Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero). But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
On
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
On
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated. The zeros of $$ c(x^2-5x+6) $$ where $c$ is any (non-zero) constant, are the same as the zeros of, $$ (x^2-5x+6) $$ What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$ There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted