Answer:
The width is 20 inches, the length is 32 inches and the height is 15 inches.
Step-by-step explanation:
Let x inches be the width of the box. If the length of the box is 12 inches greater than its width, then the length of the box is x+12 inches. If the height of the box is 5 inches less than its width, then the height of the box is x-5 inches.
The volume of box is
[tex]V=\text{length}\cdot \text{width}\cdot \text{height}\ in^3.[/tex]
Therefore,
[tex]9600=x\cdot (x+12)\cdot (x-5).[/tex]
Solve this equation:
[tex]x(x^2+12x-5x-60)=9600,\\ \\x^3+7x^2-60x-9600=0.[/tex]
The integer solutions of this equation can be among the divisors of -9600. The divisors of 9600 are:
[tex]\pm1, \pm2, \pm3, \pm4,\pm5,\pm6,\pm8,\pm10,\pm12,\pm15,\pm16,\pm20,\pm24,\pm25,\pm30,\pm32,....[/tex]
We can check all these divisors and see that x=20 is a solution, then using synthetic division
[tex]x^3+7x^2-60x-9600=(x-20)(x^2+27x+480).[/tex]
Since the quadratic factor has discriminant [tex]D=27^2-4\cdot 480=-1191<0,[/tex] we can state that the only solution is x=20 inches. The width is 20 inches, the length is 32 inches and the height is 15 inches.