What does it mean to solve or find solutions in mathematics? Something that has been really confusing me lately is that this equation has four solutions $$3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)$$
But what does that mean? Until now solutions to me has meant, what are the coordinates of $x$ when $y$ equals a given value, normally $y=0$.
But this equation has kind of thrown me off because when plotting this equation on a graph I get only two points at which the line crosses the $x$-axis.
I hope someone understands my point.
Regards
 A: In general, a solution to an equation means the values which make the equation true.
According to this definition, if $y = f(x)$, then finding the coordinates of $x$ that make $y = 0$ means finding the values that make the following equation true:
$$y = 0$$
The solution to several simultaneous equations, called a system of equations, are the values that make all equations true at the same time.
Also note that some equations have no solutions, while others might have solutions for every $x$ in the domain of the equation.
The definition is similar for inequalities.
edit: you can also find solutions to things like $y \equiv 4 \bmod 7$, which means that $y$ has a remainder of four when divided by seven. Or things like $\sqrt{2x+9} \text{ is a real number}$, or $x \text { and } y + 1 \text{ have the same absolute value}$. You're finding the values of the variables that make the statement true. 
Also note that it is incorrect to find solutions to things that cannot be called true or false, as then there is nothing to solve. For example, it doesn't make sense to find the solutions to $2$, $x + 1$, or $\text{divisible by } 5$.
A: You can think of it as two function:
$f(x) = 3x(x+1)(x^2+x+2)$
$g(x) = 16x(x+1)(2x+1)$
When do the graphical repentations of these functions intersect?
Exactly when $f(x) = g(x)$ or, 
$3x(x+1)(x^2+x+2) = 16x(x+1)(2x+1)$.
Now you can solve for $x$, to find the four $x$-values for which the two curves intersect.
A: Solving $3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)$
is equivalent in solving $3x(x+1)(x^2+x+2) - 16x(x+1)(2x+1) = 0$
This is further equivalent in solving $x(x + 1)[(Ax + B)(Cx + D)] = 0$; for some $A, B, C$ and $D$.
Solving the last equation, we therefore get 4 roots, namely $x = 0,$ or $x= -1,$ or $x = -B/A,$ or $x = …$ .
It must be clear that cancellations [of the factors $x$ and $(x + 1)$] must NOT be done to the original from the start; Otherwise we will have only 2 roots left [with the other two missed as described.]
A: Definition: Solve. We say an equation is solved if and only if we have listed the set of all objects which make the equation true. We say  we have found a solution if  we have found an element of the solution set. 
Example: If $x= 0$, then 
\begin{align}
3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)
\end{align}
becomes
\begin{align}
0 = 0
\end{align}
which is true, and hence $x =0$ solves the equation.
Example: If $x =1$, then
\begin{align}
3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)
\end{align}
becomes
\begin{align}
3\cdot 2 \cdot 4 = 16\cdot 2\cdot 3
\end{align}
which is false, and hence $x=1$ is not a solution.
A: The four roots of the problem given are $-1$, $-1/3$, $0$, and $10$. They should be visible by graphing the fourth order equation resulting from collecting terms and setting same equation to zero. That equation is $3x^4 - 26x^3 - 39x^2 - 10x$.
