Finding the dimensions

1. Sep 25, 2006

thomasrules

A box has dimensions that are linear factors of (x^3-7x^2+14x-8) cubic centimetres. What dimensions give a volume of 12 cm cubed?

I started it off by moving 12 to the left side giving the equation:

x^3-7x^2+14x-20=0

then from then I don't know....If I substitute 5, the equation is satisfied and is a factor but from there it doesn't work. Plus 5 won't really make it 12 so...i'm lost

2. Sep 25, 2006

river_rat

Okay, so $$(x-5)$$ is a factor of your cubic. Now factorise it, and see what the other factors are.

Edit: No wait - checked the other factors, are you sure your equations are correct?

Last edited: Sep 25, 2006
3. Sep 25, 2006

Data

Well, remember that the dimensions are the factors of $x^3 - 7x^2 + 14x - 8$, not of $x^3 - 7x^2 +14x - 20$.

You need to solve $x^3 - 7x^2 +14x -20 = 0$ and plug the solution for $x$ back into the factors of $x^3 - 7x^2 +14x - 8$ to get the dimensions.

By the way, these are very odd questions you're being given!

4. Sep 25, 2006

river_rat

Ahh! Didnt notice he had changed his cubics between lines - that should sort out the problem :)

5. Sep 25, 2006

thomasrules

dont know what u mean by find x because when I plug in 5 into the 2nd equation with the -20, I get = 0

From there I can't factor any further
And you say plug back into the first one? what will that do? It will only give me =12

6. Sep 25, 2006

Data

Good! So you've found one solution, x=5. That means x-5 is a factor of the cubic. Just use long division to reduce it to a quadratic and find the other roots (I'm not promising there will be any other real roots!).

Then all you have to do is to factor $x^3-7x^2+14x-8$ and plug x=5 into each of the factors to find the dimensions.

7. Sep 25, 2006

thomasrules

I did that but I used short division and got:

$$(x-5)(x^2-2x+4)=x^3-7x^2+14x-20$$

I can't get $$x^2-2x+4$$ to equal 0

and you can't factor the first equation with synthetic division

8. Sep 25, 2006

Hargoth

use $x^2 - 2x + 4 = (x^2 - 2x + 1) -1 +4 = (x-1)^2 + 3$.

9. Sep 25, 2006

thomasrules

but that doesn't do me any good hargoth

10. Sep 25, 2006

Hargoth

Whoops, I didn't remember you were looking for real solutions.
I don't see why you set $x^3-7x^2+14x-8=12$, since it isn't said that this polynomial is the volume. I'd suggest factorising $(x^3-7x^2+14x-8)$ instead, and reading off the dimensions of the box from the factorization. I think that is what is asked for.
edit: Nope that doesn't make sense either. You were of course correct.

Last edited: Sep 25, 2006
11. Sep 25, 2006

Data

He needs to factorise $x^3-7x^2+14x-8$ and solve $x^3-7x^2+14x-8=12$ (see my previous posts).

You can just use the quadratic formula on $x^2-2x+4=0$ (it doesn't have any real roots though).

All the roots of $x^3-7x^2+14x-8$ are small integers, so it should be easy for you to factor .

12. Sep 25, 2006

Hargoth

I don't think so, because, if the dimensions are linear factors of (x^3-7x^2+14x-8), say, (x-a), (x-b), (x-c), so the Volume is (x-a)(x-b)(x-c), but this is equal to the first polynomial, so setting
$(x-a)(x-b)(x-c) = x^3-7x^2+14x-8=12$ was in my eyes the right thing to do.
edit: And if this equations yield a complex solution, it doesn't help that the other linear factors are real - because the dimensions of the box you get by subtracting some real number from a complex one keep their imaginary part. This problem beats me ...

Last edited: Sep 25, 2006
13. Sep 25, 2006

Data

yes, it is. You can use that equation to solve for $x$, and then the box has dimensions $(x-a) \times (x-b) \times (x-c)$. So you need to solve $x^3-7x^2+14x-8=12$ as well as factorise $x^3-7x^2+14x-8$, as I said.

14. Sep 25, 2006

Hargoth

But this doesn't help in my eyes, see my edit.

15. Sep 25, 2006

thomasrules

but w/e i can't even factorise that though the last part

16. Sep 25, 2006

Data

You have to find the real solution(s) of $x^3-7x^2+14x-8=12$, yes, because otherwise you'll have a box with complex side lengths (in this case, there's only one real solution).

The roots of $x^3-7x^2+14x-8$ (ie. the a, b, c s.t. $(x-a)(x-b)(x-c) = x^3-7x^2+14x-8$) are all real, so everything's fine.

Last edited: Sep 25, 2006
17. Sep 25, 2006

Hargoth

Okay, data is right. You can insert your solution x=5 into the factorization of the polynomial you are told to do in the assignment to get the dimensions of the box. Sorry for causing confusion ...

18. Sep 25, 2006

thomasrules

don't worry thanks for the help anyway, any input is good

19. Sep 25, 2006

HallsofIvy

Staff Emeritus
5 is the only real solution to x3- 7x2+ 14x- 8= 12.
You also need to know that x3- 7x2+ 14x- 8 factors as (x- 1)(x- 2)(x- 4). Put x= 5 into each of those factors to get the dimensions of the box.