Question

10 you have two rectangular grids of $1\mathrm{cm}^{3}$ cubes, one measuring $n + 1$ by $n + 6\mathrm{cm}$, and the other measuring $n + 3$ by $n + 3\mathrm{cm}$, where $n$ is a positive integer.
a find and simplify an expression for the total number of $1\mathrm{cm}^{3}$ cubes in the two rectangular grids combined.
b show that if you take apart the rectangles and recombine the $1\mathrm{cm}^{3}$ cubes you will always be able to form a rectangle with no cubes left over (where the rectangle will not simply be a straight line of cubes).

Explanation:

Step1: Calculate the area of each rectangle

The area of a rectangle is given by length × width.
For the first rectangle with dimensions \((n + 1)\) and \((n+6)\), the area \(A_1=(n + 1)(n + 6)\).
Using the FOIL method: \((n + 1)(n + 6)=n^2+6n+n + 6=n^2+7n + 6\).
For the second rectangle with dimensions \((n + 3)\) and \((n + 3)\), the area \(A_2=(n + 3)^2\).
Using the formula \((a + b)^2=a^2+2ab + b^2\) (here \(a=n\), \(b = 3\)), we get \(A_2=n^2+6n+9\).

Step2: Find the total number of cubes

The total number of cubes is \(A=A_1+A_2\).
Substitute \(A_1\) and \(A_2\) into the equation:
\(A=(n^2+7n + 6)+(n^2+6n+9)\).
Combine like - terms: \(A=(n^2+n^2)+(7n+6n)+(6 + 9)\).
\(A = 2n^2+13n+15\).

Step3: Factor the quadratic expression

We want to factor \(2n^2+13n + 15\).
We need to find two numbers \(m\) and \(k\) such that \(m\times k=2\times15 = 30\) and \(m + k=13\). The numbers are \(10\) and \(3\).
Rewrite the middle term: \(2n^2+10n+3n + 15\).
Group the terms: \((2n^2+10n)+(3n + 15)\).
Factor out the common factors: \(2n(n + 5)+3(n + 5)\).
Then \(2n^2+13n + 15=(2n + 3)(n+5)\).

Answer:

a. The total number of \(1cm^3\) cubes in the two rectangular grids combined is \(2n^2+13n + 15\).
b. Since the total number of cubes can be factored as \((2n + 3)(n + 5)\) (where \(n\) is a positive integer, \(2n+3\) and \(n + 5\) are both integers greater than \(1\) for \(n\gt0\)), we can form a rectangle with length \((2n + 3)\) and width \((n + 5)\) (or vice - versa) with no cubes left over.