### Video Transcript

In this video, we’ll learn how to
perform long division with polynomials.

Polynomials — that’s an expression
that involves terms with positive whole-number powers — can sometimes be divided
using simple methods like factoring. If those methods fail, however, we
can use long division. This process looks a lot like the
process for long division with numbers. So, let’s begin by recalling how we
perform long division with numbers.

Calculate 3531 divided by 11.

In this question, the number 3531
is called the dividend and the number 11 is the divisor. The result we get when we do this
division is called the quotient. We’re going to set this problem up
using a sort of bus stop design. As we see, the dividend goes inside
the bus stop and the number 11 goes on the outside. The first thing we do is divide the
first digit of 3531 by 11. Three divided by 11 is zero. So, we place a zero above the
three, and any remainders at this stage are ignored. We then take this number, zero, and
we multiply it by the divisor. Zero multiplied by 11 is zero. So we put a zero underneath the
three.

Then, we subtract the digits in
this column. Three take away zero is three. Next, we bring down the number
five. We can, if we require, bring down
all of the digits. But the first one is
sufficient. And now we divide 35 by 11. 35 divided by 11 is three remainder
two. We ignore the remainder for now and
add the whole-number answer above five. Now, we multiply this number three
by 11. Three times 11 is 33. So, we put 33 underneath 35. And once again, we subtract. 35 minus 33 is two. Now, we bring down the three, and
we divide 23 by 11. The whole-number solution is
two. Of course, there’s a remainder, but
we do ignore it.

Now, we multiply that two by 11 to
give us 22. We write this 22 underneath the 23,
and then we subtract again. 23 minus 22 is one. And so, we bring down the final
digit. And the last thing we do is 11
divided by 11. 11 divided by 11 is one. So, we add a one above the one in
our dividend. Finally, we multiply this one by
the 11. Now, this step might not seem
necessary, but we do perform it just to check whether we have a remainder. One multiplied by 11 is 11. And when we subtract 11 from 11, we
get zero. If this answer here is zero, that
tells us there is no remainder. 11 is a factor of 3531. And when we divide 3531 by 11, we
get 321.

So, we’re now going to see what
this might look like when we perform polynomial long division; that’s dividing
algebraic expressions.

Find the quotient when two 𝑥 cubed
plus seven 𝑥 squared minus eight 𝑥 minus 21 is divided by two 𝑥 plus three.

A quotient is the result we get by
dividing one number by another. The same stands for algebraic
expressions. So, we’re going to divide this
cubic expression by this linear expression. The dividend — that’s the number
we’re going to be dividing into, and here that’s the cubic — goes inside the bus
stop. The divisor, that’s two 𝑥 plus
three, goes outside. Now, this looks a lot like dividing
with numbers. But instead of dividing digit by
digit, we divide term by term.

We’re going to begin by dividing
two 𝑥 cubed by the term with the highest power of 𝑥 in our divisor, so two 𝑥
cubed divided by two 𝑥. Let’s recap that to divide two 𝑥
cubed by two 𝑥, we simplify by dividing through by two. And then if we consider 𝑥 as being
𝑥 to the power of one, we know that when we divide algebraic terms with the same
base, we simply subtract their powers. So, we get 𝑥 to the power of three
divided by 𝑥 to the power of one as being 𝑥 squared. 𝑥 squared sits above the two 𝑥
cubed in our calculations.

Now, we take that 𝑥 squared, and
we multiply it by each part of our divisor. Two 𝑥 times 𝑥 squared is two 𝑥
cubed, and three times 𝑥 squared is three 𝑥 squared. We’re now going to subtract each
term from this expression two 𝑥 cubed plus three 𝑥 squared from two 𝑥 cubed plus
seven 𝑥 squared. Two 𝑥 cubed minus two 𝑥 cubed is
zero. And this is a good way to check
that the earlier calculation we did is likely to be correct. Seven 𝑥 squared minus three 𝑥
squared is four 𝑥 squared. Now, we don’t really need this zero
here, but what we do need to do is bring down the next term in our dividend. We’re going to bring down negative
eight 𝑥.

Now, we’re going to divide four 𝑥
squared by two 𝑥. Well, we can simplify by dividing
through by a factor of two. And then, once again, if we
consider 𝑥 as being 𝑥 to the power of one, we subtract the powers and we get two
𝑥. So, two 𝑥 goes above seven 𝑥
squared. Now, we multiply two 𝑥 by each
term in our divisor. Two 𝑥 times two 𝑥 gives us the
four 𝑥 squared. Then, three times two 𝑥 gives us
positive six 𝑥. Once again, we perform some
subtractions. Four 𝑥 squared minus four 𝑥
squared is zero. Although, of course, we don’t
really need to write that.

Then, we take negative eight 𝑥,
and we subtract positive six 𝑥. Subtracting a positive is the same
as just subtracting six 𝑥. So, we get negative 14𝑥. Finally, we bring down our very
last term. That’s negative 21. Now, we divide negative 14𝑥 by two
𝑥. This time, the 𝑥’s cancel, and we
get negative 14 divided by two, which is negative seven. And so, we put negative seven above
the negative eight 𝑥. And we’re nearly there; we’re now
going to multiply negative seven by each term in our divisor. Negative seven times two 𝑥 plus
three is negative 14𝑥 minus 21.

