386 --> 00:17:50.296 A:middle
Objective 3, Factoring
Using Long Division.

00:17:50.506 --> 00:17:53.846 A:middle
So, we can use long division
to help us factor a polynomial,

00:17:54.106 --> 00:17:57.236 A:middle
but the only way that the long
division will help us factor is

00:17:57.266 --> 00:17:59.596 A:middle
if we get a remainder of zero.

00:18:00.116 --> 00:18:02.226 A:middle
So, let's look at this
example where we want

00:18:02.226 --> 00:18:05.326 A:middle
to use long division to show
how the numerator factors.

00:18:05.906 --> 00:18:11.266 A:middle
So, in the numerator we have the
polynomial 3x squared plus 11x

00:18:11.516 --> 00:18:16.016 A:middle
minus 4, and we want to
divide that by 3x minus 1.

00:18:16.696 --> 00:18:19.766 A:middle
So, setting up our problem
the polynomial that's

00:18:19.796 --> 00:18:23.026 A:middle
in the numerator is our
dividend, so that's going

00:18:23.026 --> 00:18:25.226 A:middle
to go inside of the
long division symbol.

00:18:31.036 --> 00:18:36.726 A:middle
And the polynomial that's in
the denominator is our divisor,

00:18:36.726 --> 00:18:38.546 A:middle
so that's going to
go on the outside.

00:18:42.656 --> 00:18:45.406 A:middle
And then from here we
perform the long division.

00:18:45.556 --> 00:18:49.976 A:middle
So, focusing on the 3x and
the 3x squared you want

00:18:50.056 --> 00:18:52.686 A:middle
to ask yourself what
you need to multiply 3x

00:18:52.866 --> 00:18:54.796 A:middle
by to get to 3x squared.

00:18:55.196 --> 00:19:00.396 A:middle
Well, we would have to multiply
3x by x to get to 3x squared.

00:19:01.376 --> 00:19:05.246 A:middle
And then we multiply the
x times the 3x minus 1.

00:19:05.716 --> 00:19:10.046 A:middle
So, x times 3x gives
us 3x squared

00:19:10.046 --> 00:19:15.366 A:middle
and then x times a negative
1 will give us negative x.

00:19:17.456 --> 00:19:21.226 A:middle
From here we will subtract, so
first we will change the signs,

00:19:21.646 --> 00:19:25.616 A:middle
so this will become a
negative 3x squared plus x

00:19:25.616 --> 00:19:27.746 A:middle
and then we add straight down.

00:19:27.936 --> 00:19:32.536 A:middle
So 3x squared minus 3x
squared just gives us zero.

00:19:32.946 --> 00:19:39.136 A:middle
And then 11x plus
x will give us 12x.

00:19:39.316 --> 00:19:41.636 A:middle
From here we will bring
the next term down,

00:19:41.776 --> 00:19:44.156 A:middle
so we'll bring down
the negative 4.

00:19:44.676 --> 00:19:47.506 A:middle
And then we repeat the process.

00:19:48.046 --> 00:19:52.456 A:middle
So now focusing on the 3x and
the 12x, well, we would need

00:19:52.456 --> 00:19:57.466 A:middle
to multiply 3x by
positive 4 to get to 12x.

00:19:58.636 --> 00:20:03.016 A:middle
And then we multiply the
positive 4 by the 3x minus 1.

00:20:03.616 --> 00:20:06.916 A:middle
So, 4 times 3x gives us 12x

00:20:08.886 --> 00:20:12.556 A:middle
and then 4 times negative
1 gives us a negative 4.

00:20:13.106 --> 00:20:16.176 A:middle
Next thing we'll do is subtract,

00:20:16.176 --> 00:20:19.046 A:middle
so in order to subtract
we'll change the signs.

00:20:19.466 --> 00:20:23.296 A:middle
So, then this becomes
a negative 12x plus 4

00:20:23.666 --> 00:20:25.306 A:middle
and then we can add
straight down.

