Negative numbers are a great concept, but that’s what they are, a concept.  Negative quantities do not exist in our physical experience.  I can point to two apples or two dollars, but I can’t point to minus two apples or minus two dollars.

Math can be a source of wonder, but too often we’re in too much of a rush and it’s like that with negative numbers. We present them to students as if they’re obvious and difficulty with them in the sign of a shallow mind, when actually the opposite is true.  If they don’t seem weird, you’re probably not really thinking about them.

Here’s what some great mathematicians thought about them:

“Pascal regarded the subtraction of 4 from 0 as utter nonsense.” (Kline, Morris, Mathematical Thought form Ancient to Modern Times, Oxford University Press, New York, 1972, p 252)

Antoine Arnauld, a theologian and mathematician, invoked ratios and “questioned that  -1:1 = 1:-1 because, he said, -1 is less than +1; hence,  How could a smaller be to a greater as a greater is to a smaller?” (Kline, Morris, Mathematical Thought form Ancient to Modern Times, Oxford University Press, New York, 1972, p 252)

The person generally credited with the idea of the number line, John Wallis, “rejected as absurd the … idea of a negative number as being less than nothing, but accepted the view that it is something greater than infinity — a viewpoint shared by Swiss mathematician Leonhard Euler…” (http://en.wikipedia.org/wiki/John_Wallis)

“As recently as the 18th century, it was common practice to ignore any negative results returned by equations on the assumption that they were meaningless, just as René Descartes did with negative solutions…” (http://en.wikipedia.org/wiki/Number)

“In 1759, Francis Maseres, an English mathematician, wrote that negative numbers ‘darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple’. Because of their dark and mysterious nature, Maseres concluded that negative numbers did not exist…” (http://www.bbc.co.uk/radio4/history/inourtime/inourtime_20060309.shtml)

“In the 15th century, Nicolas Chuquet, a Frenchman, used negative numbers as exponents and referred to them as ‘absurd numbers’” http://en.wikipedia.org/wiki/Negative_and_non-negative_numbers

So, if there’s something in you that thinks negative numbers are strange, you’re in great company!

Common wisdom says you can’t use a negative base in a logarithm if you’re working with real numbers

i.e. Y=log_-b(X) is not valid, where –b is a negative number

However, consider this

For every integer a, there is some negative number –p, such that

1) (-p)^(2a+1) = -b : because 2a+1 is an odd number and a negative number to an odd power is a negative number.

P^{(2a + 1)} = b : (the relationship between positive p and positive b)

P = b^1/(2a+1) (spreadsheet line 14 solves for –p)

There is some z, such that

P^z = X  (z solved for on spreadsheet line 16)

If  c = int(z/2) (c is on line 18 of the spreadsheet) then

X = p^(z-2c) * p^2c

Because the exponent of p^2c is even, we can rewrite it as (-p)^2c

2) Let W = (-p)^2c , (spreadsheet line 22) therefore

X = p^(z-2c) * W

we know that (-p)^2c is positive because the exponent 2c is an even integer

Taking log_-b of expression 1) yields

(2a+1) log_-b(-p) = log_-b(-b) : since the log of a number to its base is 1, we have

3) Log_-b(-p) = 1/(2a+1)

Taking log_-b of expression 2 yields

Log_-b(W) = log_-b[(-p)^2c] = 2c * log_-b(-p)

Substituting 1/(2a+1) for log_-b(-p) from equation 3 yields

Log_-b(W) = 2c/(2a+1)

As a approaches infinity, p approaches one, p² approaches 1, W approaches X and log_-b(W) approaches log_-b(X), so

Log_-b(X) = 1/(2a+1) * 2c

4) Log_-b(X) = 1/(2a+1) * [2 * int( log_b(X) * (2a+1)/2)]

You can see this formula implemented step by step in the spreadsheet starting on line 27

The answer can be checked (starting on spreadsheet line 37) by the following process

Log_-2(X) = 1/(2a+1) * 2c is

X =((-b) ^{1/(2a + 1)}) ^2c

First, take the (2a + 1) root of –b which is the same as minus the (2a+1) root of b.

