Finding Zeros of Quartic Polynomials by Using Radicals

Home , Quartic equation, Resolvent cubic

June 24, 2018

Abstract is named a quartic polynomial when the degree of the polynomial, that is, the highest power of We present a technique for finding roots of a quar- the polynomial is four. Similarly, a cubic polyno- tic general polynomial equation of a single vari- mial has the highest degree three while a quadratic able by using radicals. The solution of quar- polynomial has the highest degree two. tic polynomial equations requires knowledge of Based on these elementary definitions, we state lower degree polynomial equations; therefore, we that our study will be on solving a general sin- study solving polynomial equations of degree less gle variable quartic polynomial equation by rad- than four as well. We present self-reciprocal poly- icals. By radicals, we mean expressions which nomials as a specialization and additionally solve are a combination of the sums, differences, quo- a numerical example. tients, products as well as the roots greater than or equal to two of the coefficients of the polynomial 1 Introduction [1]. Even though it is standard to learn solving quadratic polynomial equations by using radicals Solving polynomial equations have been a central in the contemporary curriculum, we don’t observe topic not only for its usage in pure mathematics a common interest in learning how to solve cubic but also in physical applications. Consequently, and quartic equations by such a method. In engi- solution of polynomial equations attracted inter- neering and applied sciences literature, we meet est peaking at the nineteenth century from mathe- other techniques for solving quartic polynomial maticians, which lead to development of modern equations; yet, we know that a general polyno- algebra. mial of degree four (or less) can be solved by radicals, which gives us an exact result with ease As we are mostly familiar with, a polynomial in of implementation, and also has the advantage a single variable is an expression involving pow- of evolving to maturity within the past centuries. ers of the variable multiplied by some coefficients. Therefore, an understanding of finding roots of a Thus, we obtain a polynomial equation when we general quartic polynomial equation by radicals equate a polynomial to zero. Solution of a polyno- would lead us an adequate means to solve prob- mial equation is essentially finding values of the lems in which these equations occur. variable such that the evaluation of the polynomial at that specific value of the variable gives us Thus, in Section 2, we discuss some back- zero. These specific values are called the zeros ground material including a theorem which proof the polynomial [1], or roots of the polynomial vides the number of zeros of a general polyno- equation. We define a general [2] (or generic [1]) mial. In Section 3, we discuss solution of the polynomial as the one whose coefficients are open quartic polynomial, starting by Tschinhaus trans- variables but not numerical values. A polynomial formation [1] and obtaining the resolvent poly-