00:20:25.756 --> 00:20:29.236 A:middle
So 12x minus 12x gives us zero

00:20:29.576 --> 00:20:32.806 A:middle
and negative 4 plus
4 gives us zero.

00:20:33.516 --> 00:20:36.016 A:middle
So, notice that we get
a remainder of zero.

00:20:36.436 --> 00:20:39.986 A:middle
So, since we get a remainder
of zero that polynomial

00:20:39.986 --> 00:20:42.766 A:middle
that we have in the
numerator will factor.

00:20:43.406 --> 00:20:47.746 A:middle
So, when you get a
remainder of zero the way

00:20:47.746 --> 00:20:50.776 A:middle
that our number will
factor, it will factor

00:20:50.776 --> 00:20:53.566 A:middle
to be the quotient
times the devisor.

00:20:53.776 --> 00:21:02.606 A:middle
So, our numerator was the
3x squared plus 11x minus 4,

00:21:03.576 --> 00:21:06.396 A:middle
and the way that will
factor will be the quotient,

00:21:06.556 --> 00:21:08.176 A:middle
which is what we got on top,

00:21:08.346 --> 00:21:14.736 A:middle
which was the x plus 4 times the
divisor which is the polynomial

00:21:14.736 --> 00:21:18.306 A:middle
that we had on the
outside, the 3x minus 1.

00:21:20.036 --> 00:21:22.206 A:middle
So, our polynomial
in the numerator,

00:21:22.206 --> 00:21:26.116 A:middle
the 3x squared plus
11x minus 4 will factor

00:21:26.116 --> 00:21:29.756 A:middle
to be x plus 4 times 3x minus 1.

00:21:33.996 --> 00:21:35.976 A:middle
Pause the video and
try these problems.

00:21:45.506 --> 00:21:47.736 A:middle
Objective 4, Synthetic Division.

00:21:48.206 --> 00:21:50.366 A:middle
So, there's actually a
shortcut that we can use

00:21:50.366 --> 00:21:53.026 A:middle
for long division called
synthetic division.

00:21:53.306 --> 00:21:56.096 A:middle
However, you can only
use synthetic division

00:21:56.176 --> 00:21:59.576 A:middle
if the divisor is of
the form x minus c,

00:22:00.076 --> 00:22:04.506 A:middle
which means that for the divisor
the exponent of x can only be 1

00:22:04.786 --> 00:22:05.826 A:middle
and then you would have plus

00:22:05.826 --> 00:22:07.996 A:middle
or minus some constant,
just some number.

00:22:08.706 --> 00:22:11.226 A:middle
So, for this example we
want to determine whether

00:22:11.226 --> 00:22:14.046 A:middle
or not we could use synthetic
division on these problems.

00:22:14.416 --> 00:22:15.906 A:middle
So again, to determine whether

00:22:15.906 --> 00:22:18.356 A:middle
or not you can use synthetic
division all you have

00:22:18.396 --> 00:22:21.326 A:middle
to do is focus on the
divisor and make sure it's

00:22:21.326 --> 00:22:25.296 A:middle
of the form x minus c. so
looking at what we have

00:22:25.296 --> 00:22:29.116 A:middle
in part a let's say we wanted
to divide these two polynomials

00:22:29.116 --> 00:22:31.966 A:middle
and we wanted to see if we
could use synthetic division.

00:22:32.136 --> 00:22:35.096 A:middle
So, you want to focus on
the divisor which is going

00:22:35.096 --> 00:22:38.176 A:middle
to be the second
polynomial, the x minus 3.

00:22:38.176 --> 00:22:42.796 A:middle
And we want to make sure it's
of the form x minus c. So,

00:22:42.796 --> 00:22:47.276 A:middle
looking at that divisor x minus
3 notice that the exponent

00:22:47.276 --> 00:22:51.566 A:middle
of x is just 1 and then you
have a minus some constant.

