Common Factors
The first thing you should always do
when factoring is to take out a common
factor. This is the simplest technique
of factoring, but it is important even
when you learn fancier techniques, because
you will make your later work much easier
if you always look for common factors
first. Taking out common factors is
using the distributive property backwards.
The distributive property says
a(b+c)=ab+ac.
The idea behind taking out a common
factor is to look for something like
the right side here where there is
a common factor, here it would be
a, and turn it into something like
the left side to factor it.
Example:
4x5+12x4-8x3=4x3(x2+3x-8)
A good trick for finding the largest
common factor when you are using this
method to factor polynomials is to
find the greatest common factor of
the numbers and the smaller power
of the variable, so here the greatest
common factor of the numbers is 4
and the smallest power of x is 3,
so we can take out 4x3
as a common factor. (See GCFs and
LCMs.)
Grouping
Grouping is a fancier technique that
is based on taking out common factors.
For grouping we split the polynomial
in two pieces and take out common factors
in each of them. If we get the same
thing left over in each piece, then
we can take that big thing out as a
common factor, and this will factor
the polynomial. It is, however, important
to note that this only works if we do
get the same thing. Simply taking out
common factors of pieces of a polynomial
is not factoring it. To factor a polynomial
we must write the whole polynomial as
a product of two polynomials.
Example:
5x3+10x2+3x+6=5x2(x+2)+3(x+2)=(5x2+3)(x+2)
Here we take out a common factor
of 5x2 from the first two
terms and a common factor of 3 from
the second two terms. This would not
be a good method of factoring except
that this polynomial is kind of special
in that what is left when taking out
the common factors in both cases is
x+2. Then what we do is take out x+2
as a common factor of the two big
chunks. The idea is that anything
we can do with x we can do with x+2,
because it represents a number too.
If it helps, blur the (x+2) or think
of it as one big strange looking letter.
General Quadratics
To learn to factor second degree polynomials,
also called quadratics we first make
it simpler by looking at ones where
the coefficient on x
2 is
1. To figure out how to factor these
polynomials we need to look at what
happens when you multiply things of
the form (x+A)(x+B), like for example
(x+3)(x+4). Applying FOIL (see
How to Add, Subtract, Multiply, and Divide
Polynomials) to this multiplication
we get
x2+4x+3x+12=x2+7x+12.
Notice that the coefficient on x
comes from adding the inners and outers,
so it will always be the sum of the
two constants. The constant term will
always come from the lasts, so it
will be the product of the two constants.
So really we could use this as a special
trick for multiplying anything of
this form. We just add the constants
for the middle coefficient and multiply
them for the last term. So to multiply
(x+5)(x-2)
we can use this shortcut and get
x2+3x-10
by just adding the 5 and the -2 to
get the 3 and multiplying the 5 and
-2 to get the -10. This means that
in order to go the other way around,
that is to factor into this kind of
form, what you need to find is two
number that multiply to the last term
and add to the the middle coefficient.
So that means that to factor the above
polynomial all you need to do is find
two numbers that multiply to -10 and
add to 3, and that would be 5 and
-2, and that tells us that it will
factor into
(x+5)(x-2).
Of course we could cheat this time,
because we already knew the answer,
but for many polynomials it wouldn't
be too hard even if we didn't, but
there is a little bit of guess work
here, so it may take a little practice
to get used to it. One thing you can
do if you get stuck is to list all
of the ways to factor the last term
and then play around with them until
you find one that adds up to the right
thing. You can even simply try all
of them, which might seem like a lot
of work, but by doing it you will
get a feel for how it works and before
too long you won't really have to
try all of them. Let's say you wanted
to factor
x2+17x+60.
This one is a little bit tricky,
because there are a lot of ways to
factor 60. To get an idea about what
happens with these, lets list them
all.
1x60
2x30
3x20
4x15
5x12
6x10
1+60=61, no good.
2+30=32, no good.
3+20=23, no good.
4+15=19, closer, but still no good.
5+12=17, eureka, we've found one.
This means that we can factor the
polynomial as
(x+5)(x+12).
