Question

Please help with Algebra II! Any help is appreciated. You will need the following materials to find the volume of a rectangular box:

A rectangular box such as a cereal or shoe box
Ruler or tape measure
Graphing technology (e.g., graphing calculator or GeoGebra)

Answer 1

Part 1:
Measure and record the length, width and height of the rectangular box you have chosen in inches. Round to the nearest whole number.

Answer 2

@jim_thompson5910

Answer 3

@hartnn

Answer 4

@satellite73

Answer 5

Um, I don't think the question can be any more clear and guiding. What do you need help with?

Answer 6

It's a project. There are a few parts. Sorry, I thought I posted the other parts

Answer 7

Part 2:
Apply the formula of a rectangular box (V = lwh) to find the volume of the object.

Now suppose you knew the volume of this object and the relation of the length to the width and height, but did not know the length. Rewriting the equation with one variable would result in a polynomial equation that you could solve to find the length.

Answer 8

Part 3:
Rewrite the formula using the variable x for the length. Substitute the value of the volume found in step 2 for V and express the width and height of the object in terms of x plus or minus a constant. For example, if the height measurement is 4 inches longer than the length, then the expression for the height will be (x + 4).

Answer 9

Part 4:
Simplify the equation and write it in standard form.

Answer 10

Part 5:
Find the solutions to this equation algebraically using the Fundamental Theorem of Algebra, the Rational Root Theorem, Descartes' Rule of Signs, and the Factor Theorem.

(Hint: If the numbers are large, graph the function first using GeoGebra to help you find one of the zeros. Use that zero to find the depressed equation which can be solved by factoring or the quadratic formula.)

Answer 11

Part 6:
Substitute 0 for the function notation and, using graphing technology, graph the function.

Answer 12

Part 7:
Answer the following questions
What does the Fundamental Theorem of Algebra indicate with respect to this equation?
What are the possible rational solutions of your equation?
How many possible positive, negative and complex solutions are there in your equation?
Graph the function. What type of function has been graphed (linear, quadratic, cubic, or quartic)? Provide your reasoning and describe the end behavior of the graph.
How do the solutions of the equation compare to the length of the rectangular object, and the x-intercept of the graph? Provide both the solutions and measurement.

Answer 13

@wolfe8

Answer 14

I'm sorry but this is a really long question for me to look at while I'm relaxing :/ But have you done what it tells you to? (i.e get boxes and measure them, etc.)

Answer 15

yes

Answer 16

Which part do you actually need help with?

Answer 17

part 3

Answer 18

@wolfe8

Answer 19

this is what i got for part 3 but idk if it's correct 8=(x-2)(x-3)

Answer 20

Alright. First of all, what were the measurements for your box?

Answer 21

my measurements are L=4 W=2 and H=1 inch

Answer 22

I think for the length expression it remains a constant, because the example given for height keeps the length constant. I am not sure, but I think it might be V = x(x-2)(x-3)

Answer 23

I'm sorry I meant, the length is always x. Since w is 2 less than x, it is x-2. Do the same for h.

Answer 24

so it would be 8=x(x-2)(x-3)?

Answer 25

@wolfe8 I also need help with part 5

Answer 26

Probably so. Ah part 5 just asks you to solve the equation you just made to find x. There are 2 ways to do that. You might want to look them up :) 1 way is by factorization, the other is by using a formula.

Answer 27

I have to use one of the given methods.... Fundamental Theorem of Algebra, the Rational Root Theorem, Descartes' Rule of Signs, and the Factor Theorem. Which will take the least time?

Answer 28

@wolfe8

Answer 29

Uh to be honest, I've never heard of those names except the Factor Theorem. I would say that is the fastest since your equation is simple.

Answer 30

okay, thanks