Solving polynomial equations
of degree larger than
two can be difficult,
and in most cases the
only way to do it is to
have a computer search
for them. There are formulas
for degree 3 and 4, but
they are complicated,
and there are no formulas
for above degree 4. But
for certain polynomials,
particularly those with
some rational solutions,
there are ways to solve
them by hand. The following
three theorems can be
helpful for this.
- The Rational Zero
Theorem
- The Upper and Lower
Bound Theorem
- Descartes Rule of
Signs
The Rational Zero Theorem
If P(x) has integer coefficients,
and p/q is a rational zero
of P(x), then p is a factor
of the leading coefficient
(the coefficient on the
highest power) and q is
a factor of the constant
coefficient.
The Upper and Lower
Bound Theorem
If a>0 and all of the
numbers in the bottom
row of the
synthetic division of P divided by x-a are positive,
then a is an upper bound
for the zeros of P.
If a<0 and the numbers
in the bottom row of the
synthetic division of P divided by x-a alternate
signs, then a is a lower
bound for the zeros of
P.
Descartes Rule of Signs
Write the polynomial P in
standard form, that is with
the terms in order of descending
powers.
The number of positive
zeros of P counting multiplicity
is equal to the number
of changes of signs of
P or that number decreased
by an even integer.
The number of negative
zeros of P counting multiplicity
is equal to the number
of positive zeros of P(-x),
so it is the number of
changes of signs of P(-x)
or that number decreased
by an even integer.
Examples
Here are some examples
of how to use these theorems.
I made up these examples
by working backwards from
the solutions that I wanted,
but other than that I have
tried to be honest and show
the working of the examples
is close as possible to
the way I would think if
I were approaching the problems.
We all think differently,
so this in no way means
that you have to do them
the same way, but I hope
it will give you a useful
example of one person's
way of thinking. For a more
formal approach see pretty
much any College Algebra
text. College Algebra and
Trigonometry by Aufmann,
Barker, and Nation is the
text that we use is the
text that we use, but there
are plenty of other good
ones.
Example 1
Solve the equation.
x4+14x3+71x2+154x+120=0
The Rational Zero Theorem
tells me that if a solution
is p/q that q must be
a factor of 1, and p must
be a factor of 120. So
here are the possibilities.
Numerator: ±1,2,3,4,5,6,8,10,12,15,20,24,30,60,120
Denominator : ±1
Possible rational solutions
: ±1,2,3,4,5,6,8,10,12,15,20,24,30,60,120
(Actually in a problem
like this, I might not
list out all the possibilities
at the beginning, because
I would figure that particularly
if it was a problem from
a textbook or on a test
that I would probably
find enough solution before
getting up to the larger
numbers. I would check
the smaller numbers and
then list more numbers
only if I had to, but
I have listed them all
at the beginning, because
it may be neater and easier
to follow that way for
those who are new at it.)
There are no changes
of signs, so by Descartes
Rule of Signs, there are
no positive solutions.
P(-x) has 4 changes of
of signs, so Descartes
Rule of Signs doesn't
much restriction on the
number of negative solutions.
That leaves a lot of possibilities,
but hopefully we will
come up with enough solutions
before going through all
of them. Start with the
easiest, which is -1.
so we can see that it
doesn't work. It is nicest
to do these on a white
board, so if you can get
a hold of one you should
try it. If I was at the
board right now I could
just erase the -1 and
the bottom two rows, but
leave up the coefficients
and try another number.
-2 is the next easiest
number to try.
It works!
Having found one solution
we can reduce the degree
of our polynomial by one,
because from the division
we get that our polynomial
factors to (x+2)(x3+12x2+47x+60).
By the principle of zero factors the only way this
can be zero is if one
of the factors is zero.
So to find other solutions
besides -2, we merely
need to find solutions
to the second factor.
A convenient way I have
found to do that is to
cross out the 0 remainder
at the end and just use
the bottom row as the
coefficients of my new
polynomials. The first
thing to check is to see
if -2 is still a zero
for this polynomial.
No, it doesn't work, but
we can see that there
could still be a smaller
(larger negative) solution,
because the signs aren't
alternating on the bottom.
(Upper and Lower Bound
Theorem) But of course
there is no guarantee
that this will be one
of ours, because it could
be irrational. But no
harm in trying -3, so
again we can erase and
try it.
And we are in luck again,
because it works. This
gives us that our polynomial
factors into (x+2)(x+3)(x
2+9x+20),
so again by the principle
of zero factors we get
that either x+2=0, x+3=0,
or x
2+9x+20=0.
The first two we have
already taken care of.
For the third factor,
we could continue with
the synthetic division,
but we don't have to,
because this is a
quadratic, and
quadratics are much easier to solve. If we can't
figure out how to factor
it, we can use the
quadratic formula. Let's see if it factors.
Can we find two numbers
whose product is 20 and
whose sum is 9? Yes, we
can, 4 and 5. So it factors
into (x+4)(x+5) and the
whole polynomial factors
into (x+2)(x+3)(x+4)(x+5)
and the solutions to the
equation are x=-2,-3,-4,-5.
Example 2
Solve the equation.
2x4+3x3-16x2+15x-4=0
Numerator: ±1,2,4
Denominator: ±1,2
Possible rational solutions:
±1,2,4,1/2
If 1 is a possibility
I always try it first
because I am lazy, and
it is easiest. Since there
are changes of signs this
time there is no reason
to exclude it.
