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Highhsoul
@Highhsoul
May 2019
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(URGENT) Bonjour j'ai besoin d'aide :
B(x)= 808x-2x^2-3200
Quelle est la forme canonique de B(x)? FActoriser B(x).
merci
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Bernie76
Bonjour,
B(x)=-2x²+808x-3200
B(x)=-2[(x²-404x) +1600]
x²-404x est le début du développement de (x-202)².
Mais (x-202)²=x²-404x+202²=x²-404x+40804
Donc : x²-404x=(x-202)²-40804 donc :
B(x)=-2[(x-202)²-40804+1600]
B(x)=-2[(x-202)²-39204]
--->mais 39204=198²
B(x)=-2 [(x-202)²-198²]
Dans les [.....] , tu as a²-b²=(a+b)(a-b).
Je te laisse factoriser.
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Thanks 1
Bernie76
(x-202)^2=x^2-404x+40804 . Passe "+40804" dans le membre de gauche et observe.
Bernie76
: )
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B(x)=-2x²+808x-3200
B(x)=-2[(x²-404x) +1600]
x²-404x est le début du développement de (x-202)².
Mais (x-202)²=x²-404x+202²=x²-404x+40804
Donc : x²-404x=(x-202)²-40804 donc :
B(x)=-2[(x-202)²-40804+1600]
B(x)=-2[(x-202)²-39204] --->mais 39204=198²
B(x)=-2 [(x-202)²-198²]
Dans les [.....] , tu as a²-b²=(a+b)(a-b).
Je te laisse factoriser.