Reduced Polynomial \\& Classical Cubic Formula

Reduced Polynomial & Classical Cubic Formula Johar M. Ashfaque A polynomial f (x) of degree n is called reduced if it has no xn−1 term i.e. f (x) = rn xn + rn−2 xn−2 + rn−3 xn−3 + .... The reduced polynomial arising from the cubic X 3 + aX 2 + bX + c takes the form h(x) = x3 + qx + r. The formula for the roots of h(x) will give the formula for the roots of the cubic. Note. The following formula is due to Scipio del Ferro and Tartaglia which first appeared in the book by Cardano in 1545. Let α be a root of h(x) and choose numbers  and γ with α =  + γ. Then α3 = ( + γ)3 = 3 + γ 3 + 3αγ. Thus 3 + γ 3 + (3γ + q)α + r = 0. So far, only one constraint has been imposed namely α =  + γ. We can impose a second constraint γ = −

q 3

so that 3 + γ 3 = −r and 3 γ 3 = −

q3 . 27

Solving these two equations for 3 and γ 3 , we have that 6 + r3 −

q3 = 0. 27

The quadratic formula then gives 3

−r +

q

 =

3

q r2 + 4 27

2

and

q . 3 Having found one root of h(x), the other two roots can easily be found as the roots of the quadratic γ=−

h(x) . (x − α)

1

Employing the cube roots of unity ω=e

2πi 3

then there are three values of . Hence the other two roots have values ω and ω 2 . The corresponding values for γ are ω 2 γ and ωγ respectively. To conclude, the roots of the cubic polynomial are given by the cubic formula  + γ, ω + ω 2 γ, ω 2  + ωγ. Simple Example. If f (x) = x3 − 15x − 126 then f (x) is reduced. In this case, q = −15, r = −126, r2 + 4 and

q3 = 15376 27

r r2 + 4

q3 = 124. 27

Therefore, 3 = 125 giving  = 5 with γ = 1. As a result, one root of the cubic is α =  + γ = 6. Employing the roots of unity, the other two roots can be seen to be 5ω + ω 2 , 5ω 2 + ω.

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