# 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.