### Video Transcript

Polynomial Long Division with
Remainder

In this video, weβre going to learn
an algorithm to help us divide one polynomial by another polynomial. Weβll discuss how this is similar
to regular long division and how to find the quotient and remainder polynomials. Finally, weβll talk about the
remainder theorem and how it relates to the factor theorem. Before we start talking about
dividing two polynomials, letβs start by talking about dividing two numbers. Consider 231 divided by five. We know a lot of different ways of
evaluating this expression. For example, if we were to just
evaluate this directly, we might write the answer as 46.2. Another way of saying this would be
46 and one-fifth or 46 plus one over five.

But remember, the process we used
to arrive at this answer is long division. Thereβs a lot of slightly different
ways of doing long division. Weβll only go through one of
these. To start, we call five our
divisor. We want to see how many times five
goes into 231. To start, we need to check how many
times five goes into 200. We know five goes into 200 40
times. So weβll write a four in the tens
column. Next, we know that 40 times five is
equal to 200, so we need to subtract this from 231. This, of course, leaves us with
31. We donβt need to do this process
again. We need to see how many times five
goes into 30.

Of course, we know five goes into
30 six times, so we need to add six to our answer of 40. Just as we did before, we need to
subtract six times five. We know this is equal to 30. And this, of course, leaves us with
one. If we were to try and do this
process again, we would need to see how many times five goes into one. And, of course, five doesnβt go
into one. So this tells us that our process
is done and we were left with one. We call this term of one our
remainder term. And we call 46 our quotient. And we can see this in what we had
before. We have 231 divided by five is
equal to our quotient 46 plus our remainder of one divided by our divisor of
five.

Another common way of seeing this
written is to multiply this equation through by five. This would give us 231 is equal to
five times 46 plus one. And one thing worth pointing out is
we can always guarantee that our remainder will be smaller than our divisor. This is because if it wasnβt, we
couldβve just increased our quotient. So now, weβre going to ask the
question, how does this relate to polynomial division? This time, instead of being given
an integer divided by an integer, weβre going to be given a polynomial divided by a
polynomial. And weβre going to find an
algorithm which will help us see how many times d of π₯ goes into π of π₯. Weβll call this polynomial long
division. And this will be very similar to
regular long division.

Just like with regular long
division, weβll find a quotient and remainder term. However, this time, because weβre
dividing polynomials, our quotient and remainder will also be polynomials. Weβll call these π of π₯ and π of
π₯. And our answer will be in exactly
the same form we got for regular long division. Weβll find polynomials π of π₯ and
π of π₯ such that π of π₯ divided by d of π₯ is equal to π of π₯ plus π of π₯
divided by d of π₯.

Finally, remember, when we were
doing regular long division, we could guarantee that our remainder would be smaller
than our divisor. And weβll have something similar
with polynomial long division. Weβll be able to guarantee that the
degree of our remainder is smaller than the degree of our divisor. And the reasoning for this is
exactly the same as it is in regular long division. We keep removing multiples of d of
π₯ until we canβt do this anymore. Letβs now go onto an example of
using polynomial long division.

Use polynomial division to simplify
three π₯ cubed plus two π₯ squared minus four π₯ plus one divided by π₯ plus
one.

The question wants us to use
polynomial division to simplify this expression. Weβll call the cubic polynomial in
our numerator, thatβs three π₯ cubed plus two π₯ squared minus four π₯ plus one, π
of π₯. And weβll call the linear
polynomial in our denominator, thatβs π₯ plus one, d of π₯. This is our divisor. Using polynomial division is very
similar to regular long division. Weβll set this up in exactly the
same way. Weβll have our divisor π₯ plus one
going into three π₯ cubed plus two π₯ squared minus four π₯ plus one.

When using regular long division,
we want to see how many times our divisor goes into the highest term. We can do exactly the same thing
here. We want to know how many times π₯
goes into the highest terminal polynomial. Thatβs three π₯ cubed. Of course, three π₯ cubed divided
by π₯ is three π₯ squared. Or alternatively, we could write
this as three π₯ squared times π₯ is equal to three π₯ cubed. Just like with regular long
division, we want to write this in our quotient. Remember though, this is a term for
π₯ squared, so weβll write this in the column for π₯ squared.

In regular long division, the next
step is to multiply this term we just added to our quotient by our divisor. We would then want to subtract this
from our polynomial. Weβll do exactly the same thing
here. First, we need to find three π₯
squared multiplied by our divisor π₯ plus one. If we evaluate this, we get three
π₯ cubed plus three π₯ squared. We then want to subtract this from
our polynomial π of π₯. Weβll evaluate this term by
term. First, three π₯ cubed minus three
π₯ cubed is equal to zero, and two π₯ squared minus three π₯ squared is equal to
negative π₯ squared.

