Introduction and Vocabulary
A one variable polynomial,
the kind you will most often
see, is an algebraic expression
made up from adding, subtracting,
and multiplying of the variable
and numbers. All such things
can be written as a sum
of products of numbers and
powers of the variable,
and when you put this in
descending order of powers
it is called the standard
form of the polynomial.
Here are some examples of
polynomials in standard
form.
x7-3x+1,
x+5, 6x3-7x2+6x-9,
5, x, 2345x17,
x5+x4+x3+x2+x+1,
8
The numbers that are
multiplied by the powers
of the variable are called
coefficients. They can
be fractions
as well as whole numbers,
and even irrational
numbers. You can have
polynomials with complex number
coefficients as well,
but most of the ones you
will see in algebra classes
will have real numbers
as coefficients. I only
used whole numbers, because
they are easier to write
down in html. Actually
not all of the numbers
I used were whole numbers,
some of them were negative,
because I said 'sum' in
the description above,
and when you are talking
about coefficients you
assume that you are thinking
of the expression as a
sum of things, so when
you have a difference
you think of it as adding
a negative, so for example
the first of those polynomials
should be thought of as
x7+-3x+1. We
just write such things
as subtraction because
that using less ink. (See
How to
Add, Subtract, Multiply,
and Divide Integers
for more about the arithmetic
of pluses and minuses.)
I will only be using integer
coefficients in this article,
but the only thing different
about using other coefficients
is that you have to know
how to do the arithmetic
for those kinds of numbers.
There are some problems
with fraction coefficients
in my MathHelp
problem set collection
Polynomials,
so you can download that
if you want some practice
with that.
The expressions being
added together are called
terms. So for the first
example here the terms
are x7, -3x,
and 1. For the second
example they are x and
5. For the third example
they are 6x3,
-7x2, 6x, and
-9. The rest are left
as exercises for the student.
The degree of a one variable
polynomial is the highest
power that occurs in it.
If the polynomial is written
in standard form, then
this will be the power
of the first term. The
first term is also sometimes
called the leading term.
The degrees of my example
polynomials are 7, 1,
3, 1, 17, 5, and 0. For
the ones and zero here
you need to remember that
anything raised to the
1 power is itself and
anything raised to the
0 power is 1. (See Integer
Exponents.) For determining
degree we regard the variable
alone as it raised to
the 1 power. If the polynomial
is just a constant, that
is a number, we can think
of it as that constant
multiplied by x0,
since anything times 1
is itself. Anyway, what
this all means is that
polynomials that have
variables, but no powers
are degree 1 and constant
polynomials, ones that
are just numbers are degree
0.
Polynomials can
be classified both by
how many terms they have
and by their degrees,
particularly when these
are small numbers. Here
are some names that you
may see used.
Number of Terms
1 - monomial
2 - binomial
3 - trinomial
4 - quadrinomial
5 - quintinomial
You will hear and see
the first three of these
quite commonly. The next
two are not quite so common.
There are more for higher
numbers as well, but they
are not very often used.
Degrees
0 - constant
1 - linear
2 - quadratic
3 - cubic
4 - quartic
5 - quintic
The same holds with these.
Up through 2, you will
hear them all the time.
People seldom talk of
degree 1 or 2 polynomials,
they almost always call
them linear or quadratic
instead. For degree 3,
cubic is pretty common
too, but beyond that I
just gave the names for
your entertainment.
Addition
To add polynomials we just
drop the parentheses and
simplify by combining like
terms. Combining like terms
is just like any other combining
like terms. It comes from
the distributive property.
(See
Getting Rid of Parentheses.) Remember that
like with any algebraic
simplifying, everything
you do comes from a small
number of properties, and
in this case those are the
properties that I have listed
in my articles
Getting Rid of Parentheses and
Simplifying Exponential Expressions. And
that's pretty much it. Anything
that can't be justified
by one of those properties
can't be done. Also remember
that whenever it looks like
there is no coefficient,
it is a 1, because 1 times
anything is itself. When
you simplify polynomials
you deal with the different
powers of the variables
just like they were different
variables, so for example,
3x
2+2x has no
like terms, so it cannot
be simplified any further,
but 3x
2+2x
2
does have like terms and
can be simplified to 5x
2
by combining like terms.
Don't be tempted to think
that the exponents add.