Now, we should always perform this
final subtraction. This will tell us if we have a
remainder or not. Negative 14𝑥 minus negative 14𝑥
is the same as negative 14𝑥 plus 14𝑥, which is zero. Negative 21 minus negative 21 is
also zero. And so, the remainder when dividing
our cubic by two 𝑥 plus three is zero. The quotient when we perform this
division is 𝑥 squared plus two 𝑥 minus seven.

In our next example, we’ll see how
polynomial division, like the one we’ve just performed, can help us to simplify
fractions.

Use polynomial division to simplify
six 𝑥 cubed plus five 𝑥 squared minus 20𝑥 minus 21 over two 𝑥 plus three.

One method we do have for
simplifying algebraic fractions is to factor where necessary. It’s not particularly
straightforward to factor this cubic on our numerator. So instead, we’re going to recall
that this line in a fraction actually just means divide. And we’re going to use polynomial
long division. Our dividend, that’s the numerator
of our fraction, goes inside the bus stop. The divisor, that’s the
denominator, goes on the outside.

And then, we remember the first
thing that we do is we take the first term in our dividend, that’s six 𝑥 cubed, and
we divide it by the first term in our divisor, that’s two 𝑥. Six divided by two is three. Then, if we consider 𝑥 as being 𝑥
to the power of one, we know that we can subtract these exponents. And 𝑥 cubed divided by 𝑥 to the
power of one is 𝑥 squared. This means that six 𝑥 cubed
divided by two 𝑥 must be three 𝑥 squared.

Our next step is to multiply three
𝑥 squared by each term in our divisor. Three 𝑥 squared times two 𝑥 is
six 𝑥 cubed. Notice that this is the same as the
first term in our dividend, so we know we’ve probably started this correctly. We then calculate three times three
𝑥 squared. Well, that’s nine 𝑥 squared. Our next step is to subtract six 𝑥
cubed plus nine 𝑥 squared from six 𝑥 cubed plus five 𝑥 squared. Six 𝑥 cubed minus six 𝑥 cubed is
zero. We don’t really need to write this
zero. And then, five 𝑥 squared minus
nine 𝑥 squared is negative four 𝑥 squared. We bring down the next term. Some people bring down all of the
terms, but I prefer to keep things a little bit simpler.

And we’re now going to divide
negative four 𝑥 squared by two 𝑥. Negative four divided by two is
negative two. 𝑥 squared divided by 𝑥 to the
power of one is 𝑥. So, negative four 𝑥 squared
divided by two 𝑥 is negative two 𝑥. And we add negative two 𝑥 above
five 𝑥 squared in our problem. Now, we multiply negative two 𝑥 by
each term in our divisor. Two 𝑥 multiplied by negative two
𝑥 is negative four 𝑥 squared, and negative two 𝑥 times three is negative six
𝑥.

We then subtract each of these
terms from negative four 𝑥 squared minus 20𝑥. Negative four 𝑥 squared minus
negative four 𝑥 squared is negative four 𝑥 plus four 𝑥 squared. So that’s zero, and we don’t really
need to write that. We then do negative 20𝑥 minus
negative six 𝑥. That’s negative 20𝑥 plus six 𝑥,
which is negative 14𝑥. We bring down negative 21. And we’re now going to divide
negative 14𝑥 by two 𝑥. 𝑥 divided by 𝑥 is just one. So, we get negative 14 divided by
two, which is negative seven. So, we add negative seven here. And once again, we divide this
number by each term in our divisor. Negative seven times two 𝑥 is
negative 14𝑥, and negative seven times three is negative 21.

We do one final subtraction, and
this is a really important step to do because it tells us whether there’s a
remainder or not. In fact, negative 14𝑥 minus 21
minus itself is just zero. And so, we’ve completed the
division. When we simplify our algebraic
fraction, we’re left with three 𝑥 squared minus two 𝑥 minus seven.

Now, at this stage, it’s really
useful just to discuss briefly how we might check our solution. We perform an inverse
operation. We take our quotient, here that’s
the solution to the division, and multiply that by the divisor, remembering, of
course, that the divisor is the algebraic expression here that we’re dividing
by. We multiply three 𝑥 squared minus
two 𝑥 minus seven by two 𝑥 plus three. And when we do, we should get the
numerator, or the dividend.

In our next example, we’ll consider
how to use polynomial long division to find missing terms.

Find the value of 𝑘 that makes the
expression 30𝑥 to the fifth power plus 57𝑥 squared minus 48𝑥 cubed minus 20𝑥 to
the fourth power plus 𝑘 divisible by five 𝑥 squared minus eight.