Itβs worth pointing out youβll
often see this term of zero omitted from working out. This is because weβll always have a
zero in this position. It doesnβt matter either way if you
prefer leaving it in or taking it out. In this instance, weβll be leaving
it out. And then just like working a long
division, we need to take down the rest of our terms. This gives us negative π₯ squared
minus four π₯ plus one. Just like with regular long
division, we now need to repeat our process. We need to see how many times π₯
goes into our highest term negative π₯ squared.

Well, we know negative π₯ squared
divided by π₯ is negative π₯. Weβll write this in our quotient in
the column we have for π₯. Now, just as we did before, we need
to multiply our divisor of π₯ plus one by negative π₯. This gives us negative π₯ times π₯
plus one. And if we evaluate this, we get
negative π₯ squared minus π₯. And remember, the next thing we
need to do is subtract this from our polynomial negative π₯ squared minus four π₯
plus one. Weβll do this term by term. First, negative π₯ squared minus
negative π₯ squared is negative π₯ squared plus π₯ squared. This is equal to zero.

Next, we have negative four π₯
minus negative π₯ is negative four π₯ plus π₯. This is equal to negative three
π₯. Now, just as we did before, we need
to take this constant of one down. So this gives us negative three π₯
plus one. Just like with regular long
division, we now need to do this again. We keep going until we canβt do
this anymore. Again, we need to see how many
times π₯ goes into our highest-order term. Thatβs negative three π₯. This time, negative three π₯
divided by π₯ is equal to negative three. Weβll add this to our quotient. The next step is to multiply
negative three by our divisor π₯ plus one and subtract this from negative three π₯
plus one.

So we want to evaluate negative
three π₯ times π₯ plus one. If we distribute this over our
parentheses, we get negative three π₯ minus three. And now, we subtract this from
negative three π₯ plus one. Weβll do this term by term. The first term, we get negative
three π₯ minus negative three π₯, which is negative three π₯ plus three π₯, which we
know is equal to zero. Next, we want one minus negative
three. Well, thatβs one plus three, which
we know is equal to four. But now, if we try and continue
this process, we get a problem. We would want to see how many times
π₯ goes into four. Well, this is just four divided by
π₯.

This is not a polynomial. In other words, we can no longer
divide this expression by our divisor. In fact, this will always happen
when weβre left with an expression with a degree strictly less than our divisor. And just like with regular long
division, weβll call this our remainder term. Weβll call this π of π₯ because
often this will be a polynomial. And just the same as regular long
division, weβll call three π₯ squared minus π₯ minus three our quotient. Weβll call this π of π₯.

And now, we can come to our answer
in exactly the same way we do with regular long division. We can use our remainder and
quotient to rewrite π of π₯ divided by d of π₯ as π of π₯ plus π of π₯ divided by
d of π₯, the quotient plus the remainder divided by the divisor. Now, all we have to do is
substitute in our expressions for π of π₯, π of π₯, and d of π₯. And by doing this, we were able to
rewrite the expression given to us in the question as three π₯ squared minus π₯
minus three plus four divided by π₯ plus one.

Letβs do another example to help us
solidify what weβve learnt.

Use polynomial long division to
find the quotient π of π₯ and the remainder π of π₯ for π of π₯ divided by d of
π₯, where π of π₯ is equal to π₯ of the seventh power plus π₯ to the sixth power
plus π₯ to the fourth power plus π₯ squared plus π₯ plus one and d of π₯ is equal to
π₯ cubed plus π₯ plus one.

The question wants us to use
polynomial long division. Weβre given our polynomial π of π₯
and our divisor d of π₯. We need to find the quotient π of
π₯ and the remainder π of π₯ when we divide π of π₯ by d of π₯. Before we start answering this
question, thereβs a couple of things we should check. For example, we should check that
both our polynomials π of π₯ and d of π₯ are written in descending exponents of
π₯. In this case, this is true, so we
can just carry on with our long division. Weβll set up our long division. We have our divisor d of π₯ is
dividing our polynomial π of π₯.

Before we start doing our long
division, thereβs one more thing we can check. If we look at our polynomial π of
π₯, we can see thereβs no term for π₯ to the fifth power and thereβs no term for π₯
cubed. In regular long division, when this
happens, we have a digit of zero in this place. However, because this is a
polynomial, we just didnβt write these terms in. Thereβs a few different ways we
could tackle this. We could just leave that as it is,
or we could add terms zero π₯ to the fifth power and zero π₯ cubed. And both of these methods work, and
you can use them if you prefer. However, in this video, weβre just
going to leave a gap where these terms are to keep our columns aligned.

