Answer
$1$ foot.
Work Step by Step
Let the uniform width of the border be $t$.
Length of the garden with rock border $=12+2t$.
Width of the garden with rock border $=10+2t$.
Area of the rectangular garden and rock border combined is
$A=(12+2t)(10+2t)$
The total area (garden + border) $A=168$ square feet.
Equate both values.
$\Rightarrow (12+2t)(10+2t)=168$
Use the FOIL method.
$\Rightarrow 120+24t+20t+4t^2=168$
Subtract $168$ from both sides.
$\Rightarrow 120+24t+20t+4t^2-168=168-168$
Simplify.
$\Rightarrow 4t^2+44t-48=0$
Factor out $4$ from each term.
$\Rightarrow 4(t^2+11t-12)=0$
Divide both sides by $4$.
$\Rightarrow t^2+11t-12=0$
Rewrite the middle term $11t$ as $12t-1t$.
$\Rightarrow t^2+12t-1t-12=0$
Group the terms.
$\Rightarrow (t^2+12t)+(-1t-12)=0$
Factor each group.
$\Rightarrow t(t+12)-1(t+12)=0$
Factor out $(t+12)$.
$\Rightarrow (t+12)(t-1)=0$
By using zero product rule set each factor equal to zero.
$\Rightarrow t+12=0$ or $t-1=0$
Isolate $t$.
$\Rightarrow t=-12$ or $t=1$
Because $t$ is a dimension, the only solution that fits is the postive one.
Hence, the width of the rock border is $1$ foot.
