>> Welcome to the Cypress College Math Review
for Dividing Polynomials.
Objective 1, Long Division Without Missing Terms.
So, for our first example we want
to divide the polynomial 2x squared plus 3x minus 5
by the polynomial x plus 1.
So, the first thing that you'll want to notice is that both
of our polynomials are written in descending order which means
that the exponents of x go from largest to smallest and both
of these polynomials do not have any missing terms.
Later on, we'll look at a couple examples
where we do have missing terms
and what we'll have to do in that case.
So, when they present you
with the problem the first polynomial, the polynomial
that you're dividing, that's going to be called the dividend
and the second polynomial that you're dividing
by is called the divisor.
So, when you're setting up your problem the first thing
that you'll do is write your long division symbol and inside
of the long division symbol should be your dividend.
Then outside of the long division symbol will be
your divisor.
So, for our example,
the dividend 2x square plus 3x minus 5 will be written
underneath the long division symbol.
And then the divisor of x plus 1 will be written outside.
So, to start dividing you want to focus on the first terms
of your two polynomials.
So, we want to focus on the x and the 2x squared.
You want to ask yourself what you need to multiple x
by to get to 2x squared.
Well, you would have to multiply x by 2x to get to 2x squared.
So, we're going to write that on top.
And then from here, we're going to multiple 2x by the x plus 1.
So, the 2x times x will give us 2x squared.
And then 2x times a positive 1 will give us plus 2x.
From here, we should subtract, but to help us
with our subtraction it makes it a little bit easier
if we change the signs.
So, then the positive 2x squared will become a negative 2x
squared and the plus 2x will become a minus 2x.
So, once you've changed your signs you can now just add
straight down.
So 2x squared minus 2x squared will just give us zero.
So, I won't write anything there.
And the 3x minus 2x will leave us with just an x. Then
from here you're going to bring down your next term.
Bring down the minus 5 and then we're just going
to repeat the process.
So, what do you have to multiply x by to get to x?
Well, you'd have to multiply x by a positive 1 to get to x
and then from here, again,
we multiply the positive 1 by the x plus 1.
So positive 1 times x gives us x.
And then positive 1 times a positive 1 will leave us
with a plus 1.
And then from here we should subtract.
But again, to help us subtract we'll change our signs.
So that'll become a negative x minus 1,
and then we can just add straight down.
So, x minus x will give us zero.
So again, I won't write anything there.
And then negative 5 minus 1 will give us a negative 6.
So, writing our final answer, we want to write it of the form,
quotient plus the remainder over the divisor.
So, your quotient is what we have on top
of the long division symbol, the 2x plus 1.
Our remainder is the negative 6.
And our divisor is x plus 1.
So, then our final answer should be written in this form,
2x plus 1, our quotient, plus our remainder negative 6
over the divisor of x plus 1.
To make it a little bit neater though,
because you have the plus and the negative 6,
we can rewrite that as just minus.
So, then we can rewrite this
as 2x plus 1 minus a 6 over x plus 1.
For our next example we want
to divide the polynomial 12x cubed minus 11x squared plus 9x
plus 18 by the polynomial 4x plus 3.
So again, notice that both our polynomials are
in descending order and both
of our polynomials do not have any missing terms.
So, to start the division, our first polynomial, the dividend,
will go inside of the long division symbol.
And then, our second polynomial, the divisor, will go outside.
To start the division, you will focus on the first terms
of the two polynomials, so we'll focus
on the 4x and the 12x cubed.
And you want to ask yourself what you need to multiply 4x
by to get to 12x cubed.
Well, you would have to multiply 4x
by 3x squared to get to 12x cubed.
And then, now we multiply the 3x squared by the 4x plus 3.
So 3x squared times 4x will give us 12x cubed,
and then 3x squared times a positive 3 will give us a plus
9x squared.
And then, from here we should subtract,
but to make the subtraction easier we'll change the signs.
So, the 12x cubed will become a negative 12x cubed
and the plus 9x squared will become a minus 9x squared.
And then from here, we can add straight down.
