d The solution you found in Part C is the value of x, not the dimension of the cube.
A
a x^3-9x^2+27x-35=0
B
b ± 1, ± 5, ± 7, ± 35
C
c See solution.
D
d 2cm* 2cm* 2cm
Practice makes perfect
a The volume of a cube is a product of the lengths of its edges. We are given that the volume is 8 cubic centimeters, and the length of the edges is x-3. Let's write an equation representing this information and rearrange it.
We got a polynomial equation that we can use to find x.
x^3-9x^2+27x-35=0
b Since the leading coefficient is 1, the Rational Root Theorem guarantees that the rational solutions of the equation in Part A are integer factors of the constant term -35. Let's list all possibilities.
± 1,± 5, ± 7,± 35
c We need to pick one of the possible solutions to try. Since x-3 is a length, only values greater than 3 are meaningful solutions in the context. The smallest such number in the list of Part B is 5, so let's use synthetic division to check it.
rl IR-0.15cm r 5 & |rr 1& -9& 27& -35 & 5& -20& -35 & c 1& -4 & 7 & 0
Since the last entry in the last line is 0, x=5 is indeed a solution.
The last line also gives us the factor form of the polynomial.
x^3-9x^2+27x-35=(x-5)(x^2-4x+7)
To show that the equation in Part A has no other real solutions, we can check the equation when the quadratic factor is 0.
x^2-4x+7=0
Let's check the discriminant.
b^2-4ac=(-4)^2-4(1)(7)=16-28=-12
Since the discriminant is negative, this quadratic has no real solutions.
This means that the only solution of the equation in Part A is x=5.
d According to the diagram, the length of the edges of the cube is x-3. Let's use the solution x=5 we got in Part C.
5-3=2
The length of all edges of the cube is 2 centimeters.
