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Terminology and Notation · Factoring Large Polynomials · Fundamental Theorem of Algebra · Rational Zeros Theorem · Example · Irreducible Expressions · Numerical Methods · Summary · Recommended Books
Terminology and Notation
First, we present some notation and definitions. A general polynomial has the form
This function is really a mathematical expression rather than an equation since the f(x) to the left of the equals sign is just a label or abbreviation for the long expression to the right of the first equals sign. The large symbol to the right of the second equals sign is called the sigma notation, and reads, "sum the product of the kth a and the kth power of x from k=1 up to k=n". This notation comes in handy when we are adding up a large number of terms that look alike.
We are really interested in the xs which satisfy the equation
These xs are called zeros of f(x) or roots of the equation f(x) = 0. The distinction between these terms is small (albeit precise) and the terms are often used interchangeably. Suppose we find the n numbers
(read this last expression as "the set of all complex x which make f(x) = 0"; the first two expressions are two different ways of listing the individual xs) that are all the possible roots of the equation. Then, we can express the polynomial in a much simpler form:
The pi notation is similar to the sigma notation described above, except that it describes a product of like terms. There are several advantages of knowing all the roots of an equation. First, we know exactly where the function becomes zero. Second, we can examine the factors (x xk) and find repeated roots, complex roots, irrational roots, etc. In short, the inner workings of the function are more exposed with this notation.
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Factoring Large Polynomials
Large polynomials (larger than quadratics, equations involving powers of x larger than x2 ) get harder to factor the bigger they get. While there are advanced techniques to directly calculate the roots of a cubic (x3) and (in some cases) a quartic (x4), these methods are quite complicated and require an advanced sophistication in algebra to be comprehensible. The reader is welcome to take a look at both of these cases to verify our opinion. We will concentrate on some theorems that offer factoring help on a less advanced basis.
To be sure, use of these theorems amounts to educated guessing, but such guessing is actually more likely to get an answer faster than the advanced solution techniques. At the least, the techniques we offer will show whether an elementary answer (like an integer or a rational number) can be expected. Failing that, we will explore a scheme for finding an answer numerically (a refinement of trial and error) using a calculator or computer. If the numerical technique is done carefully, we can sometimes use the decimal expansion calculated to guess a familiar irrational number. If all else fails, we can resort to the "big guns" and use one of the advanced techniques.
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Fundamental Theorem of Algebra
The nth degree polynomial
has exactly n roots. The roots may be repeated (i.e., not all distinct), complex (i.e., not real) or irrational, but need not be any of these (i.e., they might be integers or rational numbers). We wont bother to prove this theorem, since the proof is very involved and really does not contribute much to our problem solving techniques.
Put very simply, an nth degree polynomial has n roots.
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Rational Zeros Theorem
Suppose the coefficients
in the polynomial equation
are all integers. If
is a rational fraction in lowest terms (i.e., p and q are both integers and have no common divisors other than 1) which satisfies the equation (i.e. is a root of f(x) = 0), then p divides a0 (i.e., a0 / p is an integer) and q divides an.
Since p/q is a root of the equation, we have
Multiplying through by qn produces
Subtracting a0qn from both sides gives
Since each term on the left contains at least one p, we can factor it out:
The term in parentheses on the left is the sum of many products of integers and so is an integer. Call this integer I and we have
We already knew the as are integers, so a0qn must be an integer, too. The equation is true by assumption, so p must divide a0qn . Since p and q have no common divisors other than 1, the same must be true of p and qn, which leaves p dividing a0.
We could have subtracted anpn from the equation after multiplying through by qn, giving us
Notice that q is a common term for the left side,
From here the proof is similar, and is left as an exercise for the reader. The result is that q is proved to divide an.
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Use the rational zeros theorem to guess the possible rational roots of
Then use synthetic division to find which of the possible roots is actually a root.
According to the theorem, we are looking for numerators that divide 2 (1 and 2) and denominators that divide 3 (1 and 3). Thus, the possible roots are:
Well skip trials of each root and jump directly to the correct answer:
We should point out that even if the seven wrong answers had to be tried first, synthetic division is fast enough to go through all the potential roots in a matter of minutes.
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In the last example, the quotient polynomial is
The primitive polynomial equation
has no real solutions and is considered irreducible. It is the polynomial analogue of a prime number. If we allow complex number solutions, then the above equation has the solutions
We will normally stop factoring a polynomial when we encounter an irreducible quotient, since exceptions are normally reserved for more advanced subject matter than that we cover here.
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For readers who can use a programmable calculator, the following method can be used to get a decimal expansion of a root. If one know the decimal expansions of a few irrational numbers, guesses can be made based on the computed decimal expansion and checked. This method is a bit more cumbersome than the above guessing scheme for rational roots and should be tried only if one fails to find a rational root. The reader is advised that algebra and arithmetic errors are extremely common when learning to handle polynomials, and should be eliminated first before trying a numeric solution. Practical problems often yield messy answers, so numerical solutions are more attractive when doing mathematics for the sake of mathematics is really beside the point i.e., when solving scientific, engineering or financial problems.
We will briefly outline a numerical method for solving the cubic
First, we need to rewrite the equation so that we have a single x on one side of the equal sign and a function of x on the other side that is weaker than x (i.e., has a smaller power of x).
This last equation
is the one we will use. The basic idea is to guess an x, put it in the right hand side, calculate the function. This generates a new x on the left, which is then in turn put back into the right hand side, until the difference between the input x and the output x is "small enough".
Well do the example calculation and tabulate the results.
N = iteration number
Change in x
We could at this point presume that this is really a root, try synthetic division, and factor the polynomialThis method can be used with many, but not all, polynomials. Numerical methods can work very well, like the example above. In a wide variety of cases, this method is very frustrating because the right-hand side of the equation bounces around and does not converge nicely. Numerical analysis is a big topic, and gets technical very fast.
See the other examples on numerical methods for further tips.back to top Summary
To solve a general polynomial:
The reader is reminded that the study of general polynomials is a very complicated field. We have provided guidelines that work in a fairly large number of cases the mathematics student is likely to see, but will prove inadequate for an even larger class of problems. There are more advanced, specialized methods appropriate for different fields of study, in particular science and engineering. These advanced methods are beyond the current scope for our purposes, so we will content ourselves for now with what we have presented above.
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College Algebra (Schaum's Outlines)
The classic algebra problem book - very light on theory, plenty of problems with full solutions, more problems with answers
Schaum's Easy Outline: College Algebra
A simplified and updated version of the classic Schaum's Outline. Not as complete as the previous book, but enough for most students
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