To solve this problem algebraically using the Factor Theorem, we need to find the maximum dimensions of the smaller box when its volume is at most 2220 cm3.

Let’s denote the original dimensions of the larger box as x, y, and z, with x = 42 cm, y = 17 cm, and z = 10 cm.

We are given that the volume of the smaller box is at most 2220 cm3. Therefore, we can set up the following inequality:

(x – a)(y – a)(z – a) ≤ 2220

Here, we are subtracting the same length “a” from each dimension to create the smaller box. We need to determine the maximum value of “a” that satisfies the inequality, which will give us the maximum dimensions of the smaller box.

Expanding the left side of the inequality:

xyz – axz – ayz + a2yz ≤ 2220

Simplifying further:

xyz – (xz + yz)a + a2yz ≤ 2220

Let’s define a function f(a) = a2yz – (xz + yz)a + xyz – 2220. To find the maximum value of “a,” we need to find the roots of this function.

Now, let’s use the Factor Theorem. According to the Factor Theorem, if a polynomial f(a) has a factor (a – r), where r is a root of f(a), then f(r) = 0.

Setting f(a) = 0:

a2yz – (xz + yz)a + xyz – 2220 = 0

Now, we can write the polynomial as a quadratic equation:

a2yz – (xz + yz)a + xyz – 2220 = 0

To find the roots of this equation, we can use the quadratic formula:

a = [-(-xz – yz) ± √((-xz – yz)2 – 4yz(xyz – 2220))] / (2yz)

Simplifying this equation:

a = [(xz + yz) ± √(x2z2 + 2xyz2 + y2z2 + 4yz(xyz – 2220))] / (2yz)

Now, we need to determine the maximum value of “a.” Since the dimensions of the box cannot be negative, we only need to consider the positive root given by:

a = [(xz + yz) + √(x2z2 + 2xyz2 + y2z2 + 4yz(xyz – 2220))] / (2yz)

To find the maximum dimensions of the smaller box, we substitute the value of “a” into the expressions for the dimensions of the original box:

x – a = 42 – [(xz + yz) + √(x2z2 + 2xyz2 + y2z2 + 4yz(xyz – 2220))] / (2yz)

y – a = 17 – [(xz + yz) + √(x2z2 + 2xyz2 + y2z2 + 4yz(xyz – 2220))] / (2yz)

z – a = 10 – [(xz + yz) + √(x2z2 + 2xyz2 + y2z2 + 4yz(xyz – 2220))] / (2yz)

These equations will give us the maximum dimensions of the smaller box when its volume is at most 2220 cm3. We can solve these equations to find the numerical values of the dimensions.