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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 x’s which satisfy the equation

These x’s 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 x’s) 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 – x_k) 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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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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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 won’t 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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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 a_n.

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 a’s 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 a_npⁿ 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 a_n.

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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:

We’ll 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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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".

We’ll do the example calculation and tabulate the results.

We should point out that this number is suspiciously close to the decimal expansion of a well-known irrational number:

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.

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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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Recommended Books

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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