POLYNOMIAL EQUATIONS
Objectives: To recall how to work with equations involving sums of terms with x raised to a power greater than 2.
Recall 1: A polynomial is an expression that can be put into the form
an x^n + an-1 x^(n-1) + . . . + a0 ,
where an ,
an-1 etc. are fixed
rational
numbers,
an is different from 0 and n is
a positive integer.
For example,
6x^25 + 3x^5 + 25,
x,
x^2 - 3x +
2,
and
x - 1
are all polynomials.
However,
x^(-3) + 5(1/x) +
8,
and
1/(x^2 + 2x - 1)
are not.
The highest power n is called the degree
of
the polynomial.
The terms
ak
in front of the powers of x
are called the coefficients of
the polynomial.
Recall 2: A polynomial
equation is an equation that can
be put in the
form
anx^n + an-1x^(n-1)
+ . . . + a0 =
0,
where the left side is a
polynomial.
For example,
are all polynomial equations.
The degree of the polynomial is called the degree
of
the polynomial equation.
Note
A quadratic equation
is also a polynomial
equation of degree 2. So, anything that can
be said in this hour
about polynomial equations of degree two will also
be true for quadratic
equations. Anything said about quadratic equations
will be
true for polynomial equations of degree 2.
x^3 + x^2 - 2
factors into the product x(x - 1)(x + 2).
Therefore,
the polynomial equation
x^3 + x^2 - 2x
= 0
may be represented as
x(x - 1)(x + 2) = 0.
If we use the
fact that a product of terms is 0
if and only if one of the
terms is 0, then we may conclude
that
x = 0
This means that the
solutions to the original polynomial
equation are x = 0,
x = 1 and x = -2.
Note Allowing for complex
number
solutions, a polynomial equation of degree n has exactly
n
solutions, though some of those solutions may be coincide.
For example,
the polynomial equation
x^3 - 4x^2 +
4x = 0
has three solutions, but two of them
coincide--they are
x = 0, x = 2 and x = 2.
1. x^3 - 5x^2 + 6x =
0
is equivalent to
x(x - 2)(x - 3) = 0.
We
read off the solution from the factors, x = 0, x =
2 and x =
3.
2. x^4 - 17x^2 + 4 =
0
is equivalent to
(x^2)^2- 17x^2 + 4 = 0.
If we let y = x^2, then our equation is also
equivalent
to
y^2 - 17*y + 4 =
0,
Now replace x^2 for y in the above equation to get
(4x^2 - 1)(x^2 - 4) = 0.
The terms on the
left side of the equation factor; we
get:
(2x -
1)(2x + 1)(x - 2)(x + 2) = 0.
From this we have
our solution: x = 1/2 ,
x =-1/2 , x = 2
and x = -2.
Examples
Example 1. Find the
factors of the polynomial equation
x^4
+ x^3 + 2x + 2 = 0
Solution: In this case,
a simple
regrouping of the terms leads to a way of factoring. By
factoring an
x^3 from the first two terms and a 2 from the last two terms
we may
rewrite the equation as
x^3(x + 1) + 2(x + 1) =
0.
Now factor the (x + 1) from both terms to
get
(x + 1)(x^3 + 2) = 0.
This method of factoring is quite valuable since these
kind of
polynomials occur quite regularly.
Example 2. Here is another example of a polynomial
equation
that can be factored by looking at how to group terms.
Factor:
6x^3
+ 3x^2 - 10x - 5 = 0
Solution: 3x^2 is a common factor
in the
first two terms of the equation and -5 is a common factor in the
last
two. We therefore may factor 3x^2 from the first two terms and
-5
from the last. This gives:
(2x + 1)(3x^2 -
5) = 0.
Note You may be asking yourself
at this
minute, "Why do we consider an equation like (2x + 1)(3x^2 - 5)
= 0 fully
factored? After all, the second term is a quadratic equation,
not a
degree 1 equation." The answer is that if we were forced
to
factor (3x^2 - 5), it would factor as (sqrt(3)x - sqrt(5))(sqrt(3)x
+
sqrt(5)). But sqrt(3) and sqrt(5) are not rational numbers and
our
polynomials are defined in a way that requires the coefficients to
be
rational. Moreover, in this case it is not necessary to factor
any
further, because factoring a polynomial equation is often used only
to
solve the equation; but, it is easy to solve the equation
x^4
- 4 = 0.
Solution: Write
x^4 as (x^2)^2.
Then the equation becomes
(x^2)^2-
4 = 0,
which is the difference between two
squares. Such
a difference can be factored as
The first term is also a
difference of two squares and,
therefore, our equation may be factored
once again. We get
(x - sqrt(2))(x +
sqrt(2))(x^2 + 2) =
0.
Note
x^4 + 2x^3 + 3x^2 + 2x +
1.
This polynomial factors as (x^2 + x + 1)(x^2 +
x + 1).
But x^2 + x + 1 does not factor any further.
On the other hand, we are guaranteed that a polynomial
of any
degree, no matter how large, will factor into polynomials of degree
1 or
2. That is not to say that finding such factors is easy.
It is
sometimes impossible.
EXERCISES
Solve the
following polynomial equations for the
appropriate
variable:
1. x^2 - x - 2 =
0
2. x^4 + 2x^3 + x^2 =
0
3 x^3 - x^2 - x + 1 =
0
4 3x^3 + 5x^2 + 2x =
0
5. 16x^4 - 1 =
0
6. 3x^3 + 5x^2 + 2x =
0
7. x^2 + 3x - 5 = x +
3
Hint:
Subtract (x + 3) from both sides.
8. 3x^2 - 1 = x^2 +
x
Hint:
Collect all terms to the left side.
9. x^2 +
x - 1 = 0
10.. x^4 - 2x^3 + 2x^2 - 2x + 1 =
0
11. x^4 - 13x^2 + 36 =
0
12. 4x^4 - 25x^2 + 36 =
0
x^3 + 5x^2 +
8,
x^3 + 5x^2 + 8 = 0,
6x^25 + 3x^5 + 25 = 0,
x^2 - 3x + 2 = 0,
and
x - 1 = 0
Recall
3: If a polynomial factors into a
product of known polynomials
of degree 1 or 2, then the solutions of the
corresponding polynomial
equation are easy to find. For
example, the
polynomial
x - 1 = 0,
or x + 2 = 0.
Recall
4: You would not be expected to solve
for the unknown in an
arbitrary polynomial equation of 3 or higher because,
in general, that may
be difficult. However, there are certain types
of polynomial
equations of higher degree that factor easily. Here
are two
examples:
which is equivalent to
(4y - 1)(y - 4) = 0.
3x^2 - 5 = 0. Its solution is x =
±sqrt(5/3)
Example 3. Find the factors of
(x^2
- 2)(x^2 + 2) = 0.
The last term (x^2 + 2) cannot be factored
further.
Any
polynomial may be factored
into polynomials of degree 1 or
2. This means that there
may be some polynomials that
will not factor into polynomials of degree
less than 2. Such is the
case with