1 nomial equation, which is necessary in the solu- algebra, which we present from [4] without proof. tion procedure. The resolvent polynomial neces- The proof can be found in texts such as [4]. sary to solve a general quartic polynomial equa- Theorem 1 (Fundamental Theorem of Algebra). tion is cubic; thus, we discuss solving cubic poly- The field of complex numbers is algebraically nomial equations in Section 3.1. In Section 3.2, closed; that is, every polynomial in C[x] has a we proceed solving the quartic polynomial and zero in C. give the formulas necessary for solving by radicals. As a case study, we will discuss solving Thus, for any zero xi ∈ C of a polynomial self-reciprocal quartic polynomials by radicals in p(x), we can factor as p(x) = (x − xi)q(x) with Section 4. Self-reciprocal polynomials occur in q(x) ∈ C[x]. The consequence of the fundamen- applications which require inversion in a circle. tal theorem of algebra on the number of roots of Such an application is available in [3]. We give a a polynomial equation of degree n is presented in numerical example in Sections 4.1. the following theorem Theorem 2. If p(x) ∈ F [x] be a polynomial of 2 Number of Roots degree n in the field F , which is a subfield of C, then p(x) has exactly n zeros in C. Before formulating the solution procedure to Proof. If we take a polynomial p(x) of degree quartic polynomial equations, we discuss the n, by the fundamental theorem of algebra, we number of roots of a polynomial equation. As we must have a linear factor x1 ∈ C so that p(x) = will see in Section 3, the transformations neces- (x − x1)q(x) with p(x1) = 0. But q(x), which sary for finding the zeros of a polynomial can lead is of degree n − 1 must also have a linear fac- us to the conclusion that a polynomial of degree n tor by the fundamental theorem of algebra. Thus, can have more than n zeros. In this section, we q(x) = (x − x2)r(x) with q(x2) = 0 and x2 show that this kind of conclusion is incorrect for not necessarily different from x1. The polynomial a polynomial whose coefficients belong to a sub- p(x) can be written as p(x) = (x−x1)(x−x2)r(x) field (F ) of the complex number field (C). where r(x) has degree n − 2. Continuing this pro- Before dealing with number of roots, we intro- cedure recursively leads us to completely split the duce and discuss the quartic equations. A general polynomial into linear factors x1, x2, . . . , xn such quartic polynomial equation in a single variable is that written in the form: p(x) = (x − x1)(x − x2) ··· (x − xn) 4 3 2 a4x + a3x + a2x + a1x + a0 = 0 (1) Thus, we conclude that a polynomial of degree n has n zeros in the algebraically closed field C. with ai ∈ F , i = 0,..., 4. Additionally, the field F contains the field of rational numbers (Q). The Using Theorem 2, we can conclude that the left side of the equation 1 can be represented as quartic polynomial in Eq. 1 has four zeros in C. a polynomial p(x), which belong to a polynomial Obviously, if the zeros we seek are located in a ring F [x]. Consequently, a polynomial ring con- subset instead of C, the number of zeros that we tains any polynomial of degree n with coefficients are interested in can be less than four. in F . The rest of the discussion is developed in the literature for any polynomial of degree n, thus valid for an arbitrary polynomial in F [x]. 3 General Quartic Polynomial We will let the zeros of a polynomial be an ele- of a Single Variable ment of C even though the polynomial itself is an element of the polynomial ring F [x]. This lets us Obtaining the number of roots of the general quar- to take advantage of the fundamental theorem of tic polynomial equation in F [x], we can proceed

2 with finding its roots. The procedure we offer reduce to solving quadratic polynomial equations is based on a method outlined in [4], which is a without further effort. Thus, we proceed by intro- variation of Ferrari’s solution. This method re- ducing a variable z and complete the left side to quires solving a resolvent cubic polynomial equa- square as tion. Thus, in Section 3.1, we discuss finding the 1 2 1 zeros of a cubic polynomial before we obtain the y2 + z = −qy2 − ry − s + zy2 + z2 (6) roots of the general quartic equation in Section 2 4 3.2. We would like to complete the right side of Eq. 6 We start by considering the Eq. 1 of Section to a square, i.e. to a form (my + k)2, as well. For 2. The polynomial equation being quartic makes this reason, we collect the terms of the right hand it necessary that a4 6= 0, otherwise we would end side of Eq. 6 in the variable y to obtain up an equation of degree not equal to four. Using 1 this fact, we can manipulate Eq. 1 to make the (z − q)y2 − ry + z2 − s (7) coefficient of the fourth power term one, that is, 4 we can transform into a monic quartic polynomial Thus, we let √ equation. Thus, dividing both sides of the equa- m = z − q (8) tion 1, we obtain: Then, to complete Eq. 7 to a square in the form (my + k)2, we must have 2mk = −r. Therefore, x4 + b x3 + b x2 + b x + b = 0 (2) 3 2 1 0 we find b = a /a i = 0,..., 3 b ∈ F r r with i i 4, , so i . We use k = − = − √ (9) Eq. 2 for the rest of the discussion without losing 2m 2 z − q generality. Thus, the equation 7 becomes We proceed by applying Tschirnhaus transformation; thus, we let y = x + b3/4 so that x = r2 1 (my + k)2 − + z2 − s (10) y − b3/4. Substituting for x in Eq. 2, we obtain 4(z − q) 4 the depressed [4] quartic polynomial equation as When Eq. 10 is a complete square, we must y4 + qy2 + ry + s = 0 (3) have r2 1 − + z2 − s = 0 (11) where 4(z − q) 4 3 Obviously, if Eq. 11 holds, then we have the q = − b2 + b (4) 2 8 3 2 square (my + k) . Therefore, we need to solve 1 1 Eq. 11 to proceed finding the zeros of the quartic r = b3 − b b + b 8 3 2 3 2 1 polynomial. Since z is an open variable, we can 3 1 1 assume z 6= q. Then, we can multiply both sides s = − b4 + b2b − b b + b 256 3 16 3 2 4 3 1 0 of Eq. 11 by 4(z − q) and expand the numerator to obtain We now consider writing the polynomial equation in terms of sums of squares. To achieve this, z3 − qz2 − 4sz + (4qs − r2) = 0 (12) we first arrange the depressed quartic polynomial equation as Hence, solving Eq. 12 is a necessary step for obtaining the roots of the quartic polynomial equa- y4 = −qy2 − ry − s (5) tion by radicals. Consequently, we call the cubic polynomial in Eq. 12 the resolvent of the general We will consider the case that the right side of Eq. quartic polynomial equation. We discuss solution 5 is not a square in which case the solution would of the resolvent cubic polynomial in section 3.1.

