### Video Transcript

In this video, weβre going to learn
how to divide polynomials by monomials. Letβs begin by recalling what we
mean by a polynomial and a monomial.

A polynomial is an expression that
consists of a number of terms either being added or subtracted. Each of those individual terms
could be a constant. They could have a variable like π₯
or π¦. And they can have exponents. Those exponents, however, do need
to be positive whole numbers. And they canβt be an infinite
number of terms.

Then, a monomial is an algebraic
expression that consists of just one term. Once again, any powers do need to
be positive whole numbers. So an example of a monomial could
be two π₯ or three π₯π¦. And to divide a polynomial by a
single term, we simply divide term by term. Letβs see what that looks like at
first with a really simple example.

Simplify negative 43π₯ to the
seventh power plus 12π₯ cubed minus six π₯ squared over π₯.

In order to simplify our algebraic
fraction, letβs recall what the fraction line actually means. It tells us that these two
expressions are dividing one another. That is, weβre dividing negative
43π₯ to the seventh power plus 12π₯ cubed minus six π₯ squared by π₯. In this instance, we call the
numerator of our fraction the dividend. The denominator, that expression
that weβre going to divide by, is the divisor. And to divide our dividend by π₯,
we simply do it term by term.

One way we could represent this is
to use a bus stop method. Using a bus stop method to divide
algebraic expressions is just like using a bus stop method to divide numbers. Instead, however, of dividing digit
by digit, we divide algebraic term by algebraic term. And so weβre going to begin by
dividing negative 43 times π₯ to the seventh power by π₯. This is the same as dividing
negative 43π₯ to the seventh power by one π₯ to the power of one.

Negative 43 divided by one is
negative 43. And then we recall that to divide
terms with exponents, as long as the base is the same, we subtract the
exponents. So π₯ to the power of seven divided
by π₯ to the power of one is π₯ to the sixth power. Meaning that negative 43π₯ to the
seventh power divided by π₯ is negative 43π₯ to the sixth power.

Next, we repeat this process for
12π₯ cubed. Weβre going to divide it by π₯. This is once again the same as
dividing by one π₯ to the power of one. 12 divided by one is 12. And using the same rules for
dividing with power terms, we see that π₯ cubed divided by π₯ to the power of one is
π₯ squared. And so 12π₯ cubed divided by π₯ is
12π₯ squared.

Weβll do this one more time. Now, weβre going to divide negative
six π₯ squared by π₯. Negative six divided by one is
negative six. Then, π₯ squared divided by π₯ to
the power of one is π₯ to the power of one. But we know π₯ to the power of one
is just π₯. And so the final term in our
quotient, thatβs the result we get when we divide, is negative six π₯. And so when we simplify this
algebraic fraction, we get negative 43π₯ to the sixth power plus 12π₯ squared minus
six π₯.

But of course this wasnβt the only
way to answer the question. Letβs go back to some of our
working out. Each time, we perform the division
by considering it as a fraction. And so what we could alternatively
have done is essentially reverse the process of adding fractions. And we could have written our
fraction as negative 43π₯ to the seventh power over π₯ plus 12π₯ cubed over π₯ minus
six π₯ squared over π₯ and gone from there. Either method is perfectly
valid. Itβs very much personal
preference.

Now, this is all fine and well for
simple divisors like π₯. But what do we do if we have a
multivariate term? Thatβs a term with more than one
variable. Letβs see.

Simplify 23π₯ to the fifth power π¦
cubed plus 49π₯ to the fourth power π¦ cubed plus 41π₯ cubed π¦ cubed over negative
π₯π¦.

We recall that the fraction line
means divide. And so one way we have to simplify
this fraction is to use, say, the bus stop method and divide each term by negative
π₯π¦. Alternatively, we can essentially
reverse the process of adding fractions. And we can split our fraction
up. We write it as 23π₯ to the fifth
power π¦ cubed over negative π₯π¦ plus 49π₯ to the fourth power π¦ cubed over
negative π₯π¦ plus 41π₯ cubed π¦ cubed over negative π₯π¦.

And since weβve already established
that simplifying is just like dividing, letβs think about how we simplify fractions
where the numerator and denominator is simply a number. Well, we divide through by greatest
common factors. And so to simplify each of these
individual fractions, letβs begin by looking for greatest common factors in the
numerator and denominator.