One way to avoid simply trying them
all is to do some estimating and also
notice some patterns. One pattern
that is helpful is that when the numbers
are farther apart, like 60 and 1,
they add to bigger numbers and when
they are closer together, like 6 and
10, they add to smaller numbers. There
are some other tricks you can play
as well. One I find handy sometimes
is swapping prime factors. Suppose
in this one you guessed 6x10 and say
that it was 16, which is just a little
too small. Since it is too small,
what you want is to factor it so that
the factors are a little farther apart.
Sometimes you can come up with this
by prime factoring the two numbers
and then switching some factors.
6=2x3
10=2x5
One way to make these farther apart
would be to switch the 3 and the 2
and get
2x2=4
3x5=15,
but that gives us a sum that is too
big. The way to do it that does work
is to give the 10's 2 to the 6 and
get
2x2x3=12
5=5.
It is also important to consider
the signs of the coefficients and
what happens when you multiply and
add plus and minus numbers. The simple
case is when both coefficients are
plus like the one above where you
only need to use positive numbers.
The case where the middle coefficient
is negative and the last terms is
positive is almost as easy. What you
need then is two numbers that multiply
to a positive number and add to a
negative number. The only way to get
that is if they are both negative.
Then because when you add two negative
numbers you add and your answer is
negative, you can think of it pretty
much the same way you do when they
are both positive. The only thing
you have to do different is make them
both negative instead of both positive.
So for example, suppose you want to
factor
x2-5x+6.
Then you still just need to find
two numbers to multiply to 6 and add
to 5, which would be 2 and 3, but
since we really want -5, we would
use -2 and -3 and factor it
(x-2)(x-3).
If the last term is negative it is
a little bit different. Then regardless
of what the sign of the middle coefficient
is, one of your numbers has to be
positive an the other negative to
multiply to a negative number. That
means that when you add those numbers,
in terms of positive number arithmetic,
it is a subtraction. When you add
a positive and a negative number you
subtract and use the sign of the larger
one, so you need to think in terms
of subtracting to get the middle term,
and if the middle coefficient is positive
you want the positive one to be bigger,
but if it is negative, you want the
negative one to be bigger. For example
if we wanted to factor
x2+x-12,
we could do it by finding two numbers
to multiply to 12 and subtract to
1. That would be 4 and 3. Then since
the middle coefficient is 1 and not
-1, we would use 4 and -3 and factor
the polynomial as
(x+4)(x-3).
On the other hand, if we wanted to
factor
x2-x-12,
we could think about it the same
way, what two numbers multiply to
12 and subtract to 1 and get 4 and
3, but since this time we want a -1,
we would use -4 and 3 and get
(x-4)(x+3).
Next we turn to the more difficult
problem of when the first coefficient
is not 1. Here the guessing is a little
more complicated, because the coefficient
you get from the inners and outers
will not be just the sum of the two
constants, because they will get multiplied
by the x coefficients. For example,
lets say we want to factor
3x2+11x+10.
We need to make the product of the
firsts give us 3x2. In
this problem that isn't too difficult,
because there is only one way to do
it, and that is 3x and x, so we want
a factorization of the form
(3x
)(x
).
Now, the product of the lasts still
has to be 10, but it is not their
sum that has to be 11, it is something
a bit more complicated, because in
the sum of the inners and outers one
of the numbers will be multiplied
by 3. For example, lets suppose we
tried 2 for the first one and 5 for
the second one. Then it would be
(3x+2)(x+5).
That wouldn't work, because the sum
of the inners and outers wouldn't
be 11x. Instead it would be 15x+2x=17x.
The first and last terms would be
right, but that is not good enough.
But if we put the 5 in the first factor
and the 2 in the second factor, it
would work, since then it would be
(3x+5)(x+2)
and the sum of the inners and outers
would be 6x+5x, which is indeed 11x.
But how are you to come up with that
arrangement other than trial and error?
To some extent you can't, you just
get better at the trial and error
with practice. Again, you could if
worse came to worse try all possible
combinations. For most of the problems
you will be expected to do in algebra
classes there won't be that many possible
ways to factor the last term anyway.
