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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!