To make this easier, weβre going to
begin by adding a coefficient of one. And so negative π₯π¦ is the same as
negative one π₯π¦. And then weβre going to begin by
looking for greatest common factors in the numerical parts and then the π₯-parts and
then separately the π¦-parts. And so letβs begin by looking for
the greatest common factor of 23 and one. Well, these two numbers are
coprime. And this means that their only
common factor is one. And so we can only divide 23 and
one by one. And so the numerical part of our
numerator and denominator is just 23 over one, which is 23.

And what about the greatest common
factor of π₯ to the fifth power and π₯? We ask ourselves, whatβs the
highest power of π₯ that divides evenly into both of these terms without leaving a
remainder? Well, we can only divide both of
them by π₯. And so letβs divide π₯ to the fifth
power by π₯ and π₯ by π₯. If we start with our denominator,
we know π₯ divided by π₯ is simply one. And then we can use one of our laws
for dividing indices to divide π₯ to the fifth power by π₯ or π₯ to the power of
one.

To divide these kinds of terms, as
long as their bases are equal, we simply subtract their powers. So π₯ to the fifth power divided by
π₯ to the power of one is π₯ to the fourth power. Weβll repeat this process one more
time, this time finding the greatest common factor of π¦ cubed and π¦. Their greatest common factor is
π¦. And so we divide π¦ by π¦ to get
one. And when we divide π¦ cubed by π¦,
we get two.

Weβre not quite finished
though. We have to spot that we were
dividing a positive by a negative. And so the result is going to be a
negative term. Itβs negative 23π₯ to the fourth
power π¦ squared. Remember, we no longer need one on
the denominator. So letβs move on to our second
fraction.

Once again, 49 and one are
coprime. Then we find the greatest common
factor of π₯ to the fourth power and π₯. Well, itβs π₯. So we divide our denominator by π₯
to get one and our numerator by π₯ to get π₯ cubed. Then, we find the greatest common
factor of π¦ cubed and π¦, which we already established to be π¦. So we divide our denominator and
our numerator by π¦. And once again, spotting that weβre
dividing a positive by a negative, we get negative 49π₯ cubed π¦ squared.

We can do this one more time. 41 and one are coprime. And so we find the greatest common
factor of π₯ cubed and π₯. Itβs π₯ again. So we divide both the numerator and
denominator by π₯. We calculated the greatest common
factor of π¦ cubed and π¦ to be π¦. So we divide both the numerator and
denominator by π¦. And in our third term, we are also
dividing a positive by a negative. So our third term is negative. Itβs negative 41π₯ squared π¦
squared.

And so in simplifying our fraction,
weβve divided a polynomial by a monomial. The result is negative 23π₯ to the
fourth power π¦ squared minus 49π₯ cubed π¦ squared minus 41π₯ squared π¦
squared. Now, we also saw that we can
perform this process using the bus stop method. Once again, itβs important to
realize that either method is perfectly valid. Itβs very much personal
preference.

Weβll now consider an example where
we can simplify before dividing term by term.

Simplify 12π to the fifth power
times 11π to the 13th power π to the 13th power minus 12π to the fifth power π
to the 13th power over two π to the seventh power π squared.

Remember, this fraction line means
to divide. So essentially, here weβre dividing
a polynomial by a monomial. Our first instinct might be to
distribute the parentheses on our numerator. However, we can save ourselves a
little bit of time. Weβre going to begin by looking for
the greatest common factor of 12π to the fifth power and two π to the seventh
power π squared.

Now, weβre allowed to do this and
divide by this greatest common factor because we can see that any factor of 12π to
the fifth power must be a factor of the entire numerator. So letβs find the greatest common
factor of these two terms. Now, we can do this by first
considering the numerical parts, then considering the π-part and then the
π-part. The greatest common factor of 12
and two is two. Two is the largest number that
divides evenly into 12 and two without leaving a remainder.

Then the greatest common factor of
π to the fifth power and π to the seventh power is π to the fifth power. There is no π in 12π to the fifth
power. So weβve found the greatest common
factor. And before we do anything, weβre
going to divide the numerator and the denominator of our fraction then by two π to
the fifth power.

To divide 12π to the fifth power
by two π to the fifth power, we first divide 12 by two. Thatβs six. Then π to the fifth power divided
by π to the fifth power is one. So 12π to the fifth power divided
by two π to the fifth power is six. Similarly, we divide two π to the
seventh power π squared by two π to the fifth power. Two divided by two is one. Then, π to the seventh power
divided by π to the fifth power is π squared. π squared divided by one is π
squared. So two π to the seventh power π
squared divided by two π to the fifth power is π squared π squared.

