so... if we notice the picture, the height of the finished box is 3 then
the length is "8 longer than the width" or w + 8 then
thus [tex]\bf \textit{volume of a rectangular prism}\\\\
V=lwh\qquad
\begin{cases}
l=length\\
w=width\\
h=height\\
-------\\
l=w+8\\
h=3\\
V=315
\end{cases}\implies 315=(w+8)(w)(3)
\\\\\\
\cfrac{315}{3}=w^2+8w\implies 0=w^2+8w-105
\\\\\\
0=(w+15)(w-7)[/tex]
well, -15 is clearly not a feasible value for a width... so the value is 7 then
now... that's just the finished box... notice the picture, the width on the cardboard is really 3 + width + 3, and the length is 3 + length + 3,
so you're really adding 6 or 3+3, to each, so in this case, the width is 7+6
and the length...well, you'd know that already by now