3 3.1 Solving the Resolvent Polynomial with j = 0, 1, 2. If we let √ In this section, we solve the resolvent cubic poly- 2π 1 3 ω = cis = − + i (18) nomial, which is Eq. 12. We can make the 3 2 2 Tschinhaus transformation for the cubic polynomial; thus, we let t = z − q/3 so that y = t + q/3. then √ 4π 1 3 Substituting into Eq. 12, we obtain the depressed ω2 = cis = − − i (19) resolvent cubic polynomial equation as 3 2 2 and ω3 = 1. Consequently, for the powers of ω t3 + gt + h = 0 (13) has a cyclic order of three. Based on this notation, where we tabulate the cube roots of x as

1 2 √ g = − q − 4s (14) x = 3 r cisθ (20) 3 1 √ 2 8 x = 3 r ω cisθ h = − q3 + qs − r2 2 √ 3 2 27 3 x3 = r ω cisθ We make one more transformation [4] for t = u − g/(3u) with u being another open variable, Thus, we found a way to compute cube roots of then substitute into Eq. 13. Collecting the terms a number a0 ∈ C by solving equations in the form 3 of the resulting equation under a common denom- x − a0 = 0. inator and considering that the numerator of the We proceed solving Eq. 15. Although we seem equation is zero, we obtain to have six solutions for Eq. 15, in turn for the resolvent cubic equation, we know this is not the g3 u6 + hu3 − = 0 (15) case by Theorem 2. Thus, we need to make clean- 27 ing for obtaining the correct roots. We begin by Thus, we obtain a quadratic equation in u3. Com- presenting a condition on the roots of the resol- pleting Eq. 15 to a square and solving for u3, we vent cubic equation, which we will use to deter- obtain mine the correct roots. r 2 3 3 h h g Lemma 1. If u3 and u3 be two roots of Eq. 15, u1 = − − + (16) 1 2 2 4 27 then u u = −g/3. r 1 2 h h2 g3 u3 = − + + 3 3 2 2 4 27 Proof. Suppose u1 and u2 are roots of Eq. 15. Then, we can write (u − u3)(u − u3) = 0. Hence, We will take an interlude and discuss ﬁnding 1 2 we expand to u6 − (u3 + u3)u3 + u3u3 = 0, the roots of polynomial equations in the form 1 2 1 2 thus, comparing to Eq. 15, we must have u3u3 = x3 − a = 0 in [x]. We know that any complex 1 2 0 C −g3/27. Taking cube roots of both sides, we ob- number a can be written in the form a = r(cos θ+ tain u u = −g/3. i sin θ). As a consequence of De Moivre’s theo- 1 2 3 3 rem [5], we can write Taking the cube roots of u1 and u2 in Eq. 16, 3 2 n n we obtain three roots of u1 as u1, ωu1, and ω u1. a = r (cos (nθ) + i sin (nθ)) 3 Similarly, three cube roots of u2 are u2, ωu2, and 2 Thus, if we write a0 in the form [3] rcis(θ) = ω u2. We need to select from each of the three r(cos θ + i sin θ) n = 1/3 3 3 , and let , we obtain the cube roots of u1 and u2 to obtain the roots of three cube roots x as the depressed cubic polynomial equation. For this √ θ + 2jπ purpose, we present Theorem 3. x = 3 r cis (17) 3 2 2 √ Theorem 3. Let u1, ωu1, ω u1 and u2, ωu2, ω u2 3 2jπ θ = r cis cis be roots of Eq. 15. We compute the zeros of the 3 3

