Question 943104
Not enough info.


For example, the length could be 1 and the width could be x^2 + 7x + 12. Then the new length and width are 4 and x^2 + 7x + 15, respectively, and the new area is 4x^2 + 28x + 60.


Or, perhaps, the length and width are x+3 and x+4, and the new length and width are x+6 and x+7, and the new area is (x+6)(x+7) = x^2 + 13x + 42.


Or we could generalize this and say that the length is L and the width is (x^2 + 7x + 12)/L, then add 3 to length/width and multiply.


The problem is likely an exercise in factoring the quadratic into (x+3)(x+4), but you cannot claim that the dimensions are x+3 and x+4, and it is a poor-quality problem.