\(4ax^3-12x^2-ax+3=0\left(1\right)\)
a) Để phương trình (1) có 1 trong các nghiệm là -2
\(\Leftrightarrow4a\left(-2\right)^3-12\left(-2\right)^2-a\left(-2\right)+3=0\)
\(\Leftrightarrow-32a-48+2a+3=0\)
\(\Leftrightarrow30a=-45\)
\(\Leftrightarrow a=\dfrac{-45}{30}=-\dfrac{3}{2}\)
b) Với \(a=-\dfrac{3}{2}\)
\(\left(1\right)\Leftrightarrow4.\left(-\dfrac{3}{2}\right)x^3-12x^2-\left(-\dfrac{3}{2}\right)x+3=0\)
\(\Leftrightarrow-6x^3-12x^2+\dfrac{3}{2}x+3=0\)
\(\Leftrightarrow12x^3+24x^2-3x-6=0\)
\(\Leftrightarrow4x^3+8x^2-x-2=0\)
\(\Leftrightarrow4x^3-x+8x^2-2=0\)
\(\Leftrightarrow x\left(4x^2-1\right)+2\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{1}{4}\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
Vậy phương trình theo đề bài có 3 nghiệm \(x\in\left(-2;-\dfrac{1}{2};\dfrac{1}{2}\right)\)