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Ruffini's Rule 1 Ruffini's Rule

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Ruffini's Rule 1 Ruffini's Rule Ruffini's rule 1 Ruffini's rule In mathematics, Ruffini's rule is an efficient technique for dividing a polynomial by a binomial of the form x − r. It was described by Paolo Ruffini in 1809. Ruffini's rule is a special case of synthetic division when the divisor is a linear factor. Algorithm The rule establishes a method for dividing the polynomial by the binomial to obtain the quotient polynomial and a remainder s. The algorithm is in fact the long division of P(x) by Q(x). To divide P(x) by Q(x): 1. Take the coefficients of P(x) and write them down in order. Then write r at the bottom left edge, just over the line: | a a ... a a n n-1 1 0 | r | ----|--------------------------------------------------------- | | 2. Pass the leftmost coefficient (a ) to the bottom, just under the line: n | a a ... a a n n-1 1 0 | r | ----|--------------------------------------------------------- | a n | | = b n-1 | 3. Multiply the rightmost number under the line by r and write it over the line and one position to the right: | a a ... a a n n-1 1 0 | r | b r n-1 ----|--------------------------------------------------------- | a n | | = b n-1 | Ruffini's rule 2 4. Add the two values just placed in the same column | a a ... a a n n-1 1 0 | r | b r n-1 ----|--------------------------------------------------------- | a a +(b r) n n-1 n-1 | | = b = b n-1 n-2 | 5. Repeat steps 3 and 4 until no numbers remain | a a ... a a n n-1 1 0 | r | b r ... b r b r n-1 1 0 ----|--------------------------------------------------------- | a a +(b r) ... a +b r a +b r n n-1 n-1 1 1 0 0 | | = b = b ... = b = s n-1 n-2 0 | The b values are the coefficients of the result (R(x)) polynomial, the degree of which is one less than that of P(x). The final value obtained, s, is the remainder. As shown in the polynomial remainder theorem, this remainder is equal to P(r), the value of the polynomial at r. Uses of the rule Ruffini's rule has many practical applications; most of them rely on simple division (as demonstrated below) or the common extensions given still further below. Polynomial division by x − r A worked example of polynomial division, as described above. Let: We want to divide P(x) by Q(x) using Ruffini's rule. The main problem is that Q(x) is not a binomial of the form x − r, but rather x + r. We must rewrite Q(x) in this way: Now we apply the algorithm: 1. Write down the coefficients and r. Note that, as P(x) didn't contain a coefficient for x, we've written 0: | 2 3 0 -4 | -1 | ----|---------------------------- | | Ruffini's rule 3 2. Pass the first coefficient down: | 2 3 0 -4 | -1 | ----|---------------------------- | 2 | 3. Multiply the last obtained value by r: | 2 3 0 -4 | -1 | -2 ----|---------------------------- | 2 | 4. Add the values: | 2 3 0 -4 | -1 | -2 ----|---------------------------- | 2 1 | 5. Repeat steps 3 and 4 until we've finished: | 2 3 0 -4 | -1 | -2 -1 1 ----|---------------------------- | 2 1 -1 -3 |{result coefficients}{remainder} So, if original number = divisor×quotient+remainder, then , where and Polynomial root-finding − The rational root theorem tells us that for a polynomial f(x) = a xn + a xn 1 + ... + a x + a all of whose n n−1 1 0 coefficients (a through a ) are integers, the real rational roots are always of the form p/q, where p is an integer n 0 divisor of a and q is an integer divisor of a . Thus if our polynomial is 0 n then the possible rational roots are all the integer divisors of a (−2): 0 (This example is simple because the polynomial is monic (i.e. a = 1); for non-monic polynomials the set of possible n roots will include some fractions, but only a finite number of them since a and a only have a finite number of n 0 integer divisors each.) In any case, for monic polynomials, every rational root is an integer, and so every integer root Ruffini's rule 4 is just a divisor of the constant term. It can be shown that this remains true for non-monic polynomials, i.e. to find the integer roots of any polynomials with integer coefficients, it suffices to check the divisors of the constant term. So, setting r equal to each of these possible roots in turn, we will test-divide the polynomial by (x − r). If the resulting quotient has no remainder, we have found a root. You can