For 10 it is just 1x10 and 2x5, but
here this makes a few more possibilities
than last time, because the order
will matter. But still for this problem
there are only 4 possibilities,
(3x+10)(x+1),
(3x+1)(x+10),
(3x+2)(x+5),
(3x+5)(x+2).
In the beginning it won't hurt you
to just try them all and see what
happens. Again, this can be quite
helpful in getting a feel for what
happens with this. It sounds like
a long process, but after you do it
a few times, you will find that you
can eliminate a lot of possibilities
without fully trying them. There are
again some techniques you can use
involving estimating sizes. In general
you will get larger numbers by using
factors that are farther apart and
by placing the numbers so that the
large numbers multiply large numbers
and the small numbers multiply small
numbers and you will get smaller numbers
by doing the reverse. So for example,
here (3x+1)(x+10), where the 10 is
multiplied by the 3 and the 1 is multiplied
by the 1, gives us the largest possible
middle term and (3x+5)(x+2), where
the 5 is multiplied by the 1 and the
2 is multiplied by the 3, gives us
the smallest possible middle term.
Also in these problems you can eliminate
some possibilities right away by estimating.
If I was approaching this problem
I would see right away that I couldn't
put the 10 so that it multiplied the
3, because the outers alone would
be too big. I would even be pretty
suspicious of using the 10 at all,
because even 10 times 1 is almost
11, so there wouldn't be too much
left for the other term. So I would
guess it would be 5 times 2, and putting
the 5 where it gets multiplied by
3 gives 15 which is already too big
on its own, so there is really only
one possibility that has a chance.
With a little practice you may find
it surprising that you will be able
to guess right the first time a good
amount of the time, and when your
first guess isn't right, usually your
second one is. I find when doing these
that sometimes the right solution
just sort of appears before me magically,
but I have been doing these in front
of classes for a lot of years. Just
the same, with practice you may find
that this happens to you more and
more often.
More Examples
For these examples the instruction is
to factor the polynomial.
Example 1:
2x2+9x-5
Solution:
(2x
)(x
)
There is only one way to factor 5,
5 and 1, so those are the numbers
we are going to use. The only question
is where to put them. Here since the
last term is minus, one of the numbers
must be minus and the other plus.
That means that the inners and outers
will subtract instead of add. Since
9 is large, particularly since it
will be a difference we want to split
things unevenly, so the best guess
is to place the 5 so that it gets
multiplied by the 2 and the 1 so it
gets multiplied by 1.
(2x 1)(x
5)
Since we want to get a positive 9,
we want the one that gives us the
bigger product to be positive and
the one that gives us a smaller product
to be negative. That gives us
(2x-1)(x+5).
And that should work, because we
get 10-1=9. The firsts and lasts will
work without problems. Check again
to make sure the inners and outers
do in fact work. We get
10x-1x=9x,
so it works.
Example 2:
6x2-31x+35
Solution:
This one is a bit harder, because we
have a choice for the firsts. We can
use 6x and x, or 2x and 3x, and we don't
really know which one will work. Sometimes
it is possible to use the relative sizes
of the middle and last terms to help
us guess, but there isn't really any
way to know for sure. Mainly we just
have to try one, and see if we can find
anything for it, and if we can't try
the other. For this one, I'm not really
sure at all which one it is going to
be, so let's try 6x and x first and
see what happens.
(6x
)(x
)
We need to multiply to get 35, so
it must be 1 and 35 or 5 and 7. There
are some possibilities that we can
eliminate right away. Since the last
term is plus, the signs of the two
numbers will be the same, so for the
middle term we will be adding. The
only difference the minus on the middle
term makes is that both of the numbers
will be minus. We can't put the 7
or the 35 so they get multiplied by
the 6, since that would be already
too big. Even putting the 5 so that
it gets multiplied by the 6 doesn't
look like a good idea, because that
only leaves 1 for the inners, but
that is really the only possibilities
with this set up, since actually we
can't use the 35 at all, since it
alone is bigger than 31. If we put
the 5 with the x, so that it would
get multiplied by the 6, then we would
have to put the 7 with the 6x to be
multiplied by the x. In case you have
gotten lost, that would be
(6x+7)(x+5).