Now, letβs move on to our long
division. The leading term in π of π₯ is π₯
to the seventh power. We need to divide this by π₯
cubed. And of course, π₯ to the seventh
power divided by π₯ cubed is π₯ to the fourth power. Weβll write this in our quotient,
and weβll write this in the column for π₯ to the fourth power terms. The next thing we need to do is
multiply our divisor by the term in our quotient π₯ to the fourth power. Multiplying these together, we get
π₯ to the fourth power times π₯ cubed plus π₯ plus one. And if we distribute this and
simplify, we get π₯ to the seventh power plus π₯ to the fifth power plus π₯ to the
fourth power.

We now want to subtract this from
our polynomial π of π₯. And remember, we want to keep each
term in its respective column. Weβll start with π₯ to seventh
power. Then we need to add π₯ to the fifth
power. Finally, we add a term for π₯ to
the fourth power. Now, we can just subtract this term
by term. First, we get π₯ to the seventh
power minus π₯ to the seventh power. This is equal to zero. You can write this term of zero in
if you prefer. However, this term will always give
us zero. So weβll just leave this blank.

Next, we have π₯ to the sixth power
minus zero. Of course, this is just equal to π₯
to the sixth power. Next, we have zero minus π₯ to the
fifth power. This is negative π₯ to the fifth
power. Then in our next column, we have π₯
to the fourth power minus π₯ to the fourth power. This is equal to zero. Weβll leave this bank. And remember, we need to bring down
the rest of our terms. Itβs worth pointing out some people
prefer to leave these terms at the top until theyβre needed. But weβre going to always bring
these terms down.

Weβre now ready to find the next
term in our quotient. We need to divide π₯ to the sixth
power by π₯ cubed. And π₯ to the sixth power divided
by π₯ cubed is just equal to π₯ cubed. And remember, we write this in our
column for π₯ cubed terms. The next step in our long division
will be to multiply π₯ cubed by our divisor π₯ cubed plus π₯ plus one. This gives us π₯ cubed times π₯
cubed plus π₯ plus one. And if we distribute this and
simplify, we get π₯ to the sixth power plus π₯ to the fourth power plus π₯
cubed.

We now need to subtract this from
our polynomial. Remember, itβs important that we
write each term in the correct column. We can then subtract this term by
term. In our first column, we get π₯ to
the sixth power minus π₯ to the sixth power, which is zero. In our second column, we get
negative π₯ to the fifth power minus zero, which is just equal to negative π₯ to the
fifth power. In our next column, we get zero
minus π₯ to the fourth power, which is negative π₯ to the fourth power. We get a simpler story in our next
column. We have zero minus π₯ cubed, which
is negative π₯ cubed. Then once again, we bring down the
remaining terms.

Once again, we need to find the
next term in our quotient. We need to divide negative π₯ to
the fifth power by π₯ cubed. Of course, if we do this, we get
negative π₯ squared. Once again, we need to multiply the
newly added term to our quotient by our divisor. Distributing this and simplifying,
we get negative π₯ to the fifth power minus π₯ cubed minus π₯ squared. We then need to subtract this from
our polynomial. Remember, we want to write each
term in its correct column. We evaluate the subtraction term by
term. This time, we get negative π₯ to
the fourth power plus two π₯ squared. And then we bring the rest of our
terms down.

And once again, we need to find the
next term in π of π₯. We need to divide negative π₯ to
the fourth power by π₯ cubed. Doing this, we get negative π₯. Once again, we multiply negative π₯
by our divisor. And if we evaluate this, we get
negative π₯ to the fourth power minus π₯ squared minus π₯. Now, we need to subtract this from
our polynomial. Evaluating the subtraction and
bringing down our term of one, we get three π₯ squared plus two π₯ plus one. And now, if we were to try and find
the next term in our quotient, we would get three π₯ squared divided by π₯
cubed. This is three over π₯.

This is not a polynomial. This tells us weβre done. We can see that the polynomial
weβre left with has a lower degree than our divisor. So weβve found our quotient π of
π₯ and our remainder π of π₯. And this gives us our final
answer. We were able to show our quotient
π of π₯ is equal to π₯ to the fourth power plus π₯ cubed minus π₯ squared minus π₯
and our remainder π of π₯ is equal to three π₯ squared plus two π₯ plus one.