That's only for multiplying.
Example 1:
Add the polynomials.
(4x3+x2-2x+3)+(3x2+5x)
Solution:
4x3+1x2-2x+3+3x2+5x
=4x3+4x2+3x+3
Explanation:
Drop the parentheses and
combine like terms. There
is only one x
3
term. For the x
2
terms the first one has
no coefficient, so we give
it a 1 for a coefficient
because 1 times anything
is itself and combine it
with the 3x
2
to make 4x
2.
Then for the x terms we
have -2x and 5x, which combine
to 3x. Remember to always
regard subtraction as addition
of a negative. There is
only one constant term,
so nothing needs to be done
with it.
Subtraction
Subtraction is just a teeny
bit harder than addition,
but you must really pay
attention to that teeny
bit, because otherwise it
will get you. Just like
with other kinds of simplifying
with parentheses, see
Getting Rid of Parentheses, when you have
a minus sign in front of
a set of parentheses, to
get rid of the parentheses
you have to make sure and
change the signs of every
term in the parentheses.
You can think of this as
applying the distributive
property to a multiplication
by -1, since -1 times anything
is its opposite. After you
have changed all of the
signs of the second polynomial
everything else is just
like addition.
Example 2:
Subtract the polynomials.
(3x2-5x-6)-(4x3-5x2+2x-3)
Solution:
3x2-5x-6-4x3+5x2-2x+3
=-4x3+8x2-7x-3
Explanation:
Change
all of the
signs of the second polynomial
and drop the parentheses.
Then we just combine like
terms just like with addition.
There is nothing to go with
the -4x
3. The
3x
2 and the 5x
2
add up to the 8x
2.
The -5x and the -2x make
-7x. And the -6 and the
3 add to -3.
Multiplication
Multiplication is a bit
harder, because we have
two operations going on,
multiplication and addition.
The polynomial is a sum
of things, and we want to
multiply two of these sums
together, so we have to
be a bit more careful about
getting rid of the parentheses,
because it involves changing
the order of operation.
With the parentheses the
addition is being done first,
but without them the normal
order of operations rule
says that the multiplication
is being done first. With
addition it was all one
operation, so it didn't
matter. What you end up
doing in the end is multiplying
all possible pairs, and
in the multiplications you
have to use the rules of
exponents to simplify. (See
Simplifying Exponential Expressions.)
Example 3:
Multiply the polynomials.
(3x+5)(3x-4)
Solution:
(3x)(3x)+(3x)(-4)+(5)(3x)+(5)(-4)
=9x2-12x+15x-20
=9x2+3x-20
Explanation:
We're multiplying all the
possible pairs, where one
factor comes from the first
polynomials and the other
comes from the second
polynomials. A good way
to keep track of this is
to go through the terms
of the first polynomial
and multiply the first one
by each of the terms in
the second polynomial and
then go to the second one
and multiply it by each
of the terms of the second
polynomial, then go to the
third one, etc., etc., until
you use up all of the terms.
If the polynomials just
have two terms each (binomials)
like we have here, then
a good way to remember this
is FOIL, which stands for
firsts, outers, inners,
lasts.
To see why this multiplying
all possible pairs works,
you need to apply the
distributive property
twice, first treating
the (3x+5) like one big
strange symbol for a letter,
and then splitting it
up.
(3x+5)(3x-4)
=(3x+5)(3x)+(3x+5)(-4)
=(3x)(3x)+(5)(3x)+(3x)(-4)+(5)(-4),
so in the end we see
that this does in fact
give us all possible pairs.
We don't have to do this
double distributive property
thing every time we multiply,
though. We just have to
multiply each pair, but
it is useful for you to
see it once to see that
this too just comes from
the basic properties of
real numbers.
Then after we write down
the products of each of
the pairs, we have to
use the properties of
exponents to simplify
each term. (See Simplifying Exponential Expressions.) We
add the exponents and
multiply the numbers.
Then after that there
are often like terms to
combine. Here we need
to add the -12x and the
15x to get 3x.
Example 4:
Multiply the polynomials.
