A Box Volume Polynomial and the Meaning of a Zero Root
A small carpentry unit makes open boxes. For a particular design, the volume of the box, in cubic centimetres, is modelled by p(x) = 2x³ − 2x² − 8x + 8, where x is a length in centimetres. The supervisor knows that two of the three dimensions are (x − 1) cm and (x − 2) cm, but the third has been rubbed off the workshop card. He also notices that p(1) = 0 and p(2) = 0, and concludes that x = 1 and x = 2 are perfectly usable designs, since the volume formula gives zero at those values. A customer asks for a box of this design with x = 5 cm and wants to know how much more volume the box holds than the x = 4 cm version. The unit has to answer with a correct dimension and a sensible comment on the supervisor's conclusion.
1. Find the missing third dimension and verify the factorisation.
Read the worked answer
The product of the two known dimensions is
(x − 1)(x − 2) = x² − 3x + 2.
Divide p(x) by this quadratic. The leading term 2x³ ÷ x² gives 2x. Multiplying back:
2x(x² − 3x + 2) = 2x³ − 6x² + 4x.
Subtracting from p(x):
(2x³ − 2x² − 8x + 8) − (2x³ − 6x² + 4x) = 4x² − 12x + 8.
Now 4x² ÷ x² = 4, and 4(x² − 3x + 2) = 4x² − 12x + 8, with zero remainder. Therefore
p(x) = (x − 1)(x − 2)(2x + 4).
The third dimension is (2x + 4) cm. Verification by expansion gives 2x³ + 4x² − 6x² − 12x + 4x + 8 = 2x³ − 2x² − 8x + 8, which matches.
2. Evaluate the supervisor's conclusion that x = 1 and x = 2 are usable designs because p(1) = p(2) = 0.
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The calculations p(1) = 0 and p(2) = 0 are correct, and they show that 1 and 2 are roots of the volume polynomial. But a root of the formula is not automatically a valid physical design.
At x = 1, the dimensions are x − 1 = 0, x − 2 = −1 and 2x + 4 = 6. A length of 0 cm and a length of −1 cm cannot form a box. At x = 2, the dimensions are 1, 0 and 8, again giving a zero dimension and zero volume. The factor theorem is telling us where the volume becomes zero, not where a real box exists.
For a genuine box, every dimension must be positive. The restrictions are x − 1 > 0, x − 2 > 0 and 2x + 4 > 0, so x > 2. The supervisor has confused an algebraic zero with a usable measurement. The correct conclusion is that x = 1 and x = 2 are excluded by the physical domain, not accepted because the formula yields zero.
3. Compare the volumes for x = 4 cm and x = 5 cm and state the increase.
Read the worked answer
Using the factorised form is safer and shows the dimensions directly.
For x = 4:
- dimensions: 3 cm, 2 cm, 12 cm
- volume = 3 × 2 × 12 = 72 cm³
For x = 5:
- dimensions: 4 cm, 3 cm, 14 cm
- volume = 4 × 3 × 14 = 168 cm³
Increase = 168 − 72 = 96 cm³.
Checking with the polynomial: p(4) = 128 − 32 − 32 + 8 = 72 and p(5) = 250 − 50 − 40 + 8 = 168. The results agree. The x = 5 box holds 96 cm³ more than the x = 4 box. This comparison is only meaningful because both values lie in the valid domain x > 2.