Can someone draw this line $3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)$?

Question

Also the solutions for x are apparently (according to another question I have posted)

$x=0$, $x=-1$, $x=-\frac{1}{3}$, $x=10$

But when graphing the line my self the solution are only $x=0$ and $x=-1$

Answer 1

Normally to "solve" such equations, a common approach is to factorise. E.g. you can write the equation as $$x(x+1)\left(3(x^2+x+2)-16(2x+1) \right) = 0 \iff x(x+1)(1+3x)(x-10)=0$$

So we get $x = 0, -1, -\frac13, 10$ as solutions.