solve a word problem involving a polynomial of degree 3: find width of box

[Robin2017]

I have a test tomorrow and can't figure out how to solve this problem. Here it is: The height of a rectangular box is 4 time its width and its length is 5 more than its width. The volume of the box is 133. Find the width of the box. I have this:
V=LWH
H=4w
L=w+5

I plugged H and L in the V formula but the numbers don't look right. I tried Foiling and finding the zeros. I would really appreciate the help!!!!:-----(

 

[stapel]

I have a test tomorrow and can't figure out how to solve this problem. Here it is: The height of a rectangular box is 4 time its width and its length is 5 more than its width. The volume of the box is 133. Find the width of the box. I have this:
V=LWH
H=4w
L=w+5

I plugged H and L in the V formula but the numbers don't look right. I tried Foiling and finding the zeros.

Please reply showing your work and results, so that error-checking may be attempted. Thank you!

 

[Robin2017]

Hi Stapel!

Ok, here it is:

First attempt: 133=(w+5)(4w)(w)

 

[Robin2017]

Hi Stapel!

Here it is:

133=(w+5)(4w)(w)
4w²+20²=133
4w²+20²-133=0

and

4w³+5=133
4w³=128
w³​=32

 

[stapel]

133=(w+5)(4w)(w)
4w²+20²=133

How did you go from the first line to the second line? How did "(w)(4w)(w + 5) = (4w²)(w + 5) = 4w²(w) + 4w²(5)" turn intro 4w² + 20²?

4w³+5=133

Where did this come from? How does it relate to anything that came before?

By the way, to review how to multiply polynomials, try here.

 

[lookagain]

Here it is:

Suppose you state

Let the width = w

Then the height = 4w

And the length = w + 4

- - - - - - - - - - - - - - - -- - -- -


The next equation would be good.

133=(w+5)(4w)(w) \(\displaystyle \ \ \ \ \ \ \ \ \)

4w²+20²=133\(\displaystyle \ \ \ \ \ \ \ \ \) And see below.

4w²+20²-133=0 \(\displaystyle \ \ \ \ \ \ \ \ \)The exponent on w (first term) should be a 3, and the middle term should be 20w^2.


and

4w³+5=133
4w³=128
w³​=32\(\displaystyle \ \ \ \ \ \ \ \ \)All of these three steps are wrong.

But, I checked the real solution(s) on a computer for \(\displaystyle \ \ 4w^3 \ + \ 20w^2 \ - \ 133 \ = \ 0. \)


There is one irrational solution.

But, being as this is a cubic equation, this is either A) a problem with one or typos given by you and/or the assignment,
B) it was meant to give an estimate using a calculator/computer, or C) it's an unreasonable assignment by the instructor.

 

[Ishuda]

I have a test tomorrow and can't figure out how to solve this problem. Here it is: The height of a rectangular box is 4 time its width and its length is 5 more than its width. The volume of the box is 133. Find the width of the box. I have this:
V=LWH
H=4w
L=w+5

I plugged H and L in the V formula but the numbers don't look right. I tried Foiling and finding the zeros. I would really appreciate the help!!!!:-----(

Are you sure that 133 isn't 112 or maybe 288?

 

[Robin2017]

Thank you all for your responses! :-D:-D:-D:-D The information provided is right. The problem does say to "use a graphic calculator to solve a word problem involving polynomial of degree three", so I guess it should be an irrational number.

 

[stapel]

Thank you all for your responses! :-D:-D:-D:-D The information provided is right. The problem does say to "use a graphic calculator to solve a word problem involving polynomial of degree three", so I guess it should be an irrational number.

Thanks for letting us know. In future, kindly please provide the full text of the exercise, including the complete instructions.