Finding zeros(dimensions) of a box using the rational zero theorem

[erikj98]

Finding zeros(dimensions) of a box using the rational zero theorem:

The length is twice as long as the width. The height is 2 inches greater than the width. The volume is 192 cubic inches.

 

[mmm4444bot]

The length is twice as long as the width. The height is 2 inches greater than the width. The volume is 192 cubic inches.

What have you done or thought about, so far?

Please read and follow the forum guidelines. Thank you. :cool:

 

[tkhunny]

What is the width?

If you don't know, just call it something so you can talk about it. "w" is pretty popular.

Okay, that's about half the work required.

 

[erikj98]

What is the width?

If you don't know, just call it something so you can talk about it. "w" is pretty popular.

Okay, that's about half the work required.

i dont have the width. my work so far is
L=2w
W=w
H=w+2

2w³+4w²=192

the answer to the problem is 8x4x6 but i only know that because it is in the back of the book.

 

[Deleted member 4993]

i dont have the width. my work so far is
L=2w
W=w
H=w+2

2w³+4w²=192

the answer to the problem is 8x4x6 but i only know that because it is in the back of the book.

Great!

Now your equation is:

2w³+4w² - 192 = 0 or

f(w) = w³ + 2w² - 96

Now apply rational root (zero) theorem. What are the possible rational roots of the function [f(w)] you have deduced?

 

[erikj98]

Great!

Now your equation is:

2w³+4w² - 192 = 0 or

f(w) = w³ + 2w² - 96

Now apply rational root (zero) theorem. What are the possible rational roots of the function [f(w)] you have deduced?

i was also told to use 0 as a place holder. would that make it 2w³+4w²+0 -192 = 0 ? and then use the rational root theorem?

 

[Deleted member 4993]

i was also told to use 0 as a place holder. would that make it 2w³+4w²+0 -192 = 0 ? and then use the rational root theorem?

Can please tell us:
What is rational root theorem (as you know it)?

 

[erikj98]

-96: positive/negative 1,2,4,6,8,12,16,24,48,96
1: 1

 

[tkhunny]

i dont have the width.

That's where you are wrong. The Width is "w". You just said so.

 

[lookagain]

-96: positive/negative 1,2,4,6,8,12,16,24,48,96
1: 1

You're missing a positive/negative pair:

+/- 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

The width must be positive.

Also, you can speed up this process. Rewrite the equation:

\(\displaystyle w^3 + 2w^2 - 96 = 0\)

\(\displaystyle w^3 + 2w^2 = 96\)

Factor:


\(\displaystyle w^2(w + 2) = 96 \ \ \ \ or\)

\(\displaystyle (w)(w)(w + 2) = 96\)


The larger numbers in the list of 12 possible positive roots will quickly make the product on the left-hand side
of the equation too big. So, try the smallest roots in the product and proceed from there.



And, when you give the dimensions of the box, include the units, inches, with each one in your answer.

 

[Deleted member 4993]

[FONT=MathJax_Math-italic]w^{[FONT=MathJax_Main]2[/FONT]}[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]w[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]) [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Main]96[/FONT][FONT=MathJax_Main] →[/FONT]since only 1, 4 and 16 are possible square factors of 96, one of those possibly is a solution. Which one? [/FONT]