Question 122433
Let the length L and the width W then LW=31 (area is 31.0 cm^2)

and   L=4W-4  (The length of a rectangle measures 4 cm less than 4 times the width)

then L=4(W-1) then 4(W-1)W=31 then 4W^2-4W-31=0 then W=(4+sqrt(16+16(31)))/8
or W=(4-sqrt(16+16(31)))/8  then solutions are W=(4+16sqrt(2))/8 or W=(4-16sqrt(2))/8 as the second solution is <0 the answer is W=(4+16sqrt(2))/8 if you simplify 

 W = 1/2+2sqrt(2) and
 L = 4(W-1)= 4(-1/2+2sqrt(2)) = -2+8sqrt(2)