correction exo second degre
Les polynˆ omes du second degr´ e Exercice 1
Exercice 2
Exercice 3
Les polynˆ omes du second degr´ e Exercice 1
Les polynˆ omes du second degr´ e Enonc´ e Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
P3 (x) = −x 2 + 2x + 5
P2 (x) = x 2 + 3x + 1
P4 (x) = 3x 2 + x − 4
Les polynˆ omes du second degr´ e Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x − 1)
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x − 1)
donc
P1 (x) = 2(x 2 + 4x
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x − 1)
donc
P1 (x) = 2(x 2 + 4x + 4
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x − 1)
donc
P1 (x) = 2(x 2 + 4x + 4 − 4 − 1)
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x − 1)
donc
P1 (x) = 2(x 2 + 4x + 4 − 4 − 1)
donc
P1 (x) = 2((x + 2)2
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x − 1)
donc
P1 (x) = 2(x 2 + 4x + 4 − 4 − 1)
donc
P1 (x) = 2((x + 2)2 − 5)
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x − 1)
donc
P1 (x) = 2(x 2 + 4x + 4 − 4 − 1)
donc
P1 (x) = 2((x + 2)2 − 5)
d’o` u P1 (x) = 2(x + 2)2 − 10
Mettre sous forme canonique
P1 (x) = 2x 2 + 8x − 2
R´
eponse
on a
P1 (x) = 2x 2 + 8x − 2
donc
P1 (x) = 2(x 2 + 4x −