The polynomial x 3 + 5x 2 -57x -189 expresses the volume, in cubic inches, of a shipping box, and the width is (x+3) in. If the width of the box is 15 in., what are the other two dimensions? ( Hint: The height is greater than the depth.)
the clue is that (x+3) divides the volume. So, a little checking shows that the volume
v = (x+3)(x-7)(x+9)
so, if x+3=15, x=12 and the box is 15x5x21
Well, to solve this riddle, we'll have to do a little mathematical circus performance! Let's start by setting the width of the box, which is represented by (x+3), equal to 15. So, we have (x+3) = 15.
We can solve this equation by subtracting 3 from both sides, giving us x = 12. Now that we know x, we can substitute it back into the original polynomial to find the other two dimensions.
Using x = 12, we have:
Volume = x^3 + 5x^2 - 57x - 189
= 12^3 + 5(12^2) - 57(12) - 189
Now, let's put our calculators to good use!
Volume = 1728 + 5(144) - 684 - 189
= 1728 + 720 - 684 - 189
= 2575 cubic inches
So, the volume of the box is 2575 cubic inches. Since the height is greater than the depth, we just need to find factors of 2575 that make one dimension greater than the other.
Let's break out the juggling balls and do some factoring!
2575 = 5 x 5 x 103
Given that the height is greater than the depth, we can assign 103 as the height and one of the 5's as the depth. That leaves the other 5 as the remaining dimension.
So, the three dimensions of the shipping box are:
Height = 103 inches
Depth = 5 inches
Width = 15 inches
Voila! The circus-themed shipping box is ready for its grand appearance!
To find the other two dimensions of the shipping box, we need to first set up an equation by factoring the given polynomial.
The polynomial x^3 + 5x^2 - 57x - 189 can be factored as (x - 3)(x + 3)(x + 7).
Given that the width of the box is (x + 3) in, and it is given as 15 in, we can set up the equation:
x + 3 = 15
Solving for x, we have:
x = 15 - 3
x = 12
Now, we substitute this value of x into the factored polynomial.
(x - 3)(x + 3)(x + 7)
(12 - 3)(12 + 3)(12 + 7)
9 * 15 * 19
The dimensions of the box are 9 inches, 15 inches, and 19 inches.
To find the other two dimensions of the box, we need to solve the given polynomial equation when the width (x+3) is equal to 15 inches.
Step 1: Plug in the value of the width into the polynomial equation.
Replace the width (x + 3) with 15 in the polynomial:
(15)3 + 5(15)2 - 57(15) - 189 = 0
Step 2: Simplify the equation and solve for x.
Evaluate the equation using the given width value and calculate:
3375 + 1125 - 855 - 189 = 0
3456 = 0
Step 3: Determine the other two dimensions.
Since the equation 3456 = 0 is not true, there is no solution for the width of 15 inches. Therefore, it is not possible to determine the other two dimensions using this given information.
In summary, with the given width of 15 inches, it is not possible to determine the other two dimensions of the shipping box.