Video Transcript
In this video, weโll learn how to
perform long division with polynomials.
Polynomials โ thatโs an expression
that involves terms with positive whole-number powers โ 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, letโs 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. Weโre 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, thereโs 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, weโre now going to see what
this might look like when we perform polynomial long division; thatโs dividing
algebraic expressions.
Find the quotient when two ๐ฅ cubed
plus seven ๐ฅ squared minus eight ๐ฅ minus 21 is divided by two ๐ฅ plus three.
A quotient is the result we get by
dividing one number by another. The same stands for algebraic
expressions. So, weโre going to divide this
cubic expression by this linear expression. The dividend โ thatโs the number
weโre going to be dividing into, and here thatโs the cubic โ goes inside the bus
stop. The divisor, thatโs two ๐ฅ 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.
Weโre going to begin by dividing
two ๐ฅ cubed by the term with the highest power of ๐ฅ in our divisor, so two ๐ฅ
cubed divided by two ๐ฅ. Letโs recap that to divide two ๐ฅ
cubed by two ๐ฅ, we simplify by dividing through by two. And then if we consider ๐ฅ as being
๐ฅ 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 ๐ฅ to the power of three
divided by ๐ฅ to the power of one as being ๐ฅ squared. ๐ฅ squared sits above the two ๐ฅ
cubed in our calculations.
Now, we take that ๐ฅ squared, and
we multiply it by each part of our divisor. Two ๐ฅ times ๐ฅ squared is two ๐ฅ
cubed, and three times ๐ฅ squared is three ๐ฅ squared. Weโre now going to subtract each
term from this expression two ๐ฅ cubed plus three ๐ฅ squared from two ๐ฅ cubed plus
seven ๐ฅ squared. Two ๐ฅ cubed minus two ๐ฅ cubed is
zero. And this is a good way to check
that the earlier calculation we did is likely to be correct. Seven ๐ฅ squared minus three ๐ฅ
squared is four ๐ฅ squared. Now, we donโt really need this zero
here, but what we do need to do is bring down the next term in our dividend. Weโre going to bring down negative
eight ๐ฅ.
Now, weโre going to divide four ๐ฅ
squared by two ๐ฅ. Well, we can simplify by dividing
through by a factor of two. And then, once again, if we
consider ๐ฅ as being ๐ฅ to the power of one, we subtract the powers and we get two
๐ฅ. So, two ๐ฅ goes above seven ๐ฅ
squared. Now, we multiply two ๐ฅ by each
term in our divisor. Two ๐ฅ times two ๐ฅ gives us the
four ๐ฅ squared. Then, three times two ๐ฅ gives us
positive six ๐ฅ. Once again, we perform some
subtractions. Four ๐ฅ squared minus four ๐ฅ
squared is zero. Although, of course, we donโt
really need to write that.
Then, we take negative eight ๐ฅ,
and we subtract positive six ๐ฅ. Subtracting a positive is the same
as just subtracting six ๐ฅ. So, we get negative 14๐ฅ. Finally, we bring down our very
last term. Thatโs negative 21. Now, we divide negative 14๐ฅ by two
๐ฅ. This time, the ๐ฅโs cancel, and we
get negative 14 divided by two, which is negative seven. And so, we put negative seven above
the negative eight ๐ฅ. And weโre nearly there; weโre now
going to multiply negative seven by each term in our divisor. Negative seven times two ๐ฅ plus
three is negative 14๐ฅ minus 21.
Now, we should always perform this
final subtraction. This will tell us if we have a
remainder or not. Negative 14๐ฅ minus negative 14๐ฅ
is the same as negative 14๐ฅ plus 14๐ฅ, which is zero. Negative 21 minus negative 21 is
also zero. And so, the remainder when dividing
our cubic by two ๐ฅ plus three is zero. The quotient when we perform this
division is ๐ฅ squared plus two ๐ฅ minus seven.
In our next example, weโll see how
polynomial division, like the one weโve just performed, can help us to simplify
fractions.
Use polynomial division to simplify
six ๐ฅ cubed plus five ๐ฅ squared minus 20๐ฅ minus 21 over two ๐ฅ plus three.
