Download A SIMPLE TRICK TO HELP YOUR FACTOR A SPECIAL TYPE OF

A SIMPLE TRICK TO HELP YOUR FACTOR A SPECIAL TYPE OF POLYNOMIAL
NICOLAS SIMARD
As you know, when you want to diagonalize a matrix, you rst need to nd its eigenvalues. If the matrix
is 2 × 2, you need to nd the roots of a quadratic polynomial, which is very easy. However, it becomes harder
when your matrix is 3 × 3, because you need to nd the roots of a polynomial of the form
c(x) = x3 + ax2 + bx + c
For simplicity, suppose that a, b and c are integers (this will be the case if your matrix has integers
coecients, for instance). As for quadratic polynomials, there is a general formula to nd the roots of a cubic
polynomial. However, this formula is complicated and useless; even if one could remember it by heart, it
would express the roots in terms of radicals and complex numbers, making it almost impossible to do hand
computations.
The basic idea is that of we could nd just one root of c(x), say n, then c(x) would factor as c(x) =
(x − n)(x2 + a0 x + b0 ) for some numbers a0 and b0 . If we expand the product, we nd
c(x) = x3 + ax2 + bx + c = (x − n)(x2 + a0 x + b0 ) = x3 + (a0 − n)x2 + (b0 − a0 n)x − nb0
matching the coecients of x2 and the constant terms, we see that b0 = −c/n and a0 = a + n if n 6= 0 and
so c(x) = (x − n)(x2 + (a + n)x − c/n). Note that if n = 0, then c(x) = x(x2 + ax + b). In both cases, to nd
the other roots of c(x), we simply need to nd the roots of a quadratic polynomial, which is easy.
In brief, if we could just nd one root of c(x), we could easily nd all the roots. This is where the trick is
useful... Here it is:
Let c(x) = x + ax + bx + c be a polynomial with integer coecients. If n is an integer and a
root of c(x), then c is a multiple of n. In other words, if c(n) = 0 for some integer n, then c/n is an integer.
Theorem 1.
3
2
To nd a root of c(x) without a computer, you would probably nd it by inspection: you try many small
values and see if you nd a root. This theorem tells you that you only need to try the values that divide the
constant term c of your polynomial. This can save you a lot of time! Let's look at a few examples:
Example 1: Let's factor the polynomial
c(x) = x3 + 2x2 − 25x − 50
We see that −50 is a multiple of ±1, ±2, ±5, ..., so we need to try these possibilities. We have c(1) = −72 6= 2,
c(−1) = −24 6= 0, c(2) = −84 6= 0, but c(−2) = 0. Using the formula above (or long division, if you know
how it works), we see that x3 + 2x2 − 25x − 50 = (x + 2)(x2 − 25) = (x + 2)(x − 5)(x + 5) and we are done.
Example 2:
Let's factor the polynomial
c(x) = x3 − 3x + 2
Here, 2 is a multiple of ±1 and ±2. We compute c(1) = 0, so x3 − 3x + 2 = (x − 1)(x2 + x − 2) and the roots
of x2 + x − 2 are simply 1 and −2.
1
A SIMPLE TRICK TO HELP YOUR FACTOR A SPECIAL TYPE OF POLYNOMIAL
Example 3:
2
Let's factor the polynomial
c(x) = x3 − 7x2 − x + 7
Here, 7 is a multiple of ±1 and ±7. We compute c(1) = 0, so x3 − 7x2 − x + 7 = (x − 1)(x2 − 6x − 7) and
the roots of x2 − 6x − 7 are simply 7 and −1.
Example 4:
Let's factor the polynomial
c(x) = x3 − 5x5 + 2x − 10
Here, −10 is a multiple of ±1, ±2, ±5 and ±10. We compute c(1) = −12, c(−1) = −18 , c(2) = −18,
c(−2) = −42 and c(5) = 0. It follows that x3 − 5x5 + 2x − 10 = (x − 5)(x2 + 1). Here x2 + 1 does not have
a real root.
Example 5:
Let's factor the polynomial
c(x) = x3 − x + 1
Here, 1 is a multiple of ±1. We compute c(1) = c(−1) = 1. It follows that c(x) does not have any integral
roots (and even rational roots...). In this case, it has one real root and two complex roots. Don't worry, this
shouldn't happen in your exam!