00:22:51.986 --> 00:22:56.836 A:middle
So, since the divisor is of the
form x minus c, we would be able

00:22:56.836 --> 00:22:59.316 A:middle
to use synthetic
division for that problem.

00:23:00.696 --> 00:23:03.436 A:middle
Now let's look at the two
polynomials that we have

00:23:03.436 --> 00:23:06.256 A:middle
in part b and see if we
can use synthetic division

00:23:06.256 --> 00:23:07.236 A:middle
to divide these two.

00:23:07.236 --> 00:23:10.366 A:middle
So again, you want to
focus on the divisor

00:23:10.496 --> 00:23:14.526 A:middle
which is the second polynomial,
the x squared minus 2.

00:23:15.136 --> 00:23:19.666 A:middle
So, notice for our divisor here
the exponent on the x is 2.

00:23:20.186 --> 00:23:22.586 A:middle
So, since the exponent
on the x is 2

00:23:22.816 --> 00:23:26.766 A:middle
that means the divisor is
not of the form x minus c

00:23:26.766 --> 00:23:30.116 A:middle
so you would not be able
to use synthetic division

00:23:30.116 --> 00:23:31.786 A:middle
for part b. So, you
would just have

00:23:31.886 --> 00:23:34.526 A:middle
to use the regular long
division that we've been using.

00:23:36.236 --> 00:23:39.826 A:middle
Now looking at what we have in
part c, let's again try to see

00:23:39.826 --> 00:23:42.516 A:middle
if we would be able to use
synthetic division here.

00:23:42.986 --> 00:23:44.956 A:middle
So, you want to focus
on the divisor

00:23:45.076 --> 00:23:49.876 A:middle
which is the second polynomial,
the 3x squared minus 2.

00:23:50.366 --> 00:23:54.386 A:middle
And we'll see here that
the exponent of x is a 2.

00:23:54.866 --> 00:23:58.616 A:middle
So again, the divisor is
not of the form x minus c,

00:23:58.976 --> 00:24:00.276 A:middle
so we would not be able

00:24:00.276 --> 00:24:02.806 A:middle
to use synthetic division
for part c either.

00:24:02.876 --> 00:24:05.486 A:middle
You would again just have to
use the regular long division

00:24:05.486 --> 00:24:06.496 A:middle
that we've been using.

00:24:08.206 --> 00:24:11.466 A:middle
And then lastly, for
part d, let's say we want

00:24:11.466 --> 00:24:14.386 A:middle
to divide these two
polynomials to see if we want

00:24:14.386 --> 00:24:15.746 A:middle
to use synthetic division.

00:24:15.976 --> 00:24:18.416 A:middle
You want to focus
again on the divisor

00:24:18.416 --> 00:24:21.646 A:middle
which is the second
polynomial, the x plus 1.

00:24:22.196 --> 00:24:25.326 A:middle
So, looking at that divisor
you see that the exponent

00:24:25.406 --> 00:24:30.006 A:middle
of x is just 1 and then you have
just plus some constant number.

00:24:30.506 --> 00:24:33.026 A:middle
So, in this case,
since the divisor is

00:24:33.116 --> 00:24:36.486 A:middle
of the form x minus
c, we would be able

00:24:36.516 --> 00:24:38.146 A:middle
to use synthetic division there.

00:24:45.686 --> 00:24:47.196 A:middle
So, for this example we want

00:24:47.196 --> 00:24:50.366 A:middle
to divide the polynomials
using synthetic division.

00:24:50.876 --> 00:24:54.056 A:middle
So, let's first verify that
we can use synthetic division.

00:24:54.056 --> 00:24:57.976 A:middle
So, if you notice, our
divisor, the x minus 2 is

00:24:57.976 --> 00:25:00.236 A:middle
of the form x minus c. So,

00:25:00.236 --> 00:25:03.246 A:middle
we are able to use synthetic
division for this problem.

