Ta có: \(P\left(x\right)=x^4+10x^3+25x^2=x^2\left(x^2+10x+25\right)=x^2\left(x+5\right)^2=\left(x^2+5x\right)^2\)
\(P\left(x\right)-2Q\left(x\right)=0\Leftrightarrow\left(x^2+5x\right)^2-2\left(x^2+5x+12\right)=0\)
Đặt \(x^2+5x=a\) phương trình trên trở thành:
\(a^2-2\left(a+12\right)=0\Leftrightarrow a^2-2a-24=0\Rightarrow\left[{}\begin{matrix}a=6\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5x=6\\x^2+5x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-6=0\\x^2+5x+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-6\\x=-1\\x=-4\end{matrix}\right.\)