But without even precisely working
that out, we can tell that this would
be too big, since 7 is bigger than
1. So we have come to the end of the
line with 6x and x and must see what
we can get with 2x and 3x.
(2x
)(3x
)
Okay, we have already figured out
that there is no way that we can use
35 and 1, because the 35 alone would
be too big, so we have to use 5 and
7. Let's just try both ways to see
what happens.
(2x-5)(3x-7)
This gives us 14+15=29, close, but
that's not good enough.
(2x-7)(3x-5)
This give us 21+10=31, which works!!
Notice that without too much difficulty
we eliminated all but two possibilities,
by some simple estimating, and even
if you had to try the two remaining
possibilities simply by trial and
error, that wouldn't be that bad.
With some practice, maybe you wouldn't
even have to do that. I might guess
the right answer on this one right
away, because of seeing that if I
put the 7 to be multiplied by the
3, I will get 21, which is 10 off
from what I want to get, and I know
at least that 10 is divisible by the
other numbers, so there is a chance
that they might multiply to 10. After
you do these for a while you sort
of pick up on patterns like that.
Another thing you can do is to save
time when you are trying the numbers
out, don't bother with the signs,
and sort those out after you see that
the numbers work, and of course don't
bother with the variable either, because
you know that will be there. So in
this correct arrangement you would
simply figure 3x7 for the inners and
2x5 for the outers, and check to see
if that adds up to your middle coefficient.
Example 3:
4x2-5x-6
Solution:
For this one again we have to be careful,
because we get a choice between 4x and
x or 2x and 2x. This time I am going
to deliberately choose the wrong one,
because I know the answer to this problem,
because I cheated by multiplying the
factors to get the problem. Let's try
2x and 2x, since I know it is really
4x and x.
(2x
)(2x
)
Since the last term is minus, we
will be looking for the middle term
to be the difference of the outers
and inners. 6 can be factored 6x1
and 2x3 and we need a difference of
5, but remember, that includes multiplying
by the 2's. This time since the coefficients
on the x's are the same, it won't
matter which order we put them in.
6 and 1 will give us 12-2=10 and 3
and 2 will give us 6-4=2, so neither
will work. If you are clever about
it you don't have to go through all
that to see it, because since both
numbers are getting multiplied by
2, you will be getting the difference
of two even numbers, which must be
even, and 5 is odd. You may not be
able to precisely explain it like
that, but you still might be able
to intuitively look at it and see
that the inner outer difference just
doesn't look like it could come out
to 5. So any way we do it, we can
see that 2x and 2x, won't work, so
we will have to try the other possibility
4x and x.
(4x
)(x
)
6 and 1 just looks like it won't
work. One way the difference looks
too big and the other way it looks
too small, so lets try 2 and 3. We
are looking for a difference of 5.
If we put the 3 so it gets multiplied
by the 4, that looks like it will
make the difference too large, since
the 3x4 part is 12 and the other part
is small, so it looks like the other
way around is the best choice. Let's
try it.
(4x 3)(x
2)
Write the numbers in first without
signs, because that is quicker and
we can adjust the signs later. This
gives us 8-3=5, so that looks good.
Now adjust the signs. We want it to
be a -5, so we want the negative one
to be bigger. Now here we have to
be careful, because it is not the
bigger of the two numbers that should
be negative it is the one that is
bigger after multiplying by the x
coefficients. The 2 is smaller than
the 3, but it is getting multiplied
by the 4 for make 8 and the 3 is multiplying
by 1 to get 3, and it is these products
that we are adding together. So we
want to get -8 and +3, so that they
add up to -5, so that means that even
though the 2 is smaller, it is the
2 that will get the minus sign, so
the final answer is
(4x+3)(x-2).
Let's check that to make sure. For
the firsts we get 4x2 as
planned. For the outers we get -8x.