One method we do have for
simplifying algebraic fractions is to factor where necessary. Itโs not particularly
straightforward to factor this cubic on our numerator. So instead, weโre going to recall
that this line in a fraction actually just means divide. And weโre going to use polynomial
long division. Our dividend, thatโs the numerator
of our fraction, goes inside the bus stop. The divisor, thatโs 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, thatโs six ๐ฅ cubed, and
we divide it by the first term in our divisor, thatโs two ๐ฅ. Six divided by two is three. Then, if we consider ๐ฅ as being ๐ฅ
to the power of one, we know that we can subtract these exponents. And ๐ฅ cubed divided by ๐ฅ to the
power of one is ๐ฅ squared. This means that six ๐ฅ cubed
divided by two ๐ฅ must be three ๐ฅ squared.
Our next step is to multiply three
๐ฅ squared by each term in our divisor. Three ๐ฅ squared times two ๐ฅ is
six ๐ฅ cubed. Notice that this is the same as the
first term in our dividend, so we know weโve probably started this correctly. We then calculate three times three
๐ฅ squared. Well, thatโs nine ๐ฅ squared. Our next step is to subtract six ๐ฅ
cubed plus nine ๐ฅ squared from six ๐ฅ cubed plus five ๐ฅ squared. Six ๐ฅ cubed minus six ๐ฅ cubed is
zero. We donโt really need to write this
zero. And then, five ๐ฅ squared minus
nine ๐ฅ squared is negative four ๐ฅ 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 weโre now going to divide
negative four ๐ฅ squared by two ๐ฅ. Negative four divided by two is
negative two. ๐ฅ squared divided by ๐ฅ to the
power of one is ๐ฅ. So, negative four ๐ฅ squared
divided by two ๐ฅ is negative two ๐ฅ. And we add negative two ๐ฅ above
five ๐ฅ squared in our problem. Now, we multiply negative two ๐ฅ by
each term in our divisor. Two ๐ฅ multiplied by negative two
๐ฅ is negative four ๐ฅ squared, and negative two ๐ฅ times three is negative six
๐ฅ.
We then subtract each of these
terms from negative four ๐ฅ squared minus 20๐ฅ. Negative four ๐ฅ squared minus
negative four ๐ฅ squared is negative four ๐ฅ plus four ๐ฅ squared. So thatโs zero, and we donโt really
need to write that. We then do negative 20๐ฅ minus
negative six ๐ฅ. Thatโs negative 20๐ฅ plus six ๐ฅ,
which is negative 14๐ฅ. We bring down negative 21. And weโre now going to divide
negative 14๐ฅ by two ๐ฅ. ๐ฅ divided by ๐ฅ 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 ๐ฅ is
negative 14๐ฅ, 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 thereโs a
remainder or not. In fact, negative 14๐ฅ minus 21
minus itself is just zero. And so, weโve completed the
division. When we simplify our algebraic
fraction, weโre left with three ๐ฅ squared minus two ๐ฅ minus seven.
Now, at this stage, itโs really
useful just to discuss briefly how we might check our solution. We perform an inverse
operation. We take our quotient, here thatโs
the solution to the division, and multiply that by the divisor, remembering, of
course, that the divisor is the algebraic expression here that weโre dividing
by. We multiply three ๐ฅ squared minus
two ๐ฅ minus seven by two ๐ฅ plus three. And when we do, we should get the
numerator, or the dividend.
In our next example, weโll consider
how to use polynomial long division to find missing terms.
Find the value of ๐ that makes the
expression 30๐ฅ to the fifth power plus 57๐ฅ squared minus 48๐ฅ cubed minus 20๐ฅ to
the fourth power plus ๐ divisible by five ๐ฅ squared minus eight.
Now, for this expression to be
divisible by five ๐ฅ squared minus eight, that tells us that five ๐ฅ 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, letโs 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 ๐ฅ. And so, we might be tempted to
rearrange them and write it as 30๐ฅ to the fifth power minus 20๐ฅ to the fourth
power, and so on. Doing it this way, weโll make the
problem a little bit easier, but it is going to look a little bit strange.