So 12x cubed minus 12x cubed just gives us zero.
And negative 11x squared minus 9x squared will give us a
negative 20x squared.
And then, from here we will bring
down our next term of the plus 9x.
And then we start the process over again.
So now we'll focus on the 4x and the negative 20x squared.
So, you need to ask yourself what you need to multiply the 4x
by to get to negative 20x squared.
Well, you would have to multiply 4x by a negative 5x to get
to negative 20x squared.
And then, from here we multiply the negative 5x times the 4x
plus 3.
So negative 5x times 4x will give us a negative 20x squared
and the negative 5x times a positive 3 will give us a
negative 15x.
And then from here we should subtract,
but before we subtract, we will change the signs.
So, the negative 20x squared will become a plus 20 x squared
and the minus 15x will become a plus 15x.
And then from here, we will add straight down.
So negative 20x squared plus 20 x squared just gives us zero.
And then 9x plus 15x will give us 24x.
And then from here we'll bring
down our next term of the plus 18.
And then we restart the process again.
So, looking at the 4x and the 24x you want to ask yourself,
what do you need to multiply 4x by to get to 24x.
Well, you would have to multiply 4x
by a positive 6 to get to 24x.
And then from here we will multiply the positive 6
by the 4x plus 3.
So positive 6 times 4x will give us 24x.
And then 6 times a positive 3 will give us a positive 18.
And then from here we should subtract,
but before we subtract, we'll change our signs.
So, then this becomes a negative 24x minus 18,
and then we can add just straight
down 24x minus 24x just gives us zero.
And then 18 minus 18 also gives us zero.
So, in this case we got a remainder of zero.
So, to write our final answer in the form
of quotient plus remainder over divisor,
well our quotient is what we got on top.
The 3x squared minus 5x plus 6.
And since we got a remainder
of zero this will be our final answer.
Pause the video and try these problems.
With the previous two examples we worked on,
we always tracked two things about the polynomials.
We checked that they are written in descending order
and that there were no missing terms.
So now what we're going to look at will be what we need to do
if a polynomial does have missing terms.
So, let's consider this polynomial.
Notice that its written in descending order
since the exponents of x are written
from largest to smallest.
And notice that we do not have any missing terms
since we have an x to the fourth term, an x to the third term,
an x squared term, and x term, and a constant term.
So now let's consider this next polynomial.
Notice that this polynomial does have a couple of missing terms.
So, we start off with an x to the fourth term,
and then we have our x to the third term,
but we are missing our x squared term.
And notice that we're also missing our constant term.
So, when a polynomial has missing terms such as this one,
we need to rewrite the polynomial
so that it has placeholders.
So, rewriting this polynomial to include placeholders
for the missing x squared term
and the missing constant term we would write 7x
to the fourth minus x
to the third plus 0x squared plus 5x plus zero.
Objective 2, Long Division With Missing Terms.
So, for our next example we want to divide the polynomial 3x
to the fourth minus 12x plus 5 by the polynomial x plus 1.
So, if you notice that with our first polynomial,
we actually have two missing terms.
We're missing our x to the third term
and also our x squared term.
So, when we set up our problem,
we're going to include placeholders
to replace those two missing terms.
If you look at our second polynomial, the x plus 1,
you don't have any missing terms there so we're good to go
on the second polynomial.
So, writing our problem,
underneath the long division symbol we'll write our first
polynomial, the dividend.
Again, including the placeholders
for our two missing terms.
So, we'll write 3x to the fourth.
And since we need a placeholder for our missing x
to the third term, we'll write plus 0x to the third
and then we also need a placeholder
for our x squared term, so we'll write plus 0x squared
and then we have minus 12x plus 5.
And then, writing our second polynomial,
our divisor, on the outside.
And then from here we just follow the same process
as our previous two examples.
So, to start our division we'll focus on the first terms
of our two polynomials, so we'll focus on the x
and the 3x to the fourth.
So, you need to ask yourself what you need to multiply x
by to get to 3x to the fourth.
Well, we would have to multiply x by 3x to the third
to get to 3x to the fourth.
