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.