00:25:04.046 --> 00:25:06.906 A:middle
So, to set up our problem,
first notice that both

00:25:06.906 --> 00:25:10.196 A:middle
of our polynomials are
written in descending order

00:25:10.516 --> 00:25:12.436 A:middle
and that there are
no missing terms.

00:25:13.476 --> 00:25:16.396 A:middle
Okay, so to set up our
problem you first need

00:25:16.396 --> 00:25:19.946 A:middle
to determine what number will
go in the spot on the outside.

00:25:20.846 --> 00:25:23.276 A:middle
To determine what number
will go there you want

00:25:23.276 --> 00:25:24.356 A:middle
to look at your devisor.

00:25:24.556 --> 00:25:28.136 A:middle
And notice you have a
negative 2 in your divisor.

00:25:28.636 --> 00:25:30.026 A:middle
So, the number that
you want to put

00:25:30.026 --> 00:25:33.196 A:middle
on the outside will be the
opposite of negative 2.

00:25:33.196 --> 00:25:37.256 A:middle
So, the opposite of
negative 2 is a positive 2,

00:25:37.466 --> 00:25:39.756 A:middle
so that's the number
that will go outside.

00:25:40.566 --> 00:25:43.936 A:middle
And then, the determine the
numbers that will go inside

00:25:43.936 --> 00:25:47.136 A:middle
on the top you want to
write out the coefficients

00:25:47.136 --> 00:25:48.986 A:middle
of your first polynomial.

00:25:49.566 --> 00:25:57.616 A:middle
So that will be the 3, the
negative 5, 1, and negative 2.

00:25:59.366 --> 00:26:02.736 A:middle
So now to start the synthetic
division you'll first bring

00:26:02.736 --> 00:26:07.566 A:middle
down that first number
of 3 and then

00:26:07.566 --> 00:26:10.606 A:middle
to determine what number goes
in this spot you're going

00:26:10.606 --> 00:26:13.426 A:middle
to multiply 3 times 2.

00:26:14.036 --> 00:26:18.876 A:middle
So, 3 times 2 gives
us 6 and then

00:26:18.876 --> 00:26:20.696 A:middle
from here we're going
to add straight down.

00:26:20.836 --> 00:26:24.096 A:middle
So negative 5 plus 6 gives us 1.

00:26:24.956 --> 00:26:27.106 A:middle
And then we just keep
repeating the process.

00:26:27.626 --> 00:26:32.816 A:middle
So now we're going to multiply
1 times 2 to give us 2,

00:26:32.816 --> 00:26:38.836 A:middle
and then we add straight
down 1 plus 2 gives us 3.

00:26:40.196 --> 00:26:45.296 A:middle
And then, repeating the
process 4 times 2 gives us 6.

00:26:46.846 --> 00:26:52.306 A:middle
And then adding straight down,
negative 2 plus 6 gives us 4.

00:26:53.896 --> 00:26:56.786 A:middle
So, to write our final
answer you want to focus

00:26:56.786 --> 00:26:58.506 A:middle
on the numbers down below.

00:26:59.126 --> 00:27:03.346 A:middle
So, the 4 is going to be the
remainder and the other numbers,

00:27:03.346 --> 00:27:07.186 A:middle
the 3, the 1, and the 3 are
going to be our coefficients.

00:27:08.016 --> 00:27:10.486 A:middle
So, notice that going
back to our problem,

00:27:10.486 --> 00:27:13.736 A:middle
our first polynomial
was a degree 3.

00:27:13.876 --> 00:27:16.406 A:middle
So, its degree 3 because
the largest exponent

00:27:16.406 --> 00:27:18.966 A:middle
on the x was an exponent of 3.

00:27:19.016 --> 00:27:21.466 A:middle
So that makes it a
degree 3 polynomial.

00:27:21.986 --> 00:27:26.596 A:middle
So, our polynomial for our
answer should be one degree less

00:27:26.766 --> 00:27:27.546 A:middle
than that.

