a) a) (x+2)(x – 3) = 0 b) (x – 5)(7 – x) = 0 c) (x – 1)(x+5)(-3x + 8) = 0 d) (x – 1)(3x+1) = 0 e) (x – 1)(x+2)(x-3) = 0

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a. (x + 2)(x – 3) = 0

⇔\(\left[ \begin{array}{l}x+2=0\\x-3=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\) 

b. (x – 5)(7 – x) = 0

⇔\(\left[ \begin{array}{l}x-5=0\\7-x=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\) 

c. (x – 1)(x + 5)(-3x + 8) = 0

⇔\(\left[ \begin{array}{l}x-1=0\\x+5=0\\-3x+8=0\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=1\\x=-5\\x=\frac{8}{3}\end{array} \right.\) 

d. (x – 1)(3x + 1) = 0

⇔\(\left[ \begin{array}{l}x-1=0\\3x+1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\x=\frac{-1}{3}\end{array} \right.\) 

e. (x – 1)(x + 2)(x – 3) = 0

⇔\(\left[ \begin{array}{l}x-1=0\\x+2=0\\x-3=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\x=-2\\x=3\end{array} \right.\) 

f. x(x² – 1) = 0

⇔x(x + 1)(x – 1) = 0

⇔\(\left[ \begin{array}{l}x=0\\x+1=0\\x-1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=0\\x=-1\\x=1\end{array} \right.\)