1. ## help?

I am doing advance functions, which I think is pre-calculus. This is my question.

The volume, in cubic centimeters, of a square based box is given 9x^3 + 24x^2 - 44x + 16. Determine possible dimensions of the box if the area of the base, in square centimeters, is 9x^2 - 12x + 4. How do I find the answer?

2. Originally Posted by Barthayn
I am doing advance functions, which I think is pre-calculus. This is my question.

The