Find the dimensions of the box described. The length is three inches longer than the width, the width it two inches longer than the height, and the volume is 120 cubic inches.
To answer this problem, we’re going to need to form and solve an equation. Before we can form and solve an equation though, we need to define a variable and unknown. We’re told that the length is longer than the width and the width is longer than the height. So let’s begin by defining the height. Let the height of the box be 𝑥 inches. The width is two inches longer than the height, so this means it’s 𝑥 plus two inches. Since the length is three inches more, then this is 𝑥 plus two plus three inches, or 𝑥 plus five inches. There’s one piece of information we haven’t yet used and that’s that the volume of the box is 120 cubic inches.
We know that the volume of a cuboid is its length multiplied by its height multiplied by its width. So in this case, the volume is 𝑥 times 𝑥 plus two times 𝑥 plus five. But that’s equal to 120. So we’ve got our equation. Let’s distribute our parentheses because in order to solve this, we need to ensure it’s equal to zero. We’ll begin by multiplying 𝑥 plus two by 𝑥 plus five. And when we do, we get 𝑥 squared plus seven 𝑥 plus 10. And that’s because we began by multiplying the first two terms. 𝑥 times 𝑥 gives us 𝑥 squared. We then multiply the outer term and the inner term. And we get 𝑥 times five, which is five 𝑥, two times 𝑥, which is two 𝑥, and their sum is seven 𝑥.
Finally, we multiply the last terms. So we get two times five, which is 10. We’ll distribute again by multiplying each term by 𝑥. So we get 𝑥 cubed plus seven 𝑥 squared plus 10𝑥 equals 120. Let’s subtract 120 from both sides. And we get 𝑥 cubed plus seven 𝑥 squared plus 10𝑥 minus 120 equals zero. Now that we have a cubic expression which we’ve made equal to zero, we’re going to need to factor that expression before we can solve. To factor, we’re going to need to find a factor of 𝑥 cubed plus seven 𝑥 square plus 10𝑥 minus 120. So we recall the factor theorem. And this says that if 𝑓 of 𝑎 is equal to zero, then 𝑥 minus 𝑎 is a factor of that function 𝑓 of 𝑥. So we need to find a value that when we substitute it into the expression 𝑥 cubed plus seven 𝑥 squared plus 10𝑥 minus 120, we get zero.
We look for factors of negative 120. Those are numbers we’re going to try. And that’s because our expression is going to factor to look like 𝑥 plus 𝑎 times 𝑥 plus 𝑏 times 𝑥 plus 𝑐. And the constant term is the product of 𝑎 and 𝑏 and 𝑐. Now, in fact, 𝑥 equals three is the factor of negative 120 we’re interested in. Three cubed plus seven times three squared plus 10 times three minus 120 is zero. So 𝑥 minus three must be a factor. We’re going to clear some space and perform polynomial long division to find the other factor or factors. There are a number of ways we can achieve this, but let’s use the bus stop method.
𝑥 cubed divided by 𝑥 is 𝑥 squared. Then when we multiply squared by 𝑥 minus three, we get 𝑥 cubed minus three 𝑥 squared. We now subtract 𝑥 cubed plus three 𝑥 squared from 𝑥 cubed plus seven 𝑥 squared. And we get 10𝑥 squared. Let’s bring down the 10𝑥. We now divide 10𝑥 squared by 𝑥 to get 10𝑥. We multiply 10𝑥 by 𝑥 minus three to get 10𝑥 squared minus 30𝑥. And then we subtract these two expressions, and that’s 40 𝑥. We bring down negative 120. And now we divide 40𝑥 by 𝑥 to get 40. We multiply 40 by 𝑥 minus three, which is 40𝑥 minus 120. And then we subtract. And that gives us zero. And that’s what we would expect. We would expect no remainder because we saw the 𝑥 minus three had to be a factor of our cubic.
So we can say that 𝑥 minus three times 𝑥 squared plus 10𝑥 plus 40 is equal to zero. And in fact, we cannot factor 𝑥 square plus 10𝑥 plus 40. It’s discriminate. Remember, for a quadratic equation of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, the discriminate is 𝑏 squared minus four 𝑎𝑐. Its discriminate is 10 squared minus four times one times 40. That’s less than zero. So it means not only is it not factorable, but the equation 𝑥 square plus 10𝑥 plus 40 equals zero has no real roots. And so, for this equation to be equal to zero, 𝑥 minus three itself has to be equal to zero, which means 𝑥 is equal to three.
Remember, that was the height of the box. Its width was 𝑥 plus two. So that’s three plus two, which is five. And its length was 𝑥 plus five. That’s three plus five, which is eight. So the dimensions of the box given are three, five, and eight inches. Now, in fact, we can perform a really quick check to see if what we’ve done is correct. We find the volume of the box that we’ve described, the one that has three, five, and eight inches as its dimensions. That’s three times five times eight, which is 120 cubic inches, as we expected. So we fully answered this question. Dimensions of the box are three, five, and eight inches.