And then from here, we'll multiply the 3x
to the third times the x plus 1.
So 3x to the third times x will give us 3x to the fourth
and then 3x to the third times a positive 1 will give us plus 3x
to the third.
And then from here, we should subtract,
but before we subtract, we'll change our signs.
So, then this becomes a minus 3x to the fourth minus 3x cubed.
And then from here we can add just straight down,
so 3x to the fourth minus 3x to the fourth gives us zero.
And then zero x cubed minus 3x cubed will leave us
with a negative 3x cubed.
And then we bring down our next term of plus 0x squared.
And then we repeat the process.
So now we'll focus on the x and the negative 3x cubed.
So, you should ask yourself what you need to multiply x
by to get to negative 3x cubed.
Well, we would have to multiply by a negative 3x squared to get
to the negative 3x cubed.
And then from here we multiply negative 3x squared times the x
plus 1.
So negative 3x squared times x gives us negative 3x cubed
and then negative 3x squared times a positive 1 will give us
a negative 3x squared.
From here we should subtract, but before we subtract,
we will change our signs.
So, then this becomes a positive 3x cubed plus 3x squared.
And then we'll add down.
So negative 3x cubed plus 3x cubed gives us zero.
And then 0x squared plus 3x squared gives us 3x squared.
And then from here we will bring down our next term.
So, we'll bring down the negative 12x
and then we repeat the process.
So, focusing on the x and the 3x squared you want
to ask yourself what you need to multiply x
by to get to 3x squared.
Well, I'll have to multiply x
by a positive 3x to get to 3x squared.
So 3x times x plus 1.
So 3x times x will give us 3x squared
and then 3x times a positive 1 will give us plus 3x.
And from here we should subtract, but again,
before we subtract, we'll change our signs.
So, then this becomes a negative 3x squared minus 3x.
And then from here, we can add straight
down so 3x squared minus 3x squared gives us zero.
And then negative 12x minus 3x will give us a negative 15x.
And then from here we will drop down our next term of plus 5.
So, we'll focus on the x and the negative 15x.
So, you want to ask yourself what you have to multiply x
by to get to negative 15x.
Well, you'd have to multiply x by negative 15
to get to negative 15x.
And then we'll multiply the negative 15 by the x plus 1.
So negative 15 times x gives us negative 15x
and then negative 15 times a positive 1 gives us a
negative 15.
From here we will subtract,
but before subtracting we will change our signs
so that it becomes a positive 15x plus 15.
From here we will add straight down,
so negative 15x plus 15x gives us zero.
And then a positive 5 plus 15 will give us a remainder of 20.
And now to write our final answer, we should write it
of the form quotient plus remainder over divisor.
So, our quotient is what we got on top,
the 3x to the third minus 3x squared plus 3x minus 15,
plus our remainder which was 20 over the divisor of x plus 1.
Pause the video and try these problems.
Objective 3, Factoring Using Long Division.
So, we can use long division to help us factor a polynomial,
but the only way that the long division will help us factor is
if we get a remainder of zero.
So, let's look at this example where we want
to use long division to show how the numerator factors.
So, in the numerator we have the polynomial 3x squared plus 11x
minus 4, and we want to divide that by 3x minus 1.
So, setting up our problem the polynomial that's
in the numerator is our dividend, so that's going
to go inside of the long division symbol.
And the polynomial that's in the denominator is our divisor,
so that's going to go on the outside.
And then from here we perform the long division.
So, focusing on the 3x and the 3x squared you want
to ask yourself what you need to multiply 3x
by to get to 3x squared.
Well, we would have to multiply 3x by x to get to 3x squared.
And then we multiply the x times the 3x minus 1.
So, x times 3x gives us 3x squared
and then x times a negative 1 will give us negative x.
From here we will subtract, so first we will change the signs,
so this will become a negative 3x squared plus x
and then we add straight down.
So 3x squared minus 3x squared just gives us zero.
And then 11x plus x will give us 12x.
From here we will bring the next term down,
so we'll bring down the negative 4.