4 depressed resolvent cubic polynomial in Eq. 13 by 2. If ωu1u2 = −g/3, then the back transforma- choosing one from each set of the roots according tion equation becomes to the following cases ωu u t = u + 1 2 1. If we choose roots sucht that u1u2 = −g/3, u 2 then t1 = u1 + u2, t2 = ωu1 + ω u2, and 2 Thus, for u = u1, we obtain t3 = ω u1 + ωu2.

t1 = u1 + ωu2 2. If we choose roots such that ωu1u2 = −g/3, then t1 = u1 + ωu2, t2 = ωu1 + u2, and Similarly, z = ωu would give us 2 2 1 t3 = ω u1 + ω u2.

2 t2 = ωu1 + u2 3. If we choose roots such that ω u1u2 = −g/3, then t = u + ω2u , t = ωu + ωu , and 2 1 1 2 2 1 2 and u = ω u1 leads to t = ω2u + u . 3 1 2 ωu u t = ω2u + 1 2 = ω2u + ω2u Proof. We ﬁnd the roots of the depressed resol- 3 1 2 1 2 ω u1 vent cubic polynomial by transforming back from u to t by means of the equation Thus, t1, t2, t3 gives the solution set obtained g in case 2 of the theorem. t = u − 2 3u 3. If ω u1u2 = −g/3, then the back transfor- We can arbitrarily take the cube roots of u1 to mation equation becomes back substitute for u. We can verify that if we 2 ω u1u2 choose cube roots of u2 instead, we would obtain t = u + the same results. u 2 As a consequence of Lemma 1, we need to Thus, for u = u1, we obtain t1 = u1 + ω u2. 2 make sure compatible roots of Eq. 15 are chosen. Similarly, u = ωu1 and u = ω u1 would give u u = ω3u u = −g/3 2 The possibilities are 1 2 1 2 , us t2 = ωu1 + ωu2 and t3 = ω u1 + u2, re- 4 2 ωu1u2 = ω u1u2 = −g/3, and ω u1u2 − −g/3. spectively. Hence, we obtained the solution Therefore, we consider the following cases set presented in case 3 of the theorem. 1. If u1u2 = −g/3, then the back transforma- Thus, for arbitrary selection of two of the cube tion equation becomes 2 2 roots from u1, ωu1, ω u1 and u2, ωu2, ω u2, we u u t = u + 1 2 can calculate the roots of Eq. 13. u Thus, once we obtain t , i = 1, 2, 3 using the- Thus, for u = u , we obtain i 1 orem 3, We can back substitute for z = t + q/3

t1 = u1 + u2 to obtain the three zeros zi of the resolvent cubic polynomial in Eq. 12. Similarly, u = ωu1 would imply

u1u2 2 t2 = ωu1 + = ωu1 + ω u2 3.2 Solving the Quartic Polynomial ωu1 and u = ω2u leads to In this section, we present the formulation for 1 ﬁnding the zeros of the general quartic monic u u 2 1 2 2 polynomial in Eq. 2. t3 = ω u1 + 2 = ω u1 + ωu2 ω u1 We already solved the resolvent cubic polyno- Hence, we obtain the solution set in case 1 of mial equation to obtain the roots zi which would the theorem. complete the right hand side of Eq. 6 to a square

