SAT Question 6e7ae9fc: Nonlinear Functions answer and explanation with Desmos solution

The value of $g\left(7-w\right)$ is the value of $g\left(x\right)$ when $x=7-w$, where $w$ is a constant. Substituting $7-w$ for $x$ in the given equation yields $g\left(7-w\right)=\left(7-w\right)\left(7-w-2\right){\left(7-w+6\right)}^2$, which is equivalent to $g\left(7-w\right)=\left(7-w\right)\left(5-w\right){\left(13-w\right)}^2$. It's given that the value of $g\left(7-w\right)$ is $0$. Substituting $0$ for $g\left(7-w\right)$ in the equation $g\left(7-w\right)=\left(7-w\right)\left(5-w\right){\left(13-w\right)}^2$ yields $0=\left(7-w\right)\left(5-w\right){\left(13-w\right)}^2$. Since the product of the three factors on the right-hand side of this equation is equal to $0$, at least one of these three factors must be equal to $0$. Therefore, the possible values of $w$ can be found by setting each factor equal to $0$. Setting the first factor equal to $0$ yields $7-w=0$. Adding $w$ to both sides of this equation yields $7=w$. Therefore, $7$ is one possible value of $w$. Setting the second factor equal to $0$ yields $5-w=0$. Adding $w$ to both sides of this equation yields $5=w$. Therefore, $5$ is a second possible value of $w$. Setting the third factor equal to $0$ yields ${\left(13-w\right)}^2=0$. Taking the square root of both sides of this equation yields $13-w=0$. Adding $w$ to both sides of this equation yields $13=w$. Therefore, $13$ is a third possible value of $w$. Adding the three possible values of $w$ yields $7+5+13$, or $25$. Therefore, the sum of all possible values of $w$ is $25$.