(2x2+3x+5)(x2-2x-3)
Solution:
=(2x2)(x2)+(2x2)(-2x)+(2x2)(-3)+(3x)(x2)+(3x)(-2x)+(3x)(-3)+(5)(x2)+(5)(-2x)+(5)(-3)
=2x4-4x3-6x2+3x3-6x2-9x+5x2-10x-15
=2x4-x3-12x2-19x-15
Explanation:
Here again we multiply all
possible pairs, but since
they have more than two
terms, we can't remember
it by FOIL anymore. We just
go about systematically
starting with the first
term in the first polynomial
and multiplying it by each
of the terms in the second
polynomial and then going
to the next term and doing
the same until we get through
all of the terms. After
that we need to simplify
each term by multiplying
the numbers and adding the
exponents. Then we combine
the like terms to get the
final answer. You can see
that particularly if you
had a lot of terms in the
polynomials this could get
quite long, but unfortunately
multiplication is just hard,
and there really isn't any
way around that. But there
is a slightly better way
to keep track of things,
that we will be learning
shortly.
Polynomial Arithmetic
and Natural Number Arithmetic
There is an interesting
thing about polynomial arithmetic,
and that is that it is really
very similar to natural
number arithmetic. It turns
out that you can do all
polynomial arithmetic using
methods very similar to
those you use for numbers,
and often it makes it easy
to keep track of things.
(See
How to Add,
Subtract, Multiply, and
Divide Natural Numbers.)
Here is how you could do
the above example using
these methods.
Example 1:
Add the polynomials.
(4x3+x2-2x+3)+(3x2+5x)
Solution:
4x3+
x2-2x+3
3x2+5x
4x3+4x2+3x+3
Explanation:
Just like with several place
numbers, we add vertically.
We treat the powers of the
variables like place values
and line up the like terms,
leaving spaces for any powers
that are missing. The only
difference is that we don't
have to do any carrying.
This is a really good technique
for when you have to add
up several polynomials like
in some of the exercises
in algebra books. You can
add them up just like column
addition of numbers and
just add up the coefficients
and let the variables come
along for the ride.
Example 2:
Subtract the polynomials.
(3x2-5x-6)-(4x3-5x2+2x-3)
Solution:
3x2-5x-6
-4x3+5x2-2x+3
-4x3+8x2-7x-3
Explanation:
The same thing as with addition
except that we first change
the signs of all of the
terms in the second polynomial.
I suppose it would be more
like what we do with numbers
if we left the signs as
is and subtracted the columns,
but with plus and minus
numbers to deal with I think
you will make less mistakes
if you do it this way. Then
just like with addition,
the like terms are all nicely
lined up this way, so you
just add down the columns
and bring along the variables
for the ride.
Example 3:
Multiply the polynomials.
(3x+5)(3x-4)
Solution:
3x+5
3x-4
-12x-20
9x2+15x
9x2+ 3x-20
Explanation:
Multiplication is where
this method is really helpful.
We multiply like we do with
several place numbers. With
several place numbers you
also multiply all possible
pairs of the digits. With
polynomials we can treat
the terms just like the
digits of numbers. And remember
how when you multiply numbers,
you indent when you go to
the tens place. Well, if
you do it here, all the
like terms will nicely line
up for you to combine, and
you won't lose anything.
Multiply the -4 times the
5. Then multiply the -4
times the 3x. Then go to
the x's place and multiply
the 3x times the 5 and the
3x times the 3x. Then add
up just like with numbers
and that will take care
of the combining of like
terms.
Example 4:
Multiply the polynomials.
(2x2+3x+5)(x2-2x-3)
Solution:
2x2+ 3x+ 5
x2- 2x- 3
-6x2- 9x-15
-2x3-6x2-10x
2x4+3x3+5x2
2x4+ x3-7x2-19x-15
Explanation:
In a long multiplication
like this we can really
see the benefit of this
method. Just like with multiplying
two three place numbers,
we first multiply everything
by the -3. Then we indent
and multiply everything
by -2x. Then we indent again
and multiply everything
by the x
2, and
magically all of the like
terms line up, and we can
add to combine our like
terms.
Formulas
Since multiplication can
be long and tedious it is
helpful to have some formulas
for multiplications that
come up often. Here are
the most commonly used ones.
(A+B)(A-B)=A2-B2
(A+B)2=A2+2AB+B2
(A-B)2=A2-2AB+B2
They come up often enough
that it is worth memorizing
them, not just for multiplying,
but for later things as
well, like factoring
and completing
the square where it
is helpful to be able
to recognize forms like
these. You derive these
formulas by applying FOIL
just once for each of
the formulas and then
you will never have to
use FOIL for any expressions
of their form. Instead
you can just write down
the answer by knowing
the formula.
The reason the first
one works is that whenever
you multiply the sum and
difference of the same
thing, the inners and
the outers cancel out,
so you just get the firsts
and the lasts. It goes
like this.
A2 (firsts)
-AB (outers) +BA (inners) -B2
(lasts)
=A2-B2
Now that you know this
formula, whenever you
see a multiplication that
is the sum and difference
of the same thing, you
can do it real quickly
without going through
FOIL, because we have
already gone through it
in the general case.
Example 5:
Multiply the polynomials.
(x+5)(x-5)
Solution:
x2-25
Explanation:
This fits into the pattern
of the formula, since it
is a sum and difference
of the same things. Here
A is x and B is 5. So we
just write down the difference
of the squares for the answer.
Example 6:
(2x+3)(2x-3)
Solution:
4x2-9
Explanation:
Same thing here. A is 2x,
and B is 3. You just have
to remember that to square
a product you have to square
both factors, so when you
square 2x, you square both
the 2 and the x.
You can also impress
your friends with some
mental arithmetic tricks
by using this formula,
because formulas like
this work for numbers
as well. Suppose you want
to multiply 21x19. The
quick way to do it using
this formula is to write
it as
(20+1)(20-1).
Then it is a sum and
difference, so you can
use the formula, and it
is
202-12
Both of those computations
are easy to do mentally.
We compute 202
by computing 22
and adding two zeros,
so that gives us 400,
and since 12
is 1, the answer is 399.
You can do this to good
effect whenever you can
write the numbers as the
sum and difference of
the same numbers and those
numbers are easy to square.
The easy way to unsure
that is to use numbers
that are the same amount
up and down from a multiple
of 10, like 33x27, 55x45,
or 62x58. For 33x27, you
could see it as (30+3)(30-3),
and that would mean you
would just need to square
30 and square 3 and find
the difference, which
would be 900-9=891. With
a little bit a practice
you can get quite quick
at this. 55x45 and 62x58
are left as exercises
for the student.
The second and third
formulas are important
because they tell us how
to multiply a binomial
by itself. For this instead
of canceling out, the
outers and inners are
the same, so they double
up.
(A+B)(A+B)
=A2 (firsts)
+AB (outers) +BA (inners) +B2
(lasts)
=A2+2AB+B2
Example 7:
Multiply the polynomials.
(x+3)2
Solution:
x2+6x+9
Explanation:
Sometimes it is helpful
when dealing with formulas
like these to think about
what it says in words. What
this formula says in words
is that to square a sum,
the first term of the answer
is the square of the first
term, the second term is
double the product, and
the third term of the answer
is the square of the second
term. The square of the
first term is x
2.
The product of the two terms
is 3x, and double that is
6x. The last term squared
is 3
2=9.
Example 8:
Multiply the polynomials.
(3x2+2x)2
Solution:
9x4+12x3+4x2
Explanation:
Same thing here, the terms
are just a little more complicated
and you have to remember
that when you square a product
you square both its factors.
The product is 6x
3.
When you double that you
get 12x
3.
The third formula isn't
really totally necessary,
because subtraction is
just adding the opposite,
but it makes it easier,
because then you don't
have to use any negative
numbers. You can see that
it is true, by applying
the second formula and
realizing that the only
effect that the minus
sign will have is on the
middle term, because in
in the last term the negative
number is getting squared,
so that makes a positive
number. But for the middle
term you have a plus times
a minus, so that is a
minus. When you are apply
this formula just make
sure and not make it more
difficult than it is by
thinking of the second
number as negative. The
formula is given to allow
you to only use positive
numbers, so your A and
B should be positive when
you use it.
Example 9:
Multiply the polynomials.
(4x-3)2
Solution:
16x2-24x+9
Explanation:
Again I think it is helpful
to think about it in words.
The only difference between
this and the last formula
is that the middle term
will be minus. Here A is
4x and B is 3. The sign
of the middle term of the
answer will always match
that of the second term
in the original. For the
first term we square the
first, that's 16x
2,
for the second term it is
double the product. The
product is 12x, and then
we double that to get 24x,
but we make it minus. The
last term is the square
of the second, so it is
3
2=9.