5 in the form (my + k)2 with m and k expressed in 4 Case Study: Self-Reciprocal Eq. 8 and Eq. 9, respectively. Consequently, we can write the general quartic polynomial equation Quartic Polynomials in the form We formulated a solution algorithm to obtain 2 1 2 2 (y + zi) = (miy + ki) (21) roots of the general quartic polynomial equation 2 in Section 3. The solution included solving a cu- with zi being a root of the resolvent cubic poly- bic resolvent of the quartic polynomial. Although, nomial equation and mi, ki being the values eval- solving a cubic polynomial equation is necessary uated for the root zi. Since we have three roots for the general quartic case, the method would be from the resolvent cubic polynomial, it would be simpler if we consider specialization of the gen- a natural question to choose which one for zi. We eral quartic polynomial. One such specialization discussed in Section 2 that a quartic polynomial is a self-reciprocal quartic polynomial, which we has exactly four zeros in C. Hence, we choose study in this section. Under a suitable transfor- one of the three roots of the resolvent polynomial mation, solution of self-reciprocal quartic poly- equation arbitrarily [2] while neglecting the oth- nomial equations reduce to solving two quadratic ers. polynomials. Selecting zi, we can take the square roots of both sides of Eq. 21. Thus, we obtain the fol- Definition 1 (Self-Reciprocal Quartic Polyno- lowing two quadratic equations mial). Let F be a subfield of the field of the real 1 numbers ( ). A quartic polynomial f ∈ F [x] y2 − m y − k + z = 0 (22) R i i 2 i given by the equation 1 is self-reciprocal under 1 the condition that f(x) = f †(1/x) with f † ∈ F [x] y2 + m y + k + z = 0 i i 2 i defined as:

† 4 1 4 3 2 The zeros of the ﬁrst quadratic polynomial give us f (x) = x f( ) = c0x + c1x + c2x + c3x + c4 two of the roots of the depressed quartic polyno- x mial equation Thus, c0 = c4 and c1 = c3. r m m2 z y = i − i + k − i (23) Thus, we can write a general quartic self- 1 2 4 i 2 reciprocal polynomial equation as: and r 2 mi m zi 4 3 2 y = + i + k − (24) c0x + c1x + c2x + c1x + c0 = 0 (27) 2 2 4 i 2 while that of the second quadratic polynomial give Without loss of generality, we can divide both the other two roots of the depressed quartic poly- sides by c0 to obtain nomial equation r 4 3 2 m m2 z x + d1x + d2x + d1x + 1 = 0 (28) y = − i − i − k − i (25) 3 2 4 i 2 with d1 = c1/c0 and d2 = c2/c0. The roots of the r 2 self-reciprocal polynomial equation are a conse- mi mi zi y4 = − + − ki − (26) quence of inversion with respect to the unit circle 2 4 2 Once we obtain the roots of the depressed quartic [3]; thus, the roots are never equal to zero. Using equation, we can transform to obtain the roots of this knowledge, we can divide both sides of Eq. 28 by 1/x2 to obtain Eq. 2 by using xi = yi − b3/4. This completes the formulation to obtain the roots of a general quartic 1 1 x2 + d x + d + d + = 0 (29) polynomial equation. 1 2 1 x x2

6 Let us deﬁne v = x + 1/x, which is a transforma- 4.1 Example tion discussed in [6] . Then, collecting the terms In this section, we solve a self-reciprocal quartic with the same coefﬁcient in Eq. 29, we obtain polynomial which is available in [3]. The polynomial is below 2 1 x + d1v + d2 + = 0 (30) 213 165857 213 x2 x4 − x3 + x2 − + 1 = 0 (35) 50 25600 50 We can take squares of the transformation for v We will solve this equation by using the method and rearrange to obtain discussed in this section, then we will solve it by considering it as a general quartic polynomial. We 1 x2 + = v2 − 2 will be writing the radical expressions in decimal x2 form to save space when necessary.

Substituting above into Eq. 30, we write Solution 1

2 v + d1v + d0 = 0 (31) Comparing the example with Eq. 28, we ﬁnd d1 = −213/50 and d2 = 165857/25600. Trans- with d0 = d2 −2. The roots of the quadratic equa- forming this equation by the method discussed tion are in this section, we obtain d0 = 114657/25600. Thus, r 2 2 213 114657 d1 d1 v − v + = 0 v1 = − − − d0 (32) 50 25600 2 4 Hence, we obtain v1 = 2.37106 and v2 = 1.88894 r 2 d1 d1 by using Eq. 32. Then, we ﬁnd the zeros of v2 = − + − d0 2 4 the self-reciprocal polynomial from Eq. 34 as x1 = 0.548755, x2 = 1.82231, x3 = 0.944469 − We can transform back vi = x+1/x with i = 1, 2 i0.328601, and x4 = 0.944469 + i0.328601. to obtain the quadratic equation

