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,
x^3 + 5x^2 + 8,

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,
x^3 + 5x^2 + 8 = 0,
6x^25 + 3x^5 + 25 = 0,
x^2 - 3x + 2 = 0,
and
x - 1 = 0

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.

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^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
x - 1 = 0,
or x + 2 = 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.

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:

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,
which is equivalent to
(4y - 1)(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
3x^2 - 5 = 0.  Its solution is x = ±sqrt(5/3)

Example 3. Find the factors of

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
(x^2 - 2)(x^2 + 2) = 0.

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.
The last term (x^2 + 2) cannot be factored further.

Note
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

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