Video: Finding the Volume of a Cuboid Whose Side Lengths Are Algebraic Expressions in the Form of a Polynomial

A square has sides of length 12. Squares measuring π‘₯ + 1 by π‘₯ + 1 are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of π‘₯.

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Video Transcript

A square has sides of length 12. Squares measuring π‘₯ plus one by π‘₯ plus one are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of π‘₯.

So let’s start by modeling our square of side length 12. We’re told that squares measuring π‘₯ plus one by π‘₯ plus one are cut out of each corner. We’ve then effectively created a net for an open box. And we’re asked to find the volume of this open box. Let’s start by working out the dimensions of this open box. Here, the height of the box will correspond to the length of one of the squares in the corner, which is π‘₯ plus one. The length and width of the open box correspond to the length and width of the portion of the square that remains from the original square of side length 12.

To find this length, we take our original length 12 and subtract two lots of π‘₯ plus one. Distributing the negative across the parentheses gives us 12 minus π‘₯ minus one minus π‘₯ minus one. This then simplifies to 10 minus two π‘₯ since 12 minus one minus one gives us 10 and minus π‘₯ minus π‘₯ gives us minus two π‘₯. Now, we know that the length and width of the open box are both 10 minus two π‘₯. To find the volume of the box, we use the formula that the volume of the cuboid is equal to the length times the width times the height. So the volume in terms of π‘₯ can be written as 𝑣 and π‘₯ in parentheses, which is equal to 10 minus two π‘₯ times 10 minus two π‘₯ times π‘₯ plus one.

To begin to simplify these three parentheses multiplied, we can take any two of them and multiply them together. We can use any method that we might commonly use to expand parentheses, such as FOIL or the area method. And here we’re going to use the area or grid method. To set up the grid, we take one of our parentheses and write the terms along the columns. And for our other parentheses, we then write on the row. Filling in our grid by multiplication, in the first cell, we have 10 times 10, which will be 100. In the second cell, 10 times negative two π‘₯ will give us negative 20π‘₯. Negative two π‘₯ times 10 will give us another negative 20π‘₯. And finally, negative two π‘₯ times negative two π‘₯ will give us a positive four π‘₯ squared.

The answer to the calculation can then be found by adding the four terms. So we start with our four π‘₯ squared. We then notice that we have two terms in π‘₯, negative 20π‘₯, negative 20π‘₯. So when we add those, we get negative 40π‘₯. And we add on the remaining term of 100. We can then write this expanded form into our volume calculation. We now have two expressions in parentheses which we need to find the product of. And we can do this by using the same method. We can then set up our grid so that we can multiply the terms in each of our sets of parentheses.

In our first row, π‘₯ times four π‘₯ squared will give us four π‘₯ cubed. π‘₯ times negative 40π‘₯ gives us negative 40π‘₯ squared. And π‘₯ times 100 gives us 100π‘₯. In our second row, since each of our terms along the row is multiplied by one, we have four π‘₯ squared, negative 40π‘₯, and 100. Adding the terms that we’ve calculated then, we start with four π‘₯ cubed. We then have two terms in π‘₯ squared, negative 40π‘₯ squared and four π‘₯ squared, which simplify to negative 36π‘₯ squared. The two terms in π‘₯, 100π‘₯ and negative 40π‘₯, simplify to plus 60π‘₯. And we add on our final term of 100. And so we have expanded our two sets of parentheses.

Therefore, our final answer is that the volume as a function of π‘₯ is equal to four π‘₯ cubed minus 36π‘₯ squared plus 60π‘₯ plus 100.

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