Division
We can also divide polynomials
pretty much like we divide
numbers. We do it like long
division of several place
numbers. It is nicer in
some ways, but nastier in
others. Remember how when
you do long division, you
make an estimate by using
just the first places, and
how it sometimes doesn't
work, and you have to raise
it or lower it? Well, with
polynomials you do sort
of the same thing, except
it always works. But the
nastier thing is that with
polynomials you have to
deal with negative number
subtraction sometimes, which
makes it easy to make careless
errors. To makes things
simpler I'll first do an
example that doesn't involve
any negative numbers.
Example 10:
Divide the polynomials.
(x5+5x4+9x3+11x2+12x+13)÷(x+2)
Solution:
Explanation:
Okay, what we have here
is x
5+5x
4+9x
3+11x
2+12x+13
loaves of bread that we
want to give out to x+2
people, where x is some
not yet specified number,
and we want to figure out
how many each person gets.
(See
How to Add,
Subtract, Multiply, and
Divide Natural Numbers
and
Egyptian
Division.) We do it
just like number long division,
treating the terms like
digits. Divide, multiply,
subtract, bring down. Daddy,
mommy, sister, brother,
my 4th grade teacher taught
us. If we were dividing
12 into 159123, we might
get an idea of how many
times 12 goes into 15 by
asking how many times 1
goes into 1, and it would
be 1. Similarly, here to
get an idea of how many
times x+2 goes into x
5+5x
4,
we ask how many times does
x go into x
5.
That means what can we multiply
x by to get x
5.
That would be x
4,
so we write that down on
the top. Not all textbooks
recommend it, but I think
it is a good idea to put
it over the x
4
term to keep track of the
fact that we are really
dividing x+2 into x
5+5x
4,
just like if you were doing
it with numbers you would
put it over the 5 instead
of the 1. That's the dividing
step. Now we go to the multiplying.
We were really dividing
x+2 into x
5+5x
4,
not just x into x
5,
so we multiply by the whole
x+2 to check and find out
how many loaves we have
left after giving out x
4
of them. To multiply the
x
4 times the
x+2, we multiply it by the
2 and by the x, just like
multiplying it by a two
digit number. Then comes
the subtraction. We subtract
that from x
5+5x
4
to find out that we have
3x
4 loaves left
over from that part, but
we also have the other terms,
but we give out the bread
that they represent a bit
at a time to make sure we
have enough for everyone,
so we bring down just one
more term, again just like
it were a digit of a number.
Now we want to divide
x+2 into 3x4+9x3,
and again we judge that
by asking how many times
x goes into 3x4.
What can we multiply x
by to get 3x4
? That would be 3x3,
so we write that on top.
Next comes multiply, so
we multiply that by x+2
to get 3x4+6x3.
Subtract that and we get
3x3. Bring
down the 11x2.
That gives us 3x3+11x2.
Divide x+2 into 3x3+11x2
by asking how many times
does x go into 3x3.
That would be 3x2,
so put that on the top.
Multiply by the whole
x+2 to get 3x3+6x2.
Subtract that to get 5x2.
Bring down the 12x to
get 5x2+12x.
Divide x+2 into 5x2+12x
by asking how many times
does x go into 5x2.
That would be 5x times,
so put that on the top.
Multiply it by the whole
x+2 to get 5x2+10x.
Subtract that to get 2x.
Bring down the 13 to get
2x+13.
Divide x+2 into 2x+13
by asking how many times
does x go into 2x. That
would be 2 times, so put
that on the top. Multiply
it by the whole x+2 to
get 2x+4. Subtract that
to get 9, which is the
remainder.
So that means that if
we had x5+5x4+9x3+11x2+12x+13
loaves of bread that we
wanted to give out to
x+2 people, we could give
out x4+3x3+3x2+5x+2
loaves to each person
and have 9 left over.
It is a good idea to
check a division problem
like this by multiplying
the answer by the divisor
and adding the remainder.
You should get the dividend
(the thing you divided
into) if you do that.
