Resolving a degree 3 polynomial equation
Yohann D'ANELLO & Katia MICHIELSEN
assisted by François SAUVAGEOT
September 8, 2016
Considering a degree 3 polynomial equation of the form:
0 (E).
x3 has
x3 −ax2 +bx−c =
got voluntary no coecient, and if he had one we could divide the
entire expression in order to reduce to this expression.
Let
α, β
and
γ
be the roots of the polynomial (E) (∈
3
the equivalent equation:
βγ + γα)x − αβγ .
C). We can write
(x − α)(x − β)(x − γ) = x − (α + β + γ)x2 + (αβ +
We deduce that

 a
b

and
c
= α + β + γ,
= αβ + βγ + γα
= αβγ
Thus we can easily gure out these three sums and products getting on for
the ambiguity of the variable.
√
Let us clarify the concept of ambiguity.
What means
2
?
dene it by the positive real number such as his square be 2.
restrict ourselves to positive real numbers?
We could
But why
We could work with complex
numbers and for example say
√ that the solution of a degree 2 polynomial
−b+ b2 −4ac
2
ax + bx + c = 0 is x =
! Like this, the ambiguity of the square
2a
root gives us the two roots of the polynomial. The ± disappears, replaced
by the implicit ambiguity of the square root (that can be negative).
will simplify our calculations a lot. We are going to use the notation
to give the two solutions get by the ambiguity.
x
This
a|b = x
being ambiguous, a and b
respectively take the two possible numerical values. So a and b are dierent,
but obtained the same way.
Let
j
be an imaginary number such as
j 3 = 1.
In the complex plane, j
2π
would correspond to a (trigonometric) rotation of
. We can construct this
3
circle:
1
j=
√
3
j
1
1
j2 =
1 + j + j 2 = 0.
2iπ
number: j = e 3 .
So,
√
3
2
1
We can also express
j
with an exponential complex
Let's yet construct a geometric gure representing
α, β
and
γ.
We con-
struct a triangle. With some algebra, we can move the triangle's center of
2π
rotation would not allow a
gravity to the origin. In this conguration, a
3
superimposition of the triangle. We will have to nd another trick!
α
β
γ
Remember that we are on the complex plan and in the case of three real
roots, we will have a attened triangle on the real axis.
The idea we had is to rotate each point around the center of gravity,
j and j 2 . Thus we obtain three new triangles that are similar (they
applying
are superimposing after a rotation around the center of gravity). This allows
us to keep the ambiguity with
α, β
and
2
γ
!
γj 2
u
βj
α
β
uj 2
αj 2
αj
uj
βj 2
γ
In particular, we have
u = α + βj + γj 2
(the triangle at the top left of
2
this page). And we can also dene a number v = α + βj + γj . u and v
2
are of course dierent, but have the same properties (v , vj and vj can be
superimposed with a rotation).
From this point on we can nd a method to determine u and v.
Let's
examine their cubes!
3
3
u =u
(uj)3 = u3 · j 3 = u3 · 1 = u3
3
2
(uj 2 )3 = u3 · j 2 = u3 · j 3 = u3 · 12 = u3
We see that the order of the letters is unimportant as soon as we cube it.
We could do the same for
v.
3
γj
u3 [resp. v 3 ] = (α + βj + γj 2 )3
= α3 + β 3 + γ 3
+ 3j[resp. j 2 ](α2 β + β 2 γ + γ 2 α)
+ 3j 2 [resp. j](αβ 2 + βγ 2 + γα2 )
+ 6αβγ
Let's now concentrate on the sum and product of these cubes.
u3 + v 3 = α3 + β 3 + γ 3 + 3j(α2 β + β 2 γ + γ 2 α) + 3j 2 (αβ 2 + βγ 2 + γα2 ) + 6αβγ
+ α3 + β 3 + γ 3 + 3j 2 (α2 β + β 2 γ + γ 2 α) + 3j(αβ 2 + βγ 2 + γα2 ) + 6αβγ
= 2(α3 + β 3 + γ 3 ) + 3j(α2 β + β 2 γ + γ 2 α + αβ 2 + βγ 2 + γα2 )
+ 3j 2 (α2 β + β 2 γ + γ 2 α + αβ 2 + βγ 2 + γα2 ) + 12αβγ
= 2(α3 + β 3 + γ 3 ) + (α2 β + β 2 γ + γ 2 α + αβ 2 + βγ 2 + γα2 )(3j + 3j 2 ) + 12αβγ
= 2(α3 + β 3 + γ 3 ) + 3(α2 β + β 2 γ + γ 2 α + αβ 2 + βγ 2 + γα2 )(j + j 2 ) + 12αβγ
2
2
But 1 + j + j = 0, so j + j = −1, so
= 2(α3 + β 3 + γ 3 ) − 3(α2 β + β 2 γ + γ 2 α + αβ 2 + βγ 2 + γα2 ) + 12αβγ
= 2(α + β + γ)3 − 9(α2 β + β 2 γ + γ 2 α + αβ 2 + βγ 2 + γα2 )
= 2(α + β + γ)3 − 9(α + β + γ)(αβ + βγ + γα) + 27αβγ
= 2a3 − 9ab + 27c
So
u3 v 3 = (uv)3
uv = (α + βj + γj 2 )(α + βj 2 + γj)
= α2 + β 2 + γ 2
+ (αβ + βγ + γα)
+ j 2 (αβ + βγ + γα)
uv = α2 + β 2 + γ 2 − (αβ + βγ + γα)
= (α + β + γ)2 − 3(αβ + βγ + γα)
= a2 − 3b
u3 v 3 = (a2 − 3b)3
(a2 − 3b)3 , U be u3 and V be v 3 . So U
2
and V are the roots of the polynomial x − Sx + P . Why? Let's develop
(x − U )(x − V ). We get x2 − (U + V )x + U · V . Since, S = U + V and
P = U · V , U and V are the roots of the polynomial. So,
p
√
(U + V ) + (U + V )2 − 4U · V
S + S 2 − 4P
3 3
U |V = u |v =
=
2
2
Let
S
be
2a2 − 9ab + 27c, P
be
4
(u3 + v 3 ) +
p
(u3 + v 3 )2 − 4(u · v)3
2
Last step: extract the roots of (E) from u and v . For α,
Indeed, let's calculate a + u + v :
=
it's quite simple.