choose one of the following three methods: they will all yield the same results, with the exception that only through the second method and the third method (when applying Ruffini's rule to obtain a factorization) can you discover that a given root is repeated. (Neither method will discover irrational or complex roots.) Method 1 We try to divide P(x) by the binomial (x − each possible root). If the remainder is 0, the selected number is a root (and vice versa): | +1 +2 -1 -2 | +1 +2 -1 -2 | | +1 | +1 +3 +2 -1 | -1 -1 +2 ----|---------------------------- ----|--------------------------- | +1 +3 +2 0 | +1 +1 -2 0 | +1 +2 -1 -2 | +1 +2 -1 -2 | | +2 | +2 +8 +14 -2 | -2 0 +2 ----|---------------------------- ----|--------------------------- | +1 +4 +7 +12 | +1 0 -1 0 Method 2 We start just as in Method 1 until we find a valid root. Then, instead of restarting the process with the other possible roots, we continue testing the possible roots against the result of the Ruffini on the valid root we've just found until we only have a coefficient remaining (remember that roots can be repeated: if you get stuck, try each valid root twice): | +1 +2 -1 -2 | +1 +2 -1 -2 | | -1 | -1 -1 +2 -1 | -1 -1 +2 ----|--------------------------- ----|--------------------------- | +1 +1 -2 | 0 | +1 +1 -2 | 0 | | +2 | +2 +6 +1 | +1 +2 ------------------------- ------------------------- | +1 +3 |+4 | +1 +2 | 0 | -2 | -2 ------------------- | +1 | 0 Ruffini's rule 5 Method 3 • Determine the set of the possible integer or rational roots of the polynomial according to the rational root theorem. • For each possible root r, instead of performing the division P(x)/(x -r), apply the polynomial remainder theorem, which states that the remainder of this division is P(r), i.e. the polynomial evaluated for x = r. Thus, for each r in our set, r is actually a root of the polynomial if and only if P(r) = 0 This shows that finding integer and rational roots of a polynomial neither requires any division nor the application of Ruffini's rule. However, once a valid root has been found, call it r : you can apply Ruffini's rule to determine 1 Q(x) = P(x)/(x-r ). 1 This allows you to partially factorize the polynomial as P(x) = (x -r )·Q(X) 1 Any additional (rational) root of the polynomial is also a root of Q(x) and, of course, is still to be found among the possible roots determined earlier which have not yet been checked (any value already determined not to be a root of P(x) is not a root of Q(x) either; more formally, P(r)≠0 → Q(r)≠0 ). Thus, you can proceed evaluating Q(r) instead of P(r), and (as long as you can find another root, r ) dividing Q(r) by 2 (x-r ). 2 Even if you're only searching for roots, this allows you to evaluate polynomials of successively smaller degree, as the factorization proceeds. If, as is often the case, you're also factorizing a polynomial of degree n, then: •• if you've found p=n rational solutions you end up with a complete factorization (see below) into p=n linear factors; • if you've found p<n rational solutions you end up with a partial factorization (see below) into p linear factors and another non-linear factor of degree n-p, which, in turn, may have irrational or complex roots. Examples: Finding roots without applying Ruffini's Rule P(x) = x³ +2x² -x -2 Possible roots = {1, -1, 2, -2} • P(1) = 0 → x = 1 1 • P(-1) = 0 → x = -1 2 • P(2) = 12 → 2 is not a root of the polynomial and the remainder of (x³ +2x² -x -2)/(x-2) is 12 • P(-2) = 0 → x = -2 3 Ruffini's rule 6 Finding roots applying Ruffini's Rule and obtaining a (complete) factorization P(x) = x³ +2x² -x -2 Possible roots = {1, -1, 2, -2} • P(1) = 0 → x = 1 1 Then, applying Ruffini's Rule: (x³ +2x² -x -2) / (x -1) = (x² +3x +2) → → x³ +2x² -x -2 = (x-1)(x² +3x +2) Here, r =-1 and Q(x) = x² +3x +2 1 • Q(-1) = 0 → x = -1 2 Again, applying Ruffini's Rule: (x² +3x +2) / (x +1) = (x +2) → → x³ +2x² -x -2 = (x-1)(x² +3x +2) = (x-1)(x+1)(x+2) As it was possible to completely factorize the polynomial, it's clear that the last root is -2 (the previous procedure would have given the same result, with a final quotient of 1). Polynomial factoring Having used the "p/q" result above (or, to be fair, any other means) to find all the real rational roots of a particular polynomial, it is but a trivial step further to partially factor that polynomial using those roots.
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