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Sutra: International Journal of Mathematics Science Education

Technomathematics Research Foundation Vol. 4, No. 2, pp. 27 – 29, 2011

   

Solving Cubics by Division Method

Raghavendra G. Kulkarni Bangalore, India

Abstract

We present a division method for factoring and solving cubics. The method derives formulas in a logical sequence rather than in an intuitive way (as given in the Viesta's substitution method).

1 Introduction

It is well-known that Cardano (1539) first published his solution to the gen- eral cubic equation in his book "The great art, or, the rules of algebra" using the concept of complex numbers, which was unknown at that time. However, history pages reveal that Scipione del Ferro (1515) had solved the cubic in a mathematical contest, but didn't publish his method; and it seems Tartaglia (1535) revealed his solution to Cardano, who got it published in his name [1, 2].

In this paper, we describe a division method for factoring and subsequently solving the cubic equation, which relies on derivation of formulas in a logical fashion instead of intuitive or empirical way, the Viesta's substitution proposes [2].

2 The division method

Consider the following reduced cubic polynomial:

p(x) = x3 + al x + aO (1)

for factoring and solving by the proposed division method, where the coefficients, aO and al , are real. Let us consider the quadratic polynomial, q(x), as given below:

q(x) = x2 + (m + n)x + k (2)

where m, n, and k, are unknowns to be determined. Dividing cubic (1) by the quadratic (2), we obtain the quotient as:

L(x) = x - (m + n) (3)

and the remainder as:

R(x) = [al - k + (m + n)2 ]x + [aO + k(m + n)] (4)

If the quadratic (2) has to be a factor of cubic (1), then the remainder, R(x),has to be zero for all values of x, which means each of the coefficients of x andx0 in R(x) has to be zero, as shown below.

a1 − k + (m+ n)2 = 0 (5)

a0 + k(m+ n) = 0 (6)

Using (6) we eliminate k from the expression (5), resulting in a cubic equationin (m+ n) as shown below.

(m+ n)3 + a1(m+ n) + a0 = 0 (7)

Expanding the above expression and rearranging, we obtain:

m3 + n3 + (m+ n)(a1 + 3mn) + a0 = 0 (8)

Since we have only one equation [(8)] in two unknowns (m and n), we requireone more equation to determine the unknowns. Therefore we set n as:

n =−a13m

(9)

and use it in (8), which converts (8) into a quadratic equation in m3 as shownbelow.

m6 + a0m3 − a31

27= 0 (10)

Solving (10), we obtain two values of m3 as:

m3 =

(−a02

)± 1

2

√a20 +

4a3127

(11)

Taking the cube-root of (11), we determine m as:

m =

[(−a02

)± 1

2

√a20 +

4a3127

]1/3

(12)

Since m is already determined, the remaining unknowns [in q(x)], n and k, aredetermined from (9) and (6) respectively. Thus all unknowns in the quadraticpolynomial, q(x), are determined satisfying the condition that the remainder,R(x) = 0, for all values of x. Hence the cubic polynomial, p(x), can be expressedas product of its linear [quotient L(x)] and quadratic [divisor q(x)] factors asshown below.

p(x) = L(x)q(x) (13)

Equating each of the factors in the above cubic to zero and solving, we obtainall the three roots of given cubic (1) as shown below.

x1 = m− a13m

(14)

2

x2 =1

2

{(m− a1

3m

)[−1 +

√1 +

4a0(m− a1

3m

)3]}

(15)

x3 =1

2

{(m− a1

3m

)[−1−

√1 +

4a0(m− a1

3m

)3]}

(16)

Let us solve one numerical example using the proposed method. Consider thecubic equation: x3 − 2x+ 4 = 0; first, one value of m3 is determined from (12)as: −0.0755, and taking its cube-root m is determined as: −0.42265. From (17)one root of cubic is found out as: −2. From (18) and (19), the other two rootsare determined as: 1− 1i, and 1 + 1i, where i =

√−1.

3 Conclusions

A division method for solving cubic equation is described, wherein the formulasare derived in a logical fashion to obtain the three roots of given cubic.

Acknowledgements The author thanks the management of Bharat Elec-tronics Ltd., Bangalore for supporting this work.

References

[1] G. Birkhoff and S. MacLane, ”A survey of modern algebra”, 5th edition,A. K. Peters, 1997.

[2] Eric W. Weisstein, ”Cubic Equation”, From MathWorld-A Wolfram WebResource, http://mathworld.wolfram.com/CubicEquation.html.

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