For the inners we get 3x. And for
the lasts we get -6. Putting that
together we get
4x2-8x+3x-6=4x2-5x-6,
so it works!!
With some practice, even though this
method may seem a bit fiddling at
first, you should be able to get good
enough at it so that you can manage
most problems. But in case you have
one where both the first and last
terms can be factored a lot of different
ways and you get stuck, there is another
method based on grouping you can use.
It is longer, but involves less guesswork.
Because it involves less guesswork
it can be tempting to give up on the
guesswork and only use this method,
but I don't think this is a good idea.
It is better to save this method for
problems where you get really stuck
doing it the other way, because otherwise
you won't get the practice that it
takes to get good at doing it the
other way. In other words, if I teach
you this other way, you must promise
not to abuse it. So read this part
only if you can make that promise.
Okay, now that I know I am only talking
to those who are ready for this, here
is the trick. We multiply the first
and last coefficients and then play
the game we played for ones with a
first coefficient of 1, but with this
number instead of just the last term.
We take that number and look for two
numbers that will multiply to it and
add to the middle coefficient. Then
we use those numbers to break up the
middle term, and if we do that it
turns out that the 4 term polynomial
we get will always factor by grouping.
This might be a bit hard to grasp
without examples, so I will show you
how to use this method for the above
three examples.
Example 1:
2x2+9x-5
Alternate Solution:
Multiply the 2 and the -5 to get -10
Now find two numbers that multiply
to -10 and add to 9.
That would be 10 and -1
Now use these numbers to break up
the middle term and the polynomial
becomes
2x2+10x-x-5.
It doesn't matter what order we put
them in. We could use -x+10x as well,
and in the end we would come out with
the same answer.
Now I claim we can factor this by
the grouping method. For the first
two terms take out a common factor
of 2x and for the second two terms
take out a common factor of -1, and
we get
2x(x+5)+-1(x+5)
and look, isn't that amazing, we
get a common factor of x+5 that we
can take out and we get
(2x-1)(x+5),
the same thing we got before.
Example 2:
6x2-31x+35
Alternate Solution:
Multiply 6 times 35 to get 210.
Find two numbers to multiply to get
210 and add to get 31.
Not quite as easy as last time, but
-21 and -10 will do.
Now use these numbers to break up
the middle term and we get
6x2-21x-10x+35.
Now factor this by grouping. Take
out a common factor of 3x from the
first two terms and -5 from the second
two terms and this becomes
3x(2x-7)-5(2x-7)
Again the leftover parts are both
the same, this time 2x-7, so we can
take that out as a big common factor
and get
(3x-5)(2x-7).
Example 3:
4x2-5x-6
Alternate Solution:
Multiply the 4 and the -6 to get -24.
Then find two numbers that multiply
to -24 and add to -5.
-8 and 3 will work.
Use these numbers to break up the
middle term and we get
4x2-8x+3x-6.
Now factor this by grouping. Take
out a common factor of 4x from the
first two terms and a common factor
of 3 from the second two and we get
4x(x-2)+3(x-2)
Then take out x-2 as a factor to
get the answer of
(4x+3)(x-2).
Just make sure to keep an open mind.
It is never a good idea in mathematics
class to get stuck on one way of doing
something.
Difference of Squares
There is a special formula for factor
a difference of squares that comes from
applying a shortcut multiplication formula
backwards. See
How to Add, Subtract, Multiply, and Divide
Polynomials. It turns out that whenever
you multiply the sum and difference
of the same things, the inners and outers
cancel out and you only get two terms,
and this gives you the formula
(A+B)(A-B)=A2-B2
.
To use the formula for factoring
we need to write it backwards as
A2-B2 =(A+B)(A-B).
So if we see something that is a
difference of squares, we can factor
it as the sum and difference.
Example:
x2-9=(x+3)(x-3)
Sometimes we can use the formula
more than once.
x8-1=(x4+1)(x4-1)=(x4+1)(x2+1)(x2-1)=(x4+1)(x2+1)(x-1)(x+1)
This brings up another point. Factoring
is like prime factoring of numbers.