Letβs now talk about a special case
for our divisor d of π₯. Using polynomial long division on
π of π₯ divided by d of π₯, we know we can find polynomials π of π₯ and π of π₯
such that π of π₯ is equal to d of π₯ times π of π₯ plus π of π₯. We want to talk about the case when
weβre dividing by the linear polynomial π₯ minus π. This means we have π of π₯ is
equal to π₯ minus π times π of π₯ plus π of π₯. But remember, this means the degree
of our divisor is equal to one. And we know our remainder term must
have a degree less than our divisor term.

So in the case when weβre dividing
by a linear polynomial, we know our remainder term must have degree zero. We can just write this as the
constant π. And this gives us a useful
result. Letβs see what would happen if we
were to substitute π₯ is equal to π. Substituting π₯ is equal to π, we
get π of π is equal to π minus π times π of π plus π. Of course, π minus π is equal to
zero. So weβre just left with π of π is
equal to π. And this is a useful result to help
us find the remainder when weβre dividing by a linear polynomial. We call this the remainder
theorem. So letβs formalize what we mean by
the remainder theorem.

The remainder theorem tells us if
we divide a polynomial π of π₯ by a linear polynomial π₯ minus π, the remainder
must be the constant π evaluated at π. This is a useful result to help us
find our remainder term. Another thing worth pointing out is
what happens when π of π is equal to zero. Well, when π of π is equal to
zero, the remainder theorem tells us our remainder must be equal to zero. But what does our remainder being
equal to zero mean? Well, if our remainder is equal to
zero, then we must have that π of π₯ is equal to π₯ minus π times π of π₯.

In other words, we must have π₯
minus π is a factor of our polynomial π of π₯. And this is a well-known
theorem. We call this the factor
theorem. If π of π is equal to zero, then
π₯ minus π is a factor of π of π₯. Similarly, if π₯ minus π is a
factor of π of π₯, then its remainder upon division is equal to zero. Letβs now take a look at an example
where we use the remainder theorem.

Find the remainder when three π₯
cubed minus two π₯ squared plus four π₯ plus five is divided by three π₯ plus
four.

The question is asking us to find
the remainder term when a cubic polynomial is divided by a linear polynomial. One way of doing this is by using
polynomial long division. However, we know this is a long
process. Instead, letβs notice weβre
dividing by a linear polynomial. And weβre only asked to find the
remainder term. This should remind us of the
remainder theorem. We recall the remainder theorems
tells us when π of π₯ is divided by a linear polynomial π₯ minus π, then the
remainder is constant and equal to π evaluated at π.

We do have to be a little careful
with how we use the remainder theorem in this case. We are dividing by a linear
polynomial three π₯ plus four, but itβs not in the form π₯ minus π. So instead of actually doing our
long division, letβs call our quotient polynomial π of π₯ and our remainder
polynomial π of π₯. This means weβll have three π₯
cubed minus two π₯ squared plus four π₯ minus five is equal to three π₯ plus four
times π of π₯ plus π of π₯ for some polynomials π of π₯ and π of π₯. We can see our divisor is a linear
polynomial; it has degree one. Our remainder must have a lower
degree than our divisor. This means it must have degree
zero. In other words, itβs a
constant. Weβll call this constant π.

And at this point, thereβs two
similar ways of solving this equation. If we solve our linear factor is
equal to zero, this gives us π₯ is equal to negative four over three. One way to find the value of π is
to substitute this value directly into this expression. Doing this, we get our cubic
polynomial evaluated at negative four over three is equal to zero times π evaluated
at negative four over three plus π. But this simplifies to just give us
π. And this is a perfectly valid way
of solving this equation. However, weβre going to do this by
taking a factor of three outside of our divisor. Doing this, we can rewrite this as
three times π₯ plus four over three.

And now, weβre starting to see
something interesting. Letβs consider three as part of π
of π₯. So what do we now have? We have our cubic polynomial is
equal to π₯ plus four over three times some polynomial plus a constant. In fact, what weβve done here is
found an expression for our quotient polynomial when we divide our cubic by π₯ plus
four over three. In other words, the remainder term
when we divide by π₯ plus four over three or when we divide by three π₯ plus four
are the same. This means we can just use the
remainder theorem to find our value of π. And weβll do this by evaluating our
cubic polynomial at π₯ is equal to negative four over three. This gives us the following
expression. And then, by evaluating this
expression, we were able to show that our remainder term must be equal to negative
11.

Letβs now go over the key points of
this video. First, we were able to show, by
using a similar method to long division, we can divide two polynomials. We know dividing a polynomial by
its factor will give us remainder zero. We also know that our remainder
polynomial will always have a lower degree than our divisor polynomial. Finally, we learned about the
remainder theorem, which tells us when a polynomial π of π₯ is divided by the
linear polynomial π₯ minus π, our remainder polynomial will be constant. And it will be equal to π
evaluated at π.