2 Solution 2 x − vix + 1 = 0 (33) We use the formulation for the general quartic Solving this quadratic equation gives us the roots polynomial in Section 3 to obtain the same solu- of the quartic self-inversive polynomial as tion. We obtain the depressed quartic polynomial r 2 equation by using the transformation x = y + v1 v1 x1 = − − 1 (34) 213/200. Using Eq. 4, we compute q = 2 4 −208999/640000, r = −7921683/64000000, r 2 v1 v1 and s = −1226099903/25600000000. Thus, we x2 = + − 1 2 4 write r 2 v2 v2 4 208999 2 7921683 1226099903 x3 = − − 1 y − y − y− = 0 2 4 640000 64000000 25600000000 r v v2 We proceed with obtaining the resolvent cubic x = 2 + 2 − 1 4 2 4 polynomial by using Eq. 12 as 208999 1226099903 Thus, we can obtain the roots of a self-reciprocal z3 + z2 + z 640000 6400000000 quartic equation without having to solve a resol- 12093787004663 + = 0 vent cubic polynomial equation. 256000000000000

7 Making another transformation, which would To continue solution of the quartic polynomial be z = t − 208999/1920000, we com- equation, we arbitrarily pick z1 and calculate m = pute g = 306768959/1966080000 and h = 0.636776 + i0.328601 and k = 0.0767513 − 6560810064991/226492416000000 by using Eq. i0.0396066 by using Eq. 8 and Eq. 9, respec- 15. Thus, we obtain the depressed resolvent cubic tively. We then use Eq. 23 through Eq. 26 to ﬁnd as the zeros of the depressed quartic polynomial, and x 306768959 6560810064991 back substitute for i t3 + t + = 0 1966080000 226492416000000 213 xi = yi + i = 1,..., 4 Then, we use Eq. 16 to obtain 200 to obtain 86779 r1873891 u3 = − 1 1638400000 15 x = 0.944469 + i0.328601 6560810064991 1 − 452984832000000 x2 = 1.82231 −8 = −0.0332042 x3 = 0.548755 + i1.49012 × 10

x4 = 0.944469 − i0.328601 and

86779 r1873891 Neglecting the small imaginary number in x3 as u3 = a consequence of the Sagemath computational er- 2 1638400000 15 6560810064991 rors, the results xi are in agreement with that the − solution method 1 and also with the result avail- 452984832000000 = 0.00423715 able in [3].

We will present the rest of the solution in dec- 3 imal form to save space. We note that u1 = 5 Conclusion 3 0.0332042 cisπ and u2 = 0.00423715 cis0. Thus, 3 3 we can calculate the cubic roots of u1 and u2 by As we have studied, solving general quartic poly- using Eq. 20. If we pick u1 = 0.160707 + nomial equations by radicals is straightforward by 0.278352i and u2 = 0.161817, then we obtain means of a few transformations, so the technique does not require much computational effort. This −9 ωu1u2 = −0.0520103 − i3.72529 × 10 would allow us to do the computations with rel- ative ease instead of trying a more sophisticated which is equivalent to −g/3 = −0.0520103 if we algorithm. discard Sagemath numerical errors. Thus, we can As we studied in Section 4.1, the formulation t , t , t use the case 2 of Theorem 3 to obtain 1 2 3. for solving the general quartic polynomial equa- Substituting the values back by using tions can be used for specialized polynomials such 208999 as self-reciprocal polynomials. However, a self- z = t − i = 1, 2, 3 i i 1920000 reciprocal quartic polynomial can be transformed into a simpler form which does not require solu- Thus, we obtain the roots of the resolvent cubic tion of a cubic polynomial equation, thus make polynomial equation as the solution even easier to implement. We illus- trate the ease of solution of a self-reciprocal quar- z = −0.0290555 + i0.418490 1 tic polynomial equation by means of a suitable −8 z2 = −0.268450 − i2.98023 × 10 transformation x + 1/x in solution 1 of the ex- z3 = −0.0290554 − i0.418490 ample in section 4.

8 References

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