-
$A(x) = (3x + 1)(x + 4) - (2x - 5)(x+4)$ ;
Corrigé
\[\begin{aligned}
A(x)&=(3x+1){\color{Red}(x+4)} - (2x-5){\color{Red}(x+4)}&
\\
&={\color{Red}(x+4)}\left[(3x+1) - (2x-5)\right]&
\\
&=(x+4)(3x + 1 - 2x + 5)&
\\
&=(x+4)(x+6).&
\end{aligned}\]
-
$B(x) = (7x - 3)^2 - x(7x - 3)$ ;
Corrigé
\[\begin{aligned}
B(x) &= \underbrace{(7x-3)^2}_{{\color{Red}(7x-3)}(7x-3)} - x{\color{Red}(7x - 3)}&
\\
&={\color{Red}(7x-3)}\left[(7x-3) - x\right]&
\\
&=(7x-3)(7x-3-x)&
\\
&=(7x-3)(6x-3).&
\end{aligned}\]
-
$C(x) = 2 - x - (2-x)^2$ ;
Corrigé
\[\begin{aligned}
C(x) &=2 - x - (2 - x)^2&
\\
&=\underbrace{(2-x)}_{{\color{Red}(2-x)}\cdot 1} - \underbrace{(2-x)^2}_{{\color{Red}(2-x)}(2-x)}&
\\
&={\color{Red}(2-x)}\left[1 - (2-x)\right]&
\\
&=(2-x)(1-2+x)&
\\
&=(2-x)(-1+x).&
\end{aligned}\]
-
$D(x) = 6x(x+4) + 2(x-2)(x+4)$ ;
Corrigé
\[\begin{aligned}
D(x)&=6x{\color{Red}(x+4)} + 2(x-2){\color{Red}(x+4)}&
\\
&={\color{Red}(x+4)}\left[6x + 2(x-2)\right]&
\\
&=(x+4)(6x + 2x - 4)&
\\
&=(x+4)(8x - 4).&
\end{aligned}\]
On peut aussi continuer la factorisation par
\[\begin{aligned}
D(x)&=(x+4)(8x-4)&
\\
&=(x+4)\cdot 4(2x-1)&
\\
&=4(x+4)(2x-1).&
\end{aligned}\]
-
$E(x) = 5(x+1)^2 - (3x-2)(x+1)$.
Corrigé
\[\begin{aligned}
E(x)&=\underbrace{5(x+1)^2}_{5{\color{Red}(x+1)}(x+1)} - (3x-2){\color{Red}(x+1)}&
\\
&={\color{Red}(x+1)}\left[5(x+1) - (3x -2)\right]&
\\
&=(x+1)(5x + 5 - 3x + 2)&
\\
&=(x+1)(2x+7).&
\end{aligned}\]