Then take that number to the 2c power.

It’s based on the realization that if one of your polynomial factors can be reduced, then so can your polynomial.  Here’s why:

Assume your factors are (akx + bk) (cx + d) (the first factor can be reduced by k)

If you multiply them, you get: akcx² + akdx + bkcx + bkd, which can be reduced by the same k to

k*(acx² + adk + bcx + bd)

This is a powerful realization because it means that once you’ve reduced your polynomial, you can automatically eliminate any possible factors if either of them can be reduced.

Consider 6x² – 7x – 20

Let (ax +b) and (cx + d) be the two factors

Because the x² coefficient is 6, we know that a * c = 6.  There are four possible ways for this to occur:

1*6

2*3

3*2

6*1

Because the magnitude of the purely numeric value is 20, we know that b * d = 20 (ignoring the sign for now).  There are six possible ways for this to occur:

1*20

2*10

4*5

5*4

10*2

20*1

Each combination of the first can go with any combination of the second, which leaves us with 4 * 6 possibilities.  However, the last half are identical to the first half, with just the order changed, so we’re left with these 12 possibilities, listed here without commas instead of signs between the x and the numeric terms:

(x, 1) * (6x, 20)

(x, 2) * (6x, 10)

(x, 4) * (6x, 5)

(x, 5) * (6x, 4)

(x, 10) * (6x, 2)

(x, 20) * (6x, 1)

(2x, 1) * (3x, 20)

(2x, 2) * (3x, 10)

(2x, 4) * (3x, 5)

(2x, 5) * (3x, 4)

(2x, 10) * (3x, 2)

(2x, 20) * (3x, 1)

Instead of being forced to multiply all 12 factor pairs to see which gives us the result we want, we can IMMEDIATELY eliminate EIGHT out of TWELVE because one of the factors in them can be reduced:  They are

(x, 1) * (6x, 20) right factor can be reduced by 2

(x, 2) * (6x, 10) right factor can be reduced by 2

(x, 5) * (6x, 4) right factor can be reduced by 2

(x, 10) * (6x, 2) right factor can be reduced by 2

(2x, 2) * (3x, 10) left factor can be reduced by 2

(2x, 4) * (3x, 5) left factor can be reduced by 2

(2x, 10) * (3x, 2) left factor can be reduced 2

(2x, 20) * (3x, 1) left factor can be reduced 2

This leaves us with

(x, 4) * (6x, 5)

(x, 20) * (6x, 1)

(2x, 1) * (3x, 20)

(2x, 5) * (3x, 4)

We know each of these combinations will give us the correct coefficients for our x² and numeric terms, so let’s determine what ‘x’ coefficient each will provide.  This is the sum or difference, depending on the sign, of the products of the outer and inner terms.

Because our numeric result is negative (-20) we know the signs of the numeric terms in the factors must be opposites, so our ‘x’ coordinate will be the difference of the inner and outer products.  Calculating these is very easy, considering we only have FOUR of TWELVE left.

The only combination that gives us 7 as the x coefficient is the last one so all that’s left to do is determine which term has the negative sign and which has the positive.

(2x + 5) * (3x – 4)  = 6x² -8x + 15x – 20 = 6x² + 7x – 20

(2x – 5) * (3x + 4) =  6x² +8x – 15x – 20 = 6x² – 7x – 20

The second pair contains the correct sign for our ‘x’ coefficient so our factors are:

(2x – 5) * (3x + 4)

To summarize: There is no simple way to factor a polynomial if its x² term is not 1.  If your first few guesses don’t work and you need a systematic approach, you can frequently eliminate more than half the possibilities by first reducing the polynomial so its terms have no common factors and then immediately eliminating all possible factors where either of them can be reduced.

Pretty interesting.

Do you have any other tips for factoring polynomials?