Interactive Algebra Review

Based on the algebra reviews in Applied Calculus and Finite Mathematics and Applied Calculus

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5. Solving Polynomial Equations
Part B: Polynomial Equations

Note To review the concept of an "equation" and "solving an equation," go back to Part A: Equations.

Q What is a polyomial equation?

Polynomial Equation

A polynomial equation is an equation that can be written in the form

    axn + bxn-1 + . . . + rx + s = 0,

where a, b, . . . , r and s are constants.

We call the largest exponent of x appearing in a non-zero term of a polynomial the degree of that polynomial.

Examples
1. 3x + 1 = 0 has degree 1, since the largest power of x that occurs is x = x1. Degree 1 equations are called linear equations.

2. x2 - x - 1 = 0 has degree 2, since the largest power of x that occurs is x2. Degree 2 equations are also called quadratic equations, or just quadratics.

3. x3 = 2x2 + 1 is a degree 3 polynomial (or cubic) in disguise. It can be rewritten as x3 - 2x2 - 1 = 0, which is in the standard form for a degree 3 equation.

4. x4 - x = 0 has degree 4. It is called a quartic.

Q How do we solve these equations for x?
A This question was asked by mathematicians as early as 1600 BC. Let's look at these equations one degree at a time.

Solution of Linear Equations

By definition, a linear equation can be written in the form

Solving this is a nice mental exercise: subtract b from both sides and then divide by a, getting x = -b/a. Don't bother memorizing this formula, just go ahead and solve linear equations as they arise.

Mentally solve the following equations for x. (That is, try to solve them by writing down as little as possible.)

Q1 2x + 3 = 0   has the unique solution     x =  
Q2 3x - 3 = 1   has the unique solution    x =  
Q3 ax + b = c (a 0)   has the unique solution    x =  

Solution of Quadratic Equations

By definition, a quadratic equation has the form

The solutions of this equation are also called the roots of ax2 + bx + c. We''re assuming that you saw quadratic equations somewhere in high school but may be a little hazy as to the details of their solution. There are two ways of solving these equations -- one works sometimes, and the other works every time.

Solving Quadratic Equations by Factoring (works sometimes)

If we can factor a quadratic equation ax2 + bx + c = 0, we can solve the equation by setting each factor equal to 0.

Examples

    1.x2 + 7x + 10 = 0
    (x + 5)(x + 2) = 0     Factor the left-hand side
    x + 5 = 0 or x + 2 = 0     If a product is zero, one or both factors is zero
    Solutions: x = -5 and x =--2
    2.2x2 - 5x - 12 = 0
    (2x + 3)(x - 4) = 0Factor the left-hand side
    Solutions: x = -3/2 and x = 4

Test for Factoring
The quadratic ax2 + bx + c, with a, b, and c being integers (whole numbers), factors into an expression of the form (rx + s)(tx + u) with r, s, t and u being integers precisely when the quantity b2- 4ac is a perfect square (that is, it is the square of an integer). If this happens, we say that the quadratic factors over the integers.

Examples

    x2 + x + 1 has a = 1, b = 1, and c = 1, so b2 - 4ac = -3, which is not a perfect square. Therefore, this quadratic does not factor over the integers.

    2x2- 5x -12 has a = 2, b = -5 and c = -12, so b2 - 4ac = 121. Since 121 = 112, this quadratic does factor over the integers (we factored it above).

Here are some for you.

Solving Quadratic Equations with the Quadratic Formula (works every time)

The solutions of the general quadratic equation ax2 + bx + c = 0 (a 0) are given by

     
    x =
    - b
    b2 - 4ac

    2a
We call the quantity = b2 - 4ac the discriminant of the quadratic ( is the Greek letter delta) and we have the following general principle:
  • If is positive, there are two distinct real solutions.
  • If is zero, there is only one real solution: x = -b/2a (why?).
  • If is negative, there are no real solutions.

Examples

    1. x2 - 5x - 12 = 0 has a = 2, b = -5, and c = -12.
       
      x =
      - b
      b2 - 4ac

      2a
      		  
      =
      5
      25 + 96

      4
      		  
      =
      5
      121 

      4
      		  
      =
      5 11

      4
       
      =
      16

      4
      		   or -
      6

      4
      		   =   4 or -
      3

      2
      		 .

     
    2. 4x2 = 12x - 9 can be rewritten as 4x2 -12x + 9 = 0, which has a = 4, b = -12, and c = 9.

       
      x =
      - b
      b2 - 4ac

      2a      		  
      =
      12
      144 - 144

      8
      		  
      =
      12 0

      8
      		 
      =
       
      12

      8
      		  
      =  
       
      3

      2
      		 .