00:27:27.736 --> 00:27:30.336 A:middle
So, since it needs to be
one degree less than 3

00:27:30.686 --> 00:27:34.516 A:middle
that means our polynomial
will be a degree 2 polynomial.

00:27:35.116 --> 00:27:37.746 A:middle
So, then the first 3
will be the coefficient

00:27:37.746 --> 00:27:42.556 A:middle
of our x squared term, the
1 will be the coefficient

00:27:42.556 --> 00:27:48.006 A:middle
of our x term, and the 3 is
going to be our constant term.

00:27:48.606 --> 00:27:51.446 A:middle
So, to write our final
answer in the form

00:27:51.486 --> 00:27:55.026 A:middle
of quotient plus
remainder over divisor,

00:27:55.496 --> 00:28:03.816 A:middle
we would get 3x squared
plus x plus 3

00:28:03.816 --> 00:28:07.376 A:middle
and then plus our
remainder, which was 4,

00:28:08.146 --> 00:28:11.746 A:middle
over our divisor
which was x minus 2.

00:28:17.356 --> 00:28:21.246 A:middle
For this problem we want

00:28:21.246 --> 00:28:24.826 A:middle
to divide the two polynomials
again using synthetic division.

00:28:25.206 --> 00:28:28.706 A:middle
So, to just verify that we are
able to use synthetic division,

00:28:29.086 --> 00:28:33.196 A:middle
you can notice that with our
divisor of x plus 3 it is

00:28:33.196 --> 00:28:35.276 A:middle
of the form x minus c. So,

00:28:35.276 --> 00:28:37.716 A:middle
we are able to use
synthetic division here.

00:28:38.436 --> 00:28:40.996 A:middle
So, before you divide
any two polynomials,

00:28:41.406 --> 00:28:44.686 A:middle
whether you're using long
division or synthetic division,

00:28:45.056 --> 00:28:48.566 A:middle
you want to check that both of
your polynomials are written

00:28:48.566 --> 00:28:52.356 A:middle
in descending order and that you
don't have any missing terms.

00:28:52.846 --> 00:28:56.486 A:middle
So, you'll notice that with our
first polynomial we actually do

00:28:56.486 --> 00:28:57.656 A:middle
have a missing term.

00:28:58.086 --> 00:29:02.366 A:middle
We're missing our x squared term
so since we have a missing term,

00:29:02.536 --> 00:29:06.316 A:middle
we need to rewrite that first
polynomial using a placeholder.

00:29:06.786 --> 00:29:09.546 A:middle
So, rewriting that first
polynomial you should have it

00:29:09.636 --> 00:29:14.496 A:middle
this way, including that plus
0x squared as a placeholder

00:29:14.536 --> 00:29:16.426 A:middle
for a missing x squared term.

00:29:17.406 --> 00:29:20.246 A:middle
So now to set up the
synthetic division problem.

00:29:20.636 --> 00:29:23.966 A:middle
To figure out what number will
go on the outside you want

00:29:23.966 --> 00:29:25.496 A:middle
to take a look at the divisor.

00:29:26.166 --> 00:29:28.616 A:middle
So, since you have a positive 3

00:29:28.706 --> 00:29:31.126 A:middle
in the divisor you want
the opposite of that.

00:29:31.646 --> 00:29:34.586 A:middle
So, the opposite of
positive 3 is negative 3.

00:29:34.586 --> 00:29:37.116 A:middle
So that's what will go outside.

00:29:37.116 --> 00:29:40.406 A:middle
And then to determine
what numbers will go

00:29:40.406 --> 00:29:44.036 A:middle
on top inside you want to
look at the coefficients

00:29:44.166 --> 00:29:46.026 A:middle
of our first polynomial.

00:29:46.626 --> 00:29:56.106 A:middle
So those coefficients are
1, 5, 0, negative 6, and 3.

00:29:56.106 --> 00:30:00.166 A:middle
So to begin the synthetic
division you'll bring

00:30:00.166 --> 00:30:01.716 A:middle
that first number down.