And then we repeat the process.
So now focusing on the 3x and the 12x, well, we would need
to multiply 3x by positive 4 to get to 12x.
And then we multiply the positive 4 by the 3x minus 1.
So, 4 times 3x gives us 12x
and then 4 times negative 1 gives us a negative 4.
Next thing we'll do is subtract,
so in order to subtract we'll change the signs.
So, then this becomes a negative 12x plus 4
and then we can add straight down.
So 12x minus 12x gives us zero
and negative 4 plus 4 gives us zero.
So, notice that we get a remainder of zero.
So, since we get a remainder of zero that polynomial
that we have in the numerator will factor.
So, when you get a remainder of zero the way
that our number will factor, it will factor
to be the quotient times the devisor.
So, our numerator was the 3x squared plus 11x minus 4,
and the way that will factor will be the quotient,
which is what we got on top,
which was the x plus 4 times the divisor which is the polynomial
that we had on the outside, the 3x minus 1.
So, our polynomial in the numerator,
the 3x squared plus 11x minus 4 will factor
to be x plus 4 times 3x minus 1.
Pause the video and try these problems.
Objective 4, Synthetic Division.
So, there's actually a shortcut that we can use
for long division called synthetic division.
However, you can only use synthetic division
if the divisor is of the form x minus c,
which means that for the divisor the exponent of x can only be 1
and then you would have plus
or minus some constant, just some number.
So, for this example we want to determine whether
or not we could use synthetic division on these problems.
So again, to determine whether
or not you can use synthetic division all you have
to do is focus on the divisor and make sure it's
of the form x minus c. so looking at what we have
in part a let's say we wanted to divide these two polynomials
and we wanted to see if we could use synthetic division.
So, you want to focus on the divisor which is going
to be the second polynomial, the x minus 3.
And we want to make sure it's of the form x minus c. So,
looking at that divisor x minus 3 notice that the exponent
of x is just 1 and then you have a minus some constant.
So, since the divisor is of the form x minus c, we would be able
to use synthetic division for that problem.
Now let's look at the two polynomials that we have
in part b and see if we can use synthetic division
to divide these two.
So again, you want to focus on the divisor
which is the second polynomial, the x squared minus 2.
So, notice for our divisor here the exponent on the x is 2.
So, since the exponent on the x is 2
that means the divisor is not of the form x minus c
so you would not be able to use synthetic division
for part b. So, you would just have
to use the regular long division that we've been using.
Now looking at what we have in part c, let's again try to see
if we would be able to use synthetic division here.
So, you want to focus on the divisor
which is the second polynomial, the 3x squared minus 2.
And we'll see here that the exponent of x is a 2.
So again, the divisor is not of the form x minus c,
so we would not be able
to use synthetic division for part c either.
You would again just have to use the regular long division
that we've been using.
And then lastly, for part d, let's say we want
to divide these two polynomials to see if we want
to use synthetic division.
You want to focus again on the divisor
which is the second polynomial, the x plus 1.
So, looking at that divisor you see that the exponent
of x is just 1 and then you have just plus some constant number.
So, in this case, since the divisor is
of the form x minus c, we would be able
to use synthetic division there.
So, for this example we want
to divide the polynomials using synthetic division.
So, let's first verify that we can use synthetic division.
So, if you notice, our divisor, the x minus 2 is
of the form x minus c. So,
we are able to use synthetic division for this problem.
So, to set up our problem, first notice that both
of our polynomials are written in descending order
and that there are no missing terms.
Okay, so to set up our problem you first need
to determine what number will go in the spot on the outside.
To determine what number will go there you want
to look at your devisor.
And notice you have a negative 2 in your divisor.
So, the number that you want to put
on the outside will be the opposite of negative 2.
So, the opposite of negative 2 is a positive 2,
so that's the number that will go outside.
And then, the determine the numbers that will go inside
on the top you want to write out the coefficients
of your first polynomial.
So that will be the 3, the negative 5, 1, and negative 2.