This too is similar to
what you would do with
numbers. If you divided
59 by 7 and got 8 R 3,
you could check this by
multiplying 8 times 7
to get 56 and then adding
3 to get 59. Let's see
if that works in this
case. We can do this now
with our polynomials using
the vertical form of multiplication
that we just learned about.
x4+3x3+3x2+
5x+ 2
x+ 2
2x4+6x3+6x2+10x+4
x5+3x4+3x3+
5x2+ 2x
x5+5x4+9x3+11x2+12x+4
9
x5+5x4+9x3+11x2+12x+13
Good, it worked.
Example 11:
Divide the polynomials.
(3x5-x4-2x3+24x2-4x+3)÷(x2+2x-3)
Solution:
Explanation:
This one has some negative
numbers and a 2nd degree
polynomial for a divisor,
so it is a little harder.
But the procedure is still
the same. For the negatives
you just have to remember
your integer arithmetic.
(See
How to
Add, Subtract, Multiply,
and Divide Integers
for a review if you need
it.) The subtractions in
particular can easily lead
to careless errors, so it
is a good idea when doing
them to right away change
all of the signs of the
bottom polynomial and then
you can add instead of subtracting.
But I have often found a
problem with doing that,
that it kind of looks bad
when you want to go back
and check your steps, because
the multiplications look
wrong, so my own tendency
is to avoid it if I can.
Also sometimes I am a little
timid about suggesting the
changing the signs and adding
thing to students too soon,
because it makes it sound
too complicated, and it
is too easy to forget that
that it is really just the
same procedure that you
do with numbers. But on
the other hand, most students
don't have as much experience
with plus and minus numbers
as I do, so I find that
if I don't tell them to
change the signs and add,
a lot of careless errors
happen.
Actually no matter what
I do a lot of careless
errors happen with polynomial
long division. It is simply
that sort of operation
that is ripe with them.
I have only found one
way to deal with operations
like that, and that is
to SLOW DOWN. It's not
a race. Nobody cares how
fast you can do it. You
are not going to be asked
to do this on an assembly
line. Machines can do
this sort of thing a lot
better than people can
anyone. You are doing
this sort of thing to
get some hands on experience.
We are not trying to drill
you into being able to
do it without thought.
You are supposed to do
these things WITH thought.
You will learn a lot more
if you do problems like
polynomial long divisions
slowly and accurately,
even if it means that
you don't do as many problems.
The best kind of practice
for problems like long
division is to do a small
number of problems, slowly
and carefully. Everybody
has their own way of working,
but I find that when I
have something I am doing
that I want to make sure
will be accurate and there
are lots of places to
make careless errors in
it, it helps it I write
really neat, and sometimes
even take the time to
write the lettering pretty
and stylized--but that
can be overdone--because
it slows me down, makes
me pay attention, to some
extent maybe also keeps
me from getting bored,
and also makes it easier
to check myself when I
am done.
Anyway, getting back
to the negatives and subtraction,
after a bit of thought
I came up with a compromise,
which is that when changing
the signs to add not to
obliterate the original
signs so that you can't
tell what they were, but
instead write the new
signs for changing the
subtractions into addition
little and above the original
signs. You might even
want to try writing them
in another color like
I have done in the worked
out division above, to
make it really clear which
are the originals and
which are the changed
ones. Also you don't really
have to do it for every
subtraction. Use your
judgment. You know yourself
better than I do. If you
feel confident that you
can do the subtraction
without making sign errors,
like particularly when
everything is positive
and the top numbers are
bigger than the bottom
numbers, then just do
it as a subtraction. For
example in this problem
you might not want to
bother with the 24 and
the 21, because there
it is easy enough to subtract
24-21, but I have written
the changing of signs
in there anyway. And no
matter what, you can always
think of it as a subtraction
when you are remember
the rules for the operations,
so it really is still
the same steps as with
numbers. You just need
to remember when you think
of it this way that subtraction
is adding the opposite,
which shouldn't be a problem,
because if you know your
rules of plus and
minus arithmetic,
then you know that.
So getting down to the
details of the problem,
again we divide, multiply,
subtract, and bring down.