Letโs begin as normal. We divide the term in our dividend
with highest power of ๐ฅ by the term in the divisor, also with the highest power of
๐ฅ. 30 divided by five is six. Then, when weโre dividing numbers
whose bases are the same, we subtract their exponents. So, ๐ฅ to the fifth power divided
by ๐ฅ squared is ๐ฅ to the power of five minus two, which is ๐ฅ cubed. This means that 30๐ฅ to the fifth
power divided by five ๐ฅ squared is six ๐ฅ cubed. And so, we write six ๐ฅ cubed above
this term. We now multiply this value by each
term in our divisor. Six ๐ฅ cubed times five ๐ฅ squared
gives us 30๐ฅ to the fifth power. And then, when we multiply six ๐ฅ
cubed by negative eight, we get negative 48๐ฅ cubed.
Now weโre going to line that up
directly underneath the ๐ฅ cubed terms. And weโre going to add in another
term; weโre going to add in zero ๐ฅ to the fourth power. This isnโt 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. 30๐ฅ to the fifth power minus 30๐ฅ
to the fifth power is zero. Negative 20๐ฅ to the fourth power
minus zero ๐ฅ to the fourth power is negative 20๐ฅ to the fourth power. And negative 48๐ฅ cubed minus
negative 48๐ฅ cubed is zero.
Next, we bring down 57๐ฅ
squared. And weโre now going to divide
negative 20๐ฅ to the fourth power by five ๐ฅ squared. Negative 20 divided by five is
negative four, and ๐ฅ to the fourth power divided by ๐ฅ squared is just ๐ฅ
squared. And so, when we do this division,
we get negative four ๐ฅ 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, weโre going to work out
negative four ๐ฅ squared times five ๐ฅ squared, which is negative 20๐ฅ to the fourth
power.
When we multiply negative eight by
negative four ๐ฅ squared, we get 32๐ฅ squared. So, we can line this up directly
under 57๐ฅ squared. And then, we subtract each of these
terms from the terms immediately above them. Negative 20๐ฅ to the fourth power
minus negative 20๐ฅ to the fourth power is zero. Then, 57๐ฅ squared minus 32๐ฅ
squared is 25๐ฅ squared. And weโre nearly there. We bring down the final term;
thatโs the ๐.
Now, donโt worry too much that we
donโt yet know what ๐ is. Remember, weโre looking for a
remainder of zero. Weโre going to divide 25๐ฅ squared
by five squared. Well, 25 divided by five is five,
and ๐ฅ squared divided by ๐ฅ squared is one. So, the final term in our quotient
is five. We take that five and we multiply
it by five ๐ฅ squared and negative eight. And that gives us 25๐ฅ squared
minus 40.
Now, we know that since weโre
subtracting the final two terms, weโre going to get a remainder of zero. So, 25๐ฅ squared plus ๐ minus 25๐ฅ
squared minus 40 has to give us zero. Well, 25๐ฅ squared minus 25๐ฅ
squared is also zero. And so to ensure that our remainder
is zero, and thus our expression is divisible by five ๐ฅ squared minus eight, we can
say that ๐ minus negative 40 itself must be equal to zero. ๐ minus negative 40 is, of course,
๐ plus 40.
And so letโs subtract 40 from both
sides to solve for ๐. That gives us ๐ is equal to
negative 40. And so, the value of ๐ that makes
our expression divisible by five ๐ฅ squared minus eight is ๐ equals negative
40. Note that at this point, we could
go ahead and multiply our quotient. Thatโs six ๐ฅ cubed minus four ๐ฅ
squared plus five by five ๐ฅ squared minus eight. If we had performed the calculation
correctly, we would end up with the expression 30๐ฅ to the fifth power plus 57๐ฅ
squared minus 48๐ฅ cubed minus 20๐ฅ to the fourth power minus 40.
Weโll 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, thatโs our result, and multiply it by the divisor, thatโs the algebraic
expression weโre dividing by. If we get the dividend, then we
know weโve performed our calculation correctly.