And this time we are in
luck and it works. We
might as well be really
lazy and try it again,
and as before we can just
continue on with the new
coefficients at the bottom,
because of the principle
of zero products.
And what do you know,
it worked again. So now
we are home free, because
we have reduced it down
to a
quadratic again. The original polynomial then
factors to (x-1)
2(2x
2+7x-4),
so to finish the solution
we just need to solve
2x
2+7x-4=0.
Let's see if this one
factors. To factor it
we need something of the
form
(2x
)(x
).
If we use 4 and 1 and
put the 4 so that it multiplies
by the 2 and the 1 so
that it multiplies by
1, then we get 8-1=7 for
the middle coefficient,
so that works, and we
get (2x-1)(x+4), so the
original polynomial factors
to (x-1)2(2x-1)(x+4),
giving us solutions of
1,1/2, and -4. But wait
a minute, there's something
funny here. There are
3 changes of signs in
the original polynomials,
so by Descartes Rule of
Signs there should be
3 or 1 positive solution
and there are 2. What's
wrong? To figure out what's
wrong with have to look
back at Descartes Rule
of Signs and see that
it says "counting multiplicity".
Since x-1 is a factor
twice, 1 is a zero of
multiplicity 2, so counting
muliplicity, there are
3 positive zeros.
Example 3
Solve the equation.
15x5+37x4+97x3+119x2-88x+12=0
Numerator: ±1,2,3,4,6,12
Denominator: ±1,3,5,15
Possible rational solutions:
±1,2,3,4,6,12,1/3,2/3,4/3,1/5,2/5,3/5,4/5,6/5,12/5,1/15,2/15,4/15
Try 1.
It is already clear from
this much that not only
is 1 not a solution but
that 1 is an upper bound
on the solutions, so it
is not necessary to check
2,3,4,6, or 12. Since
fractions are a pain,
it is probably best to
start with the negative
numbers. According to
Descartes rule of Signs
there are 2 or 0 positive
solutions, but even if
there are 2 there is no
guarantee that they will
be rational. Try -1.
It doesn't work. Try -2.
It works. Will it work
again?
No, but from here we can
tell that -2 is a lower
bound, since the signs
will alternate. We can
tell without doing the
addition that -47+-258
is going to be negative
and when you multiply
that negative by -2 it
will be positive and adding
that to 6 that will be
positive. So there is
no point in checking any
more negative whole numbers,
so we are faced with having
to check fractions. Probably
the easiest to try is
1/3, so let's erase the
second -2 and try it.
It works. Try it again.
It doesn't work, but it
is clear from here that
it is an upper bound since
even 90 times 1/3 is bigger
than 18, so the bottom
row numbers will all be
positive. So there is
no point in trying 2/3
or anything else bigger
than 1/3. Probably the
best thing to check now
is 1/5.
Eureka! It works, and
now we have it reduced
to a quadratic, so we
can solve it. The original
polynomial then factors
to (x+2)(x-1/3)(x-1/5)(15x
2+15x+90).
Factoring out 15, this
last factor becomes
x2+x+6.
This doesn't factor,
but we can use the quadratic formula to solve it.
b2-4ac=1-4(6)=-23
so
All together then the
solutions to the equations
are
.
Example 4
Solve the equation. x
6+
5x
5+ x
4
-25x
3 -24x
2+
30x+ 36=0
Numerator: ±1,2,3,4,6,9,12,18
Denominator: ±1
Possible rational solutions:
±1,2,3,4,6,9,12,18
By adding up the coefficients
we can see that 1 won't
work. There are two changes
of signs, so Descartes
Rule of Signs says that
there could be 2 positive
solutions, so let's try
2.
2 doesn't work and it
also doesn't show 2 as
an upper bound, so how
about 3.
It is clear from here
that it is not going to
work and furthermore that
it is an upper bound to
the solutions, since 3
times 50 will be much
bigger than 24 and from
then all everything will
be positive. So it looks
like there aren't going
to be any positive rational
solutions. According to
Descartes Rule of Signs
there are two possibilites
here. There could be indeed
be no positive solutions,
because the number of
solutions can differ be
a multiple of two from
the number of changes
of signs, but it also
could be that there are
2 irrational positive
solutions. In either case
we don't have any tools
to help us find them,
so it is time to concentrate
on the negative solutions.
Try -1.
Nope, it doesn't work.
Close. Maybe -2 will work.
Yes, we've got it. 1 down,
5 to go. :-( Oh well,
we must perservere. Will
it work again?
No such luck. How about
-3?
Yes!!! And look at the
interesting polynomial
we get when dividing it
out. Those zeros are going
to make it much easier
to find the remaining
solutions. Our original
polynomial factors into
(x+2)(x+3)(x
4-5x
2+6).
This 3rd polynomial is
sort of a
quadratic in disguise. It can be factored by
treating x
2
like it were the variable
and factoring the same
way you would factor x
2-5x+6,
find two numbers whose
product is 6 and whose
sum is -5. The two numbers
are -2 and -3, so it factors
to (x
2-2)(x
2-3).
Setting these equal to
0 we get
as solutions, so all together
the solutions to the original
equation are