Always factor completely, that is
keep factoring the factors until nothing
more will factor. The ones that can't
factor come along for the ride. Sums
of squares can't be factored, so they
come along for the ride.
Perfect Squares
Sometimes we can also use the squaring
formulas (see
How to Add, Subtract, Multiply, and Divide
Polynomials),
(A+B)2=A2+2AB+B2
(A-B)2=A2-2AB+B2
in reverse for factoring. If you
see a polynomial with three terms
and the first and last terms are perfect
squares, you can check and see if
the middle term is right for this
kind of factorization.
Example:
x2+14x+49
The first and last terms are perfect
squares. If this were to fit into
the pattern of the first of these
formulas, it would have to be
(x+7)2
The first and last terms are right
for that. Is the middle term right.
The middle term should be twice the
product of 7 and x, which is 14x,
and yes, that is what we have, so
this factorization works. For this
problem we could have also used the
method above for general quadratics,
but for some, particularly like ones
with fractions, we can get it easier
this way.
Sum and Difference of Cubes
There are special formulas that need
to be memorized for factoring the sum
and difference of cubes.
A3+B3=(A+B)(A2-AB+B2)
A3-B3=(A-B)(A2+AB+B2)
Notice that although it isn't possible
to factor a sum of squares, it is
possible to factor a sum of cubes.
It turns out in general that it is
always possible to factor sums of
odd powers, but not even powers. The
way I always remember these formulas
is that the first factor is the original
without the cubes and the second factor
is almost like the squaring formulas,
but without the 2 and the sign on
the second term of it is always the
opposite of the sign in the original
and in the first factor. In other
words these formulas almost look like
A3+B3=(A+B)(A-B)2
A3-B3=(A-B)(A+B)2
except that there is no 2 on the
middle term. Also when you are factoring,
once you have written down the first
factor, you have A and B handy, so
you can easily read them off from
it for writing down the second factor.
Example 1:
x3+8
Solution:
Since the first term is x cubed and
the second term is 2 cubed, this is
a sum of cubes. For the first factor
we just remove the cubes and write down
(x+2).
Then for the second factor, just
take these two terms that are written
down in the first factor and plug
them into the formula for the second
factor, which is just like the formula
for the square of x-2, except that
you don't double for the middle term,
the middle term is just the product
of the two terms, so we square the
x and then take the product of 2 and
x and then square the 2 and get
(x+2)(x2-2x+4).
Example 2:
x3-27
Solution:
This one is a difference of cubes, where
A is x and B is 3, so for the first
factor we get
(x-3).
Now for the second factor we just
plug these two numbers into the formula
for the second factor. Square the
x for the first term, multiply 3 times
x for the second term and square the
3 for the last term and we get
(x-3)(x2+3x+9).
Sum and Difference of Higher Powers
There are formulas for sums and differences
of higher powers, but mostly people
don't bother to learn them. You can
either look them up, or work them out
by long division. (See
Synthetic
Division.) But some higher powers
can be done by using the difference
of square and sum and difference of
cubes formulas, namely any power that
is a multiple of 2 or 3. Any power that
is a multiple of 2 is a perfect square
and any power that is a multiple of
3 is a perfect cubes. This is true because
powers to powers multiply. (See
Simplifying Exponential Expressions.) So
for example x
16 is a perfect
square, because it is (x
8)
2
and x
24 is a perfect square
and a perfect cube, since it is (x
12)
2
and (x
8)
3. So
you can factor differences of even powers
by using the difference of squares formula
and sums and differences of powers that
are multiples of 3 by using the sum
and difference of cubes formulas. And
sometimes you can factor more than once
this way. Sometimes when you approach
it two different ways it can look like
you get different answers. An interesting
example of that is when you have a difference
of 6th powers where you can treat it
either as a difference of squares or
a difference of cubes.
Example:
x6-1
Method 1:
Look at it as a difference of squares.