a + u + v = α + β + γ + α + βj + γj 2 + α + βj 2 + γj
= 3α + β(1 + j + j 2 ) + γ(1 + j + j 2 )
= 3α
a+u+v
So α =
3
uj 2 = α2 +β+γj and vj = αj+β+γj 2 , we get a+uj 2 +vj =
a + uj + vj 2 = 3γ . We can write Cardan's equalities:
But if we take
3β
! De même,
α=
a+u+v
3
You may notice that
β=
α, β
a + uj 2 + vj
3
and
γ=
a + uj + vj 2
3
γ
are exceptionally distinguished: for you
2
2
2
to understand it easier. If we had written S = u + v|uj + vj|uj + vj , we
a+S
, but it wouldn't have so clear.
could write α|β|γ =
3
The extraction of our roots is nished!
To test the eciency of this method, let's use it on an example :
x3 −
6x2 + 11x − 6.
a = 6, b = 11 and c = 6.
3
3
So u v = (6 − 3 · 11) = 3 = 27
3
3
3
and u + v = 2 · 6 − 9 · 6 · 11 + 27 · 6 = 432 − 594 + 162 = 0.
2
3
3
Let's now resolve the equation x + 27 = 0 where u and v are solutions.
√
3 3
−27 =
No need to rack your brains vainly: we already see that u |u =
√
27i. u and v are complex numbers.
We have here
3 3
2
We can nd all roots. For readability reasons, we suppose that the square
roots are positive.
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a+u+v
3
a + uj 2 + vj
β=
3
a + uj + vj 2
γ=
q 3 q
√
3 √
3
u+v =
27i + − 27i
√
√
π
π
1
1
= ( 27 · e 2 i ) 3 + (− 27 · e 2 i ) 3
√
√
−π
π
= 3 · e6i + 3 · e 6 i
√ π
π
= 3(e 6 i + e 6 i )
√
π
= 3 · 2<(e 6 i )
√
√
3
=2 3·
2
=3
α=
q
3 √
q
√
3
uj + vj =
27ij + − 27ij
√
√
−4π
−π
4π
π
= 3 · e6i · e 6 i + 3 · e 6 i · e 6 i
√
√
−π
π
= 3 · e 2 i + 3 · e2i
√
√
= 3i − 3i
=0
2
2
q
√
3
uj + vj =
27ij + − 27ij 2
√
√
−π
−4π
4π
π
= 3 · e6i · e 6 i + 3 · e 6 i · e 6 i
√
√
−5π
5π
= 3·e 6 i+ 3·e 6 i
√ 5π
5π
= 3(e 6 i + e 6 i )
√
5π
= 3 · 2<(e 6 i )
√
√
3
= 2 3 · (−
)
2
= −3
2
q
3 √
6
6+3
9
a+u+v
=
= =3
3
3
3
a + uj 2 + vj
6+0
6
β=
=
= =2
3
3
3
2
6−3
3
a + uj + vj
=
= = 1.
And γ =
3
3
3
So,
α=
x3 − 6x2 + 11x − 6
x − 6x + 11x − 6 = (x − 1)(x − 2)(x − 3). CQFD
The roots of the polynomial
3
2
are 1, 2 and 3.
So,
Of course it would have been much easier to test small values (such as 1,
2, 3) to nd some evident roots of the polynomial. This is in fact sucient
for most of the exams.
But since the mathematical world is far to be as
simple as these articial exercises, we oer you a method of resolution that
gives all the roots of ANY degree 3 polynomial.
Even if it isn't the must optimal one, it is a beautiful demonstration,
using geometry and complex numbers. According to us, it is this beauty and
the diversity of the methods that make maths.
We hope you enjoyed and understood this long and bright demonstration
and that you will never see these polynomials the same way again.
We would like to thank especially François Sauvageot for having lead us
to this nice demonstration. (It took much longer to write than to explain!)
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