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\begin{center}\section*{\textcolor{red}{Solution des exercices}}\end{center}

\subsection*{Solutions}

\begin{solution}
$A(x)=-6 (5 x-4) (5 x-1)$.
\end{solution}

\begin{solution}
$A(x)=10 (x+2)$.
\end{solution}

\begin{solution}
$A(x)=-(5 x+3) (11 x-1)$.
\end{solution}


\begin{solution}
$(-x-9) (4 x-4)-(4 x-4) (5 x-7)=-8 (x-1) (3 x+1).$
\end{solution}

\begin{solution}
$(-4 x-1) (-2 x-8)-(-4 x-1) (-x-4)=(x+4) (4 x+1).$
\end{solution}

\begin{solution}
$3 (3 x+2) (9 x-2)+(7 x+1) (9 x-2)=(9 x-2) (16 x+7).$
\end{solution}

\begin{solution}
$-4 (x+1) (10 x+7)-(9 x-1) (10 x+7)=-(10 x+7) (13 x+3).$
\end{solution}



\begin{solution}
$f(x)=64 x^2+144 x+81=(8 x+9)^2$.
\end{solution}

\begin{solution}
$f(x)=49 x^2-126 x+81=(7 x-9)^2$
\end{solution}

\begin{solution}
$f(x)=4 x^2-32 x+64=4 (x-4)^2$
\end{solution}

\begin{solution}
$f(x)=25 x^2-100 x+100=(5x-10)^2=25 (x-2)^2$
\end{solution}

\begin{solution}
$f(x)=81 x^2-4=(9 x-2) (9 x+2)$
\end{solution}

\begin{solution}
$f(x)=\frac{x^2}{4}+\frac{7 x}{2}+\frac{49}{4}=\left(\frac{x}{2}+\frac{7}{2}\right)^2$ ou $f(x)=\frac{1}{4} (x+7)^2$
\end{solution}

\begin{solution}
$f(x)=\frac{25 x^2}{9}-10 x+9=\left(\frac{5 x}{3}-3\right)^2=\frac{1}{9} (5 x-9)^2$
\end{solution}

\begin{solution}
$f(x)=81 x^2-\frac{1}{64}=\left(9 x+\frac{1}{8}\right) \left(9x-\frac{1}{8}\right)$
\end{solution}

\begin{solution}
$A(x)=-8 (5 x+1) (9 x+11)$.
\end{solution}

\begin{solution}
$A(x)=(4 x+9) (11 x-1)$.
\end{solution}

\begin{solution}
$A(x)=-6 (3 x-1) (31 x-5)$.
\end{solution}
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