We see how many times
x2+2x-3 goes
into 3x5-x4-2x3
by asking ourselves how
many times does x2
goes into 3x5,
just like you would do
if you were dividing 123
into 312443 and were judging
how many times 123 goes
into 312 by asking how
many times 1 goes into
3. In the number case
this wouldn't quite work
this time, so you see
that polynomials are in
this way easier than numbers,
because the 'estimate'
always works here. The
reason for that, by the
way, is that numbers are
more restrictive about
what can be digits. The
digits of numbers can
only be 0 through 9, but
coefficients can be anything,
so all that matters with
the polynomials is that
the degree gets reduced,
and that will happen as
long as the first term
is right. So we ask how
many times x2
goes into 3x5,
and that is 3x3,
because 3x3
is the expression that
you can multiply by x2
to get 3x5,
so we write this on top
and go on to the multiplying
phase. Again like in the
last example, we don't
just multiply it by x2
, we multiply it by the
whole divisor term by
term to get 3x5+6x4-9x3.
Then we subtract that
from 3x5-x4-2x3.
The safest way to subtract
when you have a lot of
minuses involved is to
change all of the signs
of the bottom polynomial
and then add. I have indicated
the new signs in blue,
so we add using those
signs and get -7x4+7x3.
Then we bring down the
24x2.
Now we start the process
again. Divide. How many
times does x2
go into -7x4.
That would be -7x2
times, so we put that
on top. Multiply. Multiply
-7x2 times
x2+2x-3 to
get -7x4-14x3+21x2.
Subtract. To subtract
this we change all of
its signs and add it,
which gives us 21x3+3x2.
Bring down. Bring down
the next term and we are
ready for the next round.
Divide. How many times
does x2 go
into 21x3 ?
21x times, so we put that
on the top. Multiply.
We multiply that times
x2+2x-3 to
get 21x3+42x2-63x.
Subtract. To subtract,
change the signs of the
bottom one and add. We
get -39x2+59x.
Bring down. Bring down
the 3 and we are ready
to do it one last time.
Divide. How many times
does x2 go
into -39x2.
-39 times, so we put that
on the top. Multiply.
Multiply that by x2+2x-3
to get -39x2-78x+117.
Subtract. To subtract
we change all the signs
of the bottom one and
add, which gives us 137x-114.
Nothing more to bring
down, so that is our remainder.
Notice here that we have
a remainder that is a
linear expression, a 1st
degree polynomial, instead
of just a number. The
rule is that the remainder
must have degree less
that the degree of the
divisor, so this is okay
since our divisor was
degree 2.
Again we should check
our answer by multiplying.
3x3-7x2+21x-39
x2+
2x- 3
-9x3+21x2-63x+117
6x4-14x3+42x2-78x
3x5-7x4+21x3-39x2
3x5- x4-
2x3+24x2-141x+117
137x-114
3x5- x4-
2x3+24x2-
4x+
3
and it checks.
Example 12:
Divide the polynomials
(x6+x5-x2+2)÷(x3+x+2)
Solution:
Explanation:
Here what's new is that
there are some 'missing'
terms, that is not all of
the powers are there. When
you do any of these operations
with polynomials that have
missing terms you need to
either fill them in with
terms with 0 coefficients
or at least leave blanks
for them and treat them
as if they were terms with
0 coefficients. I have filled
in with zeros for this one,
but as you get more familiar
with it, you may be able
to get away with just leaving
spaces. If you imagine what
it would look like without
the green parts, that is
what it would look like
if you did it with blanks.
Then as usual you divide,
multiply, subtract, and
bring down, and again
you can do your subtractions
by changing all of the
signs of the bottom one
and adding and I have
again indicated the new
signs for that in blue.
This time the remainder
is degree 2, which is
okay, because the divisor
was degree 3.
Again we should always
check, and with the multiplication
we are also going to need
spaces. For variety I
will do it with spaces
instead of zeros, since
multiplication is a bit
easier than division.
x3+ x2-
x-3
x3
+ x+2
2x3+2x2-2x-6
x4+ x3-
x2- 3x
x6+ x5-
x4-3x3
x6+ x5
+ x2-5x-6
-2x2+5x+8
x6+ x5
+ x2
+2
Good, it worked!
A Shorter Way
In spite of these more complicated
examples I have given, really
the most common use of long
division is for simpler
problems like the Example
5, but possibly involving
some negatives. For problems
like that there is shortcut
method that you can check
out if you'd like called
synthetic division. You
can find out about it by
reading my article
Synthetic
Division.
More Examples and Practice
To see some more examples
and get more practice with
the four operations of polynomial
arithmetic see my
MathHelp
problem set collection
Polynomials.
