Answer:
x³ - 24x² + 188x - 375 = 0;  x = 3
Step-by-step explanation:
Data:
l = 10 in; w = 8 in; h = 6 in
Calculations:
(a) Polynomial
Let x = amount to be removed from each dimension
Then, the volume of the new box is
V = lwh
 = (10 - x)(8 - x)(6 - x)
 = 105
Here's the polynomial by long multiplication:
10 Â Â Â - Â Â x
 8    -   x
80 Â Â - Â 8x
     - 10x  +  x²
80    - 18x  +  x²
 6   -    x          Â
480  - 108x + 6x²
    -  80x + 18x²  - x³
480 - 188x + 24x² - x³
So,
 480 - 188x + 24x² - x³ = 105   Subtract 105 from each side Â
 x³ + 24 x² - 188x + 375 = 0     Multiply each side by -1
x³  - 24x² + 188x  - 375 = 0     This is the polynomial equation
(b) Solution to the equation
We can use the rational root theorem to help find a root.
The general formula for a third-degree polynomial is
f(x) = ax³ + bx² + cx + d
Your polynomial is Â
f(x) = x³  - 24x² + 188x  - 375 = 0
a = 1; d = -375 Â
Factors of a = ±1
Factors of d = ±1, ±3. ±5, ±15, ±25, ±75, ±125, ±375
Possible roots are x = ±1, ±3, ±5, ±15, ±25, ±75, ±125, ±375
Now, it's a matter trial and error in seeing which value of x is a zero of the polynomial.
f(3) =  3³ - 24×3² + 188×3 - 375
f(3) = 27 - 24×9  +  564  -375
f(3) = 27 - Â 216 Â Â + Â 564 - 375
f(3) = 0
So, x = 3 is one root of the polynomial, and there are no other real roots.
A graph of the polynomial shows one zero at x = 3.
Stefan should take 3 in from each dimension of the box.
Then, the dimensions of the box are
 l = 10 - 3 = 7 in
w = Â 8 - 3 = 5 in
h = Â 6 - 3 = 3 in
Check:
V = lwh = 7 × 5 × 3 = 105 in³
