Question

Answer 1

a) x³ + 27 + (x+3)(x-9) = 0

⇔ x³ + 27 + x² - 9x +3x - 27 = 0

⇔ x³ + x² - 6x = 0

⇔ x( x² + x - 6 ) = 0

⇔ \(\left[{}\begin{matrix}x=0\\x^2+x-6=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\end{matrix}\right.\)

b) 3x² - 12x = 0

⇔ x(3x -12) = 0     

⇔\(\left[{}\begin{matrix}x=0\\3x-12=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)