Now, for this expression to be
divisible by five 𝑥 squared minus eight, that tells us that five 𝑥 squared minus
eight is a factor of it. If this is the case, then when we
perform the division, we should have no remainder, or a remainder of zero. So, let’s perform the long
division. Now, if we look carefully at our
expression, we see that the terms appear to be in a bit of a funny order. They are usually written in
decreasing powers of 𝑥. And so, we might be tempted to
rearrange them and write it as 30𝑥 to the fifth power minus 20𝑥 to the fourth
power, and so on. Doing it this way, we’ll make the
problem a little bit easier, but it is going to look a little bit strange.

Let’s begin as normal. We divide the term in our dividend
with highest power of 𝑥 by the term in the divisor, also with the highest power of
𝑥. 30 divided by five is six. Then, when we’re dividing numbers
whose bases are the same, we subtract their exponents. So, 𝑥 to the fifth power divided
by 𝑥 squared is 𝑥 to the power of five minus two, which is 𝑥 cubed. This means that 30𝑥 to the fifth
power divided by five 𝑥 squared is six 𝑥 cubed. And so, we write six 𝑥 cubed above
this term. We now multiply this value by each
term in our divisor. Six 𝑥 cubed times five 𝑥 squared
gives us 30𝑥 to the fifth power. And then, when we multiply six 𝑥
cubed by negative eight, we get negative 48𝑥 cubed.

Now we’re going to line that up
directly underneath the 𝑥 cubed terms. And we’re going to add in another
term; we’re going to add in zero 𝑥 to the fourth power. This isn’t entirely necessary, but
it can make it a little bit easier to follow what happens next. Our next step is to divide each of
these three terms by the corresponding terms above. 30𝑥 to the fifth power minus 30𝑥
to the fifth power is zero. Negative 20𝑥 to the fourth power
minus zero 𝑥 to the fourth power is negative 20𝑥 to the fourth power. And negative 48𝑥 cubed minus
negative 48𝑥 cubed is zero.

Next, we bring down 57𝑥
squared. And we’re now going to divide
negative 20𝑥 to the fourth power by five 𝑥 squared. Negative 20 divided by five is
negative four, and 𝑥 to the fourth power divided by 𝑥 squared is just 𝑥
squared. And so, when we do this division,
we get negative four 𝑥 squared. And this is the next term in our
quotient. Remember, each time we find a term
in our quotient, we multiply it by each part of the divisor. So, we’re going to work out
negative four 𝑥 squared times five 𝑥 squared, which is negative 20𝑥 to the fourth
power.

When we multiply negative eight by
negative four 𝑥 squared, we get 32𝑥 squared. So, we can line this up directly
under 57𝑥 squared. And then, we subtract each of these
terms from the terms immediately above them. Negative 20𝑥 to the fourth power
minus negative 20𝑥 to the fourth power is zero. Then, 57𝑥 squared minus 32𝑥
squared is 25𝑥 squared. And we’re nearly there. We bring down the final term;
that’s the 𝑘.

Now, don’t worry too much that we
don’t yet know what 𝑘 is. Remember, we’re looking for a
remainder of zero. We’re going to divide 25𝑥 squared
by five squared. Well, 25 divided by five is five,
and 𝑥 squared divided by 𝑥 squared is one. So, the final term in our quotient
is five. We take that five and we multiply
it by five 𝑥 squared and negative eight. And that gives us 25𝑥 squared
minus 40.

Now, we know that since we’re
subtracting the final two terms, we’re going to get a remainder of zero. So, 25𝑥 squared plus 𝑘 minus 25𝑥
squared minus 40 has to give us zero. Well, 25𝑥 squared minus 25𝑥
squared is also zero. And so to ensure that our remainder
is zero, and thus our expression is divisible by five 𝑥 squared minus eight, we can
say that 𝑘 minus negative 40 itself must be equal to zero. 𝑘 minus negative 40 is, of course,
𝑘 plus 40.

And so let’s subtract 40 from both
sides to solve for 𝑘. That gives us 𝑘 is equal to
negative 40. And so, the value of 𝑘 that makes
our expression divisible by five 𝑥 squared minus eight is 𝑘 equals negative
40. Note that at this point, we could
go ahead and multiply our quotient. That’s six 𝑥 cubed minus four 𝑥
squared plus five by five 𝑥 squared minus eight. If we had performed the calculation
correctly, we would end up with the expression 30𝑥 to the fifth power plus 57𝑥
squared minus 48𝑥 cubed minus 20𝑥 to the fourth power minus 40.

We’ll now recap some of the key
points from this lesson. In this video, we saw that to
perform polynomial long division, we perform a similar process to working with
numbers. Instead of dividing each digit in
turn, we actually divide each term in turn. We also saw that if we get a
remainder of zero, in other words, our very final subtraction yields zero, that the
divisor is a factor of the dividend. In other words, the dividend is
exactly divisible by the divisor. And finally, we saw that we can
check any division calculations by using inverse operations. In other words, we take the
quotient, that’s our result, and multiply it by the divisor, that’s the algebraic
expression we’re dividing by. If we get the dividend, then we
know we’ve performed our calculation correctly.