00:30:02.046 --> 00:30:03.626 A:middle
So, we'll bring down the 1.

00:30:05.456 --> 00:30:06.656 A:middle
And then from here we're going

00:30:06.656 --> 00:30:12.856 A:middle
to multiply negative 3
times 1 gives us negative 3,

00:30:12.856 --> 00:30:18.056 A:middle
and then we'll add straight down
5 plus a negative 3 gives us 2.

00:30:19.076 --> 00:30:22.686 A:middle
And then, we repeat the process,

00:30:23.356 --> 00:30:28.166 A:middle
so negative 3 times 2
gives us a negative 6.

00:30:30.176 --> 00:30:35.176 A:middle
Adding straight down, 0 plus
negative 6 gives us negative 6.

00:30:36.516 --> 00:30:38.146 A:middle
Repeating the process again,

00:30:38.506 --> 00:30:44.476 A:middle
negative 3 times negative 6
will give us a positive 18.

00:30:45.576 --> 00:30:50.486 A:middle
Adding straight down negative
6 plus 18 will give us 12

00:30:51.426 --> 00:30:53.746 A:middle
and then repeating the
process one more time,

00:30:53.986 --> 00:30:59.096 A:middle
negative 3 times 12 will
give us negative 36.

00:30:59.096 --> 00:31:01.296 A:middle
And then adding straight

00:31:01.296 --> 00:31:07.326 A:middle
down 3 plus a negative 36 will
leave us with a negative 33.

00:31:09.036 --> 00:31:13.006 A:middle
So, to write our final answer
you want to focus on the numbers

00:31:13.006 --> 00:31:14.526 A:middle
that are at the very bottom.

00:31:15.036 --> 00:31:18.686 A:middle
So, the negative 33 is
going to be our remainder

00:31:19.216 --> 00:31:22.616 A:middle
and the other numbers are
going to be our coefficients.

00:31:23.686 --> 00:31:25.366 A:middle
So, going back to our problem,

00:31:25.506 --> 00:31:30.396 A:middle
notice that our first polynomial
is a degree 4 polynomial

00:31:30.516 --> 00:31:33.056 A:middle
since the largest
exponent on x is 4.

00:31:33.656 --> 00:31:37.466 A:middle
So, for our answer we want it
to be one degree less than that.

00:31:37.646 --> 00:31:41.006 A:middle
So, our answer will be
a degree 3 polynomial.

00:31:42.036 --> 00:31:43.786 A:middle
So, looking at the
numbers down below,

00:31:44.086 --> 00:31:48.026 A:middle
the 1 will be the coefficient
of the x to the third term,

00:31:48.556 --> 00:31:52.786 A:middle
the 2 will be the coefficient
of the x squared term,

00:31:53.406 --> 00:31:57.486 A:middle
the negative 6 will be the
coefficient of our x term,

00:31:57.846 --> 00:32:00.636 A:middle
and then the 12 will
be our constant.

00:32:01.516 --> 00:32:04.426 A:middle
So, to write our final
answer in the form

00:32:04.426 --> 00:32:06.686 A:middle
of quotient plus remainder

00:32:07.016 --> 00:32:19.816 A:middle
over divisor we will get x cubed
plus 2x squared minus 6x plus 12

00:32:20.636 --> 00:32:28.286 A:middle
plus our remainder of negative
33 over our divisor or x plus 3.

00:32:29.476 --> 00:32:32.256 A:middle
And to clean it up just a little
bit since we have that plus

00:32:32.346 --> 00:32:35.516 A:middle
and negative 33, we can
write that as a minus.

00:32:35.516 --> 00:32:38.606 A:middle
So then, we'll rewrite
our final answer

00:32:38.906 --> 00:32:49.016 A:middle
as x cubed plus 2x squared
minus 6x plus 12 minus a 33

00:32:49.726 --> 00:32:51.966 A:middle
over x plus 3.

00:32:58.396 --> 00:33:00.976 A:middle
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