Tu poses P(x)=ax^3+bx²+cx+d

Tu dérives P(x)V(x-1), et tu choisis a, b, c et d en sorte que ça fasse f(x).

P(x) = ax³ + bx² + cx + d

F(x) = (ax³ + bx² + cx + d).V(x-1)

F'(x) = (3ax² + 2bx + c).V(x-1) + (ax³ + bx² + cx + d)/(2V(x-1))

F'(x) = [(3ax² + 2bx + c).2.(x-1) + (ax³ + bx² + cx + d)]/(2V(x-1))

F'(x) = (6ax³ + 4bx² + 2cx - 6ax² - 4bx - 2c + ax³ + bx² + cx + d)/(2V(x-1))

F'(x) = (7ax³ + (5b-6a)x² + (3c-4b)x - 2c + d)/(2V(x-1))

Et on doit avoir: f(x) = F'(x) -->

x².V(x-1)  = (7ax³ + (5b-6a)x² + (3c-4b)x - 2c + d)/(2V(x-1))

x².V(x-1).2.V(x-1) = (7ax³ + (5b-6a)x² + (3c-4b)x - 2c + d)

2x²(x-1) = 7ax³ + (5b-6a)x² + (3c-4b)x - 2c + d

2x³ - 2x² = 7ax³ + (5b-6a)x² + (3c-4b)x - 2c + d

On identifie les coefficients de même puissance en x des 2 membres, on a alors le système:

2 = 7a
-2 = 5b-6a
0 = 3c-4b
-2c+d = 0

--> a = 2/7 ; b = -2/35 ; c = -8/105 ; d = -16/105

P(x) = (2/7)x³ - (2/35).x² - (8/105)x - (16/105)

et F(x) = [(2/7)x³ - (2/35).x² - (8/105)x - (16/105)].V(x-1)

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Sauf disraction.