So now to start the synthetic division you'll first bring
down that first number of 3 and then
to determine what number goes in this spot you're going
to multiply 3 times 2.
So, 3 times 2 gives us 6 and then
from here we're going to add straight down.
So negative 5 plus 6 gives us 1.
And then we just keep repeating the process.
So now we're going to multiply 1 times 2 to give us 2,
and then we add straight down 1 plus 2 gives us 3.
And then, repeating the process 4 times 2 gives us 6.
And then adding straight down, negative 2 plus 6 gives us 4.
So, to write our final answer you want to focus
on the numbers down below.
So, the 4 is going to be the remainder and the other numbers,
the 3, the 1, and the 3 are going to be our coefficients.
So, notice that going back to our problem,
our first polynomial was a degree 3.
So, its degree 3 because the largest exponent
on the x was an exponent of 3.
So that makes it a degree 3 polynomial.
So, our polynomial for our answer should be one degree less
than that.
So, since it needs to be one degree less than 3
that means our polynomial will be a degree 2 polynomial.
So, then the first 3 will be the coefficient
of our x squared term, the 1 will be the coefficient
of our x term, and the 3 is going to be our constant term.
So, to write our final answer in the form
of quotient plus remainder over divisor,
we would get 3x squared plus x plus 3
and then plus our remainder, which was 4,
over our divisor which was x minus 2.
For this problem we want
to divide the two polynomials again using synthetic division.
So, to just verify that we are able to use synthetic division,
you can notice that with our divisor of x plus 3 it is
of the form x minus c. So,
we are able to use synthetic division here.
So, before you divide any two polynomials,
whether you're using long division or synthetic division,
you want to check that both of your polynomials are written
in descending order and that you don't have any missing terms.
So, you'll notice that with our first polynomial we actually do
have a missing term.
We're missing our x squared term so since we have a missing term,
we need to rewrite that first polynomial using a placeholder.
So, rewriting that first polynomial you should have it
this way, including that plus 0x squared as a placeholder
for a missing x squared term.
So now to set up the synthetic division problem.
To figure out what number will go on the outside you want
to take a look at the divisor.
So, since you have a positive 3
in the divisor you want the opposite of that.
So, the opposite of positive 3 is negative 3.
So that's what will go outside.
And then to determine what numbers will go
on top inside you want to look at the coefficients
of our first polynomial.
So those coefficients are 1, 5, 0, negative 6, and 3.
So to begin the synthetic division you'll bring
that first number down.
So, we'll bring down the 1.
And then from here we're going
to multiply negative 3 times 1 gives us negative 3,
and then we'll add straight down 5 plus a negative 3 gives us 2.
And then, we repeat the process,
so negative 3 times 2 gives us a negative 6.
Adding straight down, 0 plus negative 6 gives us negative 6.
Repeating the process again,
negative 3 times negative 6 will give us a positive 18.
Adding straight down negative 6 plus 18 will give us 12
and then repeating the process one more time,
negative 3 times 12 will give us negative 36.
And then adding straight
down 3 plus a negative 36 will leave us with a negative 33.
So, to write our final answer you want to focus on the numbers
that are at the very bottom.
So, the negative 33 is going to be our remainder
and the other numbers are going to be our coefficients.
So, going back to our problem,
notice that our first polynomial is a degree 4 polynomial
since the largest exponent on x is 4.
So, for our answer we want it to be one degree less than that.
So, our answer will be a degree 3 polynomial.
So, looking at the numbers down below,
the 1 will be the coefficient of the x to the third term,
the 2 will be the coefficient of the x squared term,
the negative 6 will be the coefficient of our x term,
and then the 12 will be our constant.
So, to write our final answer in the form
of quotient plus remainder
over divisor we will get x cubed plus 2x squared minus 6x plus 12
plus our remainder of negative 33 over our divisor or x plus 3.
And to clean it up just a little bit since we have that plus
and negative 33, we can write that as a minus.
So then, we'll rewrite our final answer
as x cubed plus 2x squared minus 6x plus 12 minus a 33
over x plus 3.
Pause the video and try these problems.