(x
3)
2=x
6,
1
2=1.
x6-1=(x3-1)(x3+1),
but we can go further, because each
of these factor can be factored using
the sum and difference of cubes formulas.
x3-1=(x-1)(x2+x+1)
and
x3+1=(x+1)(x2-x+1),
so putting this all together, we
get
x6-1=(x3-1)(x3+1)=(x-1)(x2+x+1)(x+1)(x2-x+1).
Method 2:
Look at it as a difference of cubes.
(x
2)
3=x
6,
1
3=1.
x6-1=(x2-1)(x4+x2+1)
The first factor can be factor further
as a difference of squares, and we
get
(x-1)(x+1)(x4+x2+1),
but it looks like that is all we
can do, because it is hard to see
how to factor the other factor.
But we should get the same answers
with both methods, because after all,
they are equal, so it must be true
that
x4+x2+1=(x2+x+1)(x2-x+1),
but it is not at all clear how you
would figure that out if you did it
that way. For this reason it is generally
better when you have problems like
this where there is a choice to look
at it first as a difference of squares.
But there is a way you can get the
complete factorization the other way,
and it involves a fancier factorization
that you can use for some other problems
as well. I don't know any official
name for it, but I call it completing
the square from the inside, because
it involves adding and subtracting
something that will change the middle
term so that the first chunk is a
perfect square, and then factoring
it as a big difference of squares.
Completing the Square from the Inside
(Extra for Experts)
This is a method of factoring that we
can use when we have a 4th degree polynomial
that has no odd terms, and we can't
factor simply by treating x
2
as the variable. For some polynomials
of this sort we can factor them much
easier by treating x
2 as
the variable. For example if we want
to factor x
4-5x
2+6,
we can do that. We just ask ourselves
what two numbers will multiply to 6
and add to -5, and since that is -2
and -3, we can factor it to (x
2-2)(x
2-3),
just like if it were x
2-5x+6.
But it turns out that if we have one
like that where we can't find numbers
to multiply to the last term and add
to the middle coefficient, we can often
still factor them by using this method
of completing the square from the inside.
Example:
x4+x2+1
We can't factor this one like the
last one, because we can't find two
numbers to multiply to 1 and add to
1. What we do with a problem like
this is try to find something to add
to the middle term so that the polynomial
becomes a perfect square, and we add
and subtract it. By doing this we
can turn it into a difference of squares.
In this case if there was a 2 coefficient
on the x2, we would have
x4+2x2+1, which
would be a perfect square, so what
we do is add x2 and subtract
it, to get
x4+2x2+1-x2=(x2+1)2-x2
If we couldn't see by guessing what
the coefficient would need to be for
this we could get it by doing the
opposite of what we normally do for
completing the square (see Quadratic
Equations), take the square root
of the constant term, and then double
that, so here this would be
sqrt(1)=1
2·1=2.
Then to get that coefficient we figure
out what multiple of x2
we need to add to the existing term
to get it. In this case since we already
have 1, we need to add another 1 to
get 2.
After we do this we get a big difference
of squares, so we can use the difference
of squares formula to factor it and
get
[(x2+1)+x][(x2+1)-x].
Then getting rid of the extra parentheses
and writing things in standard order
this becomes
(x2+x+1)(x2-x+1),
just like we got from the other method.
Here are a couple more examples of
this method.
Example 1:
x4+2x2+9
Solution:
sqrt(9)=3
2·3=6
So here the middle term needs
to be 6. Since it is 2 now, we need
to add and subtract 4x2.
Then we get
x4+2x2+9=x4+6x2+9-4x2=(x2+3)2-4x2=[(x2+3)+2x][(x2+3)-2x]=(x2+2x+3)(x2-2x+3).
Example 2:
x4-19x2+25
Solution:
Here if the middle coefficient were
a -10, then it would be a perfect square,
so add and subtract 9x
2.
sqrt(25)=5
2·5=10
x4-19x2+25=x4-10x2+25-9x2=(x2-5)2-9x2=[(x2-5)+3x][(x2-5)-3x]=(x2+3x-5)(x2-3x-5).
Tips
- First look for a common factor.
- Always factor completely. This
is like prime factoring. See How
to Prime Factor a Number.