This means we can rewrite our
fraction as six times 11π to the 13th power π to the 13th power minus 12π to the
fifth power π to the 13th power over π squared π squared. Now, weβre ready to distribute our
parentheses. We do so by multiplying six by the
first term and then six by the second. Six times 11π to the 13th power π
to the 13th power is 66π to the 13th power π to the 13th power. And then six times negative 12π to
the fifth power π to the 13th power is negative 72π to the fifth power π to the
13th power. And this is all over π squared π
squared.

Now, there are a number of ways we
can evaluate the division at this point. Since that fraction line actually
means divide, we could use the bus stop or we can reverse the process for adding
fractions. And we can separate this into two
individual fractions. Then, much as we did earlier, weβll
divide by the greatest common factor of our numerator and denominator. And if we wish, we can do this by
the numerical part, then by the π-part, then by the π-part separately.

Letβs make the denominator one π
squared π squared. And then the greatest common factor
of 66 and one is one. So we leave these for now. Then, what about the greatest
common factor of π to the 13th power and π squared? Well, itβs π squared. So we divide both the numerator and
denominator by π squared. This leaves π to the 11th power on
our numerator and then simply one on our denominator.

Similarly, the greatest common
factor of π squared and π to the 13th power is π squared. So when we divide through by π
squared, weβre simply left with π to the 11th power on our numerator. And so our first term is 66π to
the 11th power π to the 11th power over one, or simply 66π to the 11th power π to
the 11th power.

Letβs repeat this process for our
second fraction. 72 and one have a greatest common
factor of one, so we leave them. Then, the greatest common factor of
π to the fifth power and π squared is π squared. Dividing through, and weβre left
with π cubed on the numerator. Then, the greatest common factor of
π squared and π to the 13th power is π squared. And as before, that leaves us with
π to the 11th power. And so we see that weβre left with
66π to the 11th power π to the 11th power minus 72π cubed π to the 11th
power.

In our final example, weβll see how
we can link dividing polynomials by monomials to geometrical questions.

The area of a triangle is 12π₯
squared plus four π₯ square centimeters, and its base is four π₯ centimeters. Write an expression for its
height.

Letβs begin by recalling how we
find the area of a triangle. For a triangle whose base is π
units and whose perpendicular height is β units, its area is a half base times
height. And the area will be given in
square units. Now, it doesnβt matter that weβre
working with algebraic expressions. We can still substitute these into
this formula.

Letβs let the height be equal to β
or β centimeters. Weβre told the area is 12π₯ squared
plus four π₯ and its base is four π₯. So we can write 12π₯ squared plus
four π₯ equals a half times four π₯ times β. Now, because we can find a half of
four π₯ quite easily, we should. This will make the next step a
little bit easier. But if we couldnβt, what we could
do is multiply both sides by two.

Weβre not going to do that
though. Weβre going to write the right-hand
side as two π₯ times β. And then since weβre looking to
find the value of β or certainly an expression for β, we solve for β by dividing
through by two π₯. So β is 12π₯ squared plus four π₯
all over two π₯.

There are a number of ways we can
simplify this fraction or divide the numerator by the denominator. One way is to use the bus stop
method. So letβs see what that looks
like. Since weβre dividing by a monomial,
we donβt need to use long division. Weβre simply going to divide each
term in our dividend, thatβs the quadratic here, by the divisor, thatβs two π₯. So 12 divided by two is six, and π₯
squared divided by π₯ is π₯. So we see that 12π₯ squared divided
by two π₯ is six π₯. Then we divide four π₯ by two
π₯. Well, four divided by two is two,
and π₯ divided by π₯ is one. So when we divide 12π₯ squared plus
four π₯ by two π₯, we get six π₯ plus two. And so thatβs our expression for
β.

It is, of course, worth noting that
weβre working in centimeters and square centimeters. So the units for the height β are
in centimeters also. The height is six π₯ plus two
centimeters.

In this video, we saw that to
divide a polynomial by a monomial, we divide term by term. One way we have to do this is to
use a bus stop style method. Another method we have to do this
is to split the problem up into fractions whose numerators are also monomials. And then we simplify by dividing
both the numerator and denominator by their greatest common factor.