A slice 1 unit thickness is removed from one see of a cube. Use the rational zeroes theorem and synthetic division to find the original dimensions of the cube, if the remaining volume is 48 cm^3
v= x^3
v=x*x*x
x=(x-1)*x*x
48=(x-1)*x*x
carry on ...
x^2(x-1) = 48
x^3 - x^2 - 48 = 0
some quick mental gymnastic will show that
x = 4 works. (64-16 = 48)
so it factors into
(x-4)(x^2 +3x+12) = 0
some trivial calculations will show the remaining quadratic has no real roots, so
x = 4
the original was 4 by 4 by 4
To find the original dimensions of the cube, let's start by solving the equation:
48 = (x - 1) * x * x
To determine the possible rational solutions, we can use the Rational Zeros Theorem. According to this theorem, the possible rational solutions are given by the factors of the constant term (48) divided by the factors of the leading coefficient (1). In this case, since the leading coefficient is 1, the possible rational solutions are the factors of 48.
The factors of 48 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.
Now, we can use synthetic division to test each of these rational solutions until we find one that results in a remainder of zero.
Let's start with x = 1:
Using synthetic division, we set up the table:
1 │ 1 0 -1 48
────────
1
────────
1 1 0 48
The remainder is 48, which is not equal to zero. So, x = 1 is not a solution.
Let's move on to the next possible rational solution, x = -1:
-1 │ 1 0 -1 48
────────
-1
────────
1 -1 0 48
The remainder is 48, which is still not equal to zero. So, x = -1 is not a solution either.
We need to continue this process until we find a rational solution that results in a remainder of zero.
Testing each of the remaining possible solutions, we find that x = 2 is a solution:
2 │ 1 0 -1 48
────────
2
────────
1 2 3 48
The remainder is 0, which means x = 2 is a solution.
So, one of the solutions to the equation is x = 2. This means that the original dimensions of the cube were each 2 units in length.
To verify that this is correct, we can calculate the volume using x = 2:
V = (2 - 1) * 2 * 2 = 4 cubic units.
Therefore, the original dimensions of the cube were 2 units in length, 2 units in width, and 2 units in height.