     
    3. x2 + 2x - 1 = 0 has a = 1, b = 2, and c = -1.

       
      x =
      - b
      b2 - 4ac

      2a      		  
      =
      -2
      8

      2
      		  
      =
      -2 2
      2

      2
      		  
      = -1

      2
    The two solutions are -1 +
    2    		   0.414   and   -1 -
    2    		   -2.414

     
    4. x2 + x + 1 = 0 has a = 1, b = 1, and c = 1. Since = -3 is negative, there are no real solutions.

Here are some for you. When there are two solutions, use a comma to separate them. You can also use "sqrt" for square root. Press here for some examples of how to enter formulas.

Q1        x2 - 3x + 4
has solution(s)
has no solutions.		  
 
Q2        x2 + 3x + 1 = 0
has solution(s)
has no solutions.		  
 
Q3        4x2 - 12x + 9 = 0
has solution(s)
has no solutions.		  
 
Q4        x2 + x - 1 = 0
has solution(s)
has no solutions.		  
 

Solution of Cubic Equations

By definition, a cubic equation can be written in the form

Now we get into something of a bind. While there is a perfectly respectable formula for the solutions, it is very complicated and involves the use of complex numbers rather heavily1. So we discuss instead a much simpler method that sometimes works nicely. Here is the method in a nutshell.

Solving Cubics by Finding One Factor

Start with a given cubic equation ax3 + bx2 + cx + d = 0

Step 1
By trial-and-error, find one solution x = s. If a, b, c, and d are integers, the only possible rational solutions  are those of the form s = (factor of d)/(factor of a).

Step 2
It will now be possible to factor the cubic as

    ax3 + bx2 + cx + d = (x - s)(ax2 + ex + f) = 0.

To find ax2+ex+f, divide the cubic by x-s using long division. 

Step 3
The factored equation says that either x - s = 0 or ax2 + ex + f = 0. We already know that s is a solution, and now we see that the other solutions are the roots of the quadratic. Note that this quadratic may or may not have any real solutions, as usual.

Example
To solve the cubic x3 - x2 + x -1 = 0, we first find a single solution. Here, a = 1 and d = -1. Since the only factors of 1 are 1, the only possible rational solutions are x = 1. By substitution, we see that x = 1 is a solution. Thus (x- 1) is a factor. Dividing by (x - 1) yields the quotient (x2 + 1). Thus,

    x3 - x2 + x - 1 = (x - 1)(x2 + 1) = 0,

so that either x - 1 = 0 or x2 + 1 = 0. Since the discriminant of the quadratic x2 + 1 is negative, we don't get any real solutions from x2 + 1 = 0, so the only real solution is x = 1.

Possible Outcomes When Solving a Cubic Equation
If you consider all the cases, there are three possible outcomes when solving a cubic equation:

    1. Three real solutions (see the next example)
    2. One real solution (as in the example above)
    3. Two real solutions (try, for example, x3 + x2 - x - 1 = 0)

 

In the following, try to find one factor of the given cubic.

By factoring the left-hand side find all possible solutions for x in the equations in the next group of exercises. If there are three solutions, enter each solution in a different box. If there is only one solution, enter it in the first box, and leave the other boxes blank.

Solution of Higher-Order Polynomial Equations

Logically speaking, our next step should be a discussion of quartics, then quintics, and so on forever. Well, we've got to stop somewhere, and cubics may be as good a place as any. On the other hand, since we've gotten so far, we ought to at least tell you what is known about higher order polynomials.

Quartics Just as in the case of cubics, there is a formula to find the solutions of quartic.  

Quintics and Beyond All good things must come to an end, we're afraid. It turns out that there is no "quintic formula." In other words, there is no single algebraic formula or collection of algebraic formulas that will give the solutions to all quintics. This question was settled by the Norwegian mathematician Niels Henrik Abel in 1824 after almost 300 years of controversy about this question. (In fact, several notable mathematicians had previously claimed to have devised formulas for solving the quintic, but these were all shot down by other mathematicians-this being one of the favorite pastimes of practitioners of our art.) The same negative answer applies to polynomial equations of degree 6 and higher. It's not that these equations don't have solutions, just that they can't be found using algebraic formulas.   However, there are certain special classes of polynomial equations that can be solved with algebraic methods. The way of identifying such equations was discovered around 1829 by the French mathematician ƒvariste Galois.  

Now try some of the exercises in Section A.5 of the Algebra Review in Applied Calculus and Finite Mathematics and Applied Calculus.

Last Updated: August, 2001
Copyright © 2001 Stefan Waner and Steven R. Costenoble

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