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Factoriser chacune des expressions suivantes en reconnaissant un facteur commun :
-
$A(x) = 4(x-3) + x(x-3)$
Corrigé
\[\begin{aligned}
A(x)&=4{\color{Red}(x-3)} + x{\color{Red}(x-3)}&
\\
&={\color{Red}(x-3)}(4+x)&
\end{aligned}\]
-
$B(x) = x^2 - 4x$
Corrigé
\[\begin{aligned}
B(x)&=\underbrace{x^2}_{{\color{Red}x}\cdot x} - 4{\color{Red}x}&
\\
&={\color{Red}x}(x-4).&
\end{aligned}\]
-
$C(x) = 2x(x+1) - 4(x+1)$
Corrigé
\[\begin{aligned}
C(x)&=2x{\color{Red}(x+1)} - 4{\color{Red}(x+1)}&
\\
&={\color{Red}(x+1)}(2x-4).&
\end{aligned}\]
-
$D(x) = 5x^2 - x$
Corrigé
\[\begin{aligned}
D(x)&=5x^2 - x&
\\
&= 5x\cdot{\color{Red}x} - {\color{Red}x}\cdot 1&
\\
&= {\color{Red}x}(5x - 1)&
\end{aligned}\]
-
$E(x) = (x+2)(x-1) + (3x+5)(x+2)$
Corrigé
\[\begin{aligned}
E(x)&=(x+2)(x-1) + (3x+5)(x+2)&
\\
&={\color{Red}(x+2)}(x-1) + (3x+5){\color{Red}(x+2)}&
\\
&={\color{Red}(x+2)}\left[(x-1) + (3x + 5)\right]&
\\
&=(x+2)(x-1+3x+5)&
\\
&=(x+2)(4x + 4)&
\end{aligned}\]
-
$F(x) = 7(5-x) - (2x+3)(5-x)$
Corrigé
\[\begin{aligned}
F(x)&=7(5-x) - (2x+3)(5-x)&
\\
&=7{\color{Red}(5-x)} - (2x+3){\color{Red}(5-x)}&
\\
&={\color{Red}(5-x)}\left[7 - (2x +3)\right]&
\\
&=(5-x)(7-2x-3)&
\\
&=(5-x)(4-2x)&
\end{aligned}\]
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