The number of real solutions of the equation $x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0$ is _________.

<p>The given equation is $x(x^2+3|x|+5|x-1|+6|x-2|)=0$, which can be solved by analyzing it in parts. It can be broken down into: $x=0$ and $x^2+3|x|+5|x-1|+6|x-2|=0$. </p><p>For $x=0$, it's clear that it is a solution to the equation since it makes the entire expression equal to zero.</p><b>Case (I)</b> <br/><br/>$x<0$ <br/><br/>$$ \begin{aligned} & x^2-3 x-5(x-1)-6(x-2)=0 \\\\ & x^2-14 x+17=0 \end{aligned} $$ <br/><br/>$\because$ All roots are positive $\Rightarrow$ no solution <br/><br/><b>Case (II)</b> <br/><br/>$$ \begin{aligned} & 0 < x < 1 \\\\ & x^2+3 x-5(x-1)-6(x-2)=0 \\\\ & x^2-8 x+17=0 \\\\ & \because D < 0 \Rightarrow \text { no solution } \end{aligned} $$ <br/><br/><b>Case (III)</b> <br/><br/>$1 < x < 2$ <br/><br/>$x^2+3 x+5(x-1)-6(x-2)=0$ <br/><br/>$x^2+2 x+7=0$ <br/><br/>$\Rightarrow$ no solution <br/><br/><b>Case (IV)</b> <br/><br/>$$ \begin{aligned} & x > 2 \\\\ & x^2+3 x+5(x-1)+6(x-2)=0 \\\\ & x^2+14 x-19=0 \end{aligned} $$ <br/><br/>All roots less than 2 <br/><br/>$\Rightarrow$ no solution <br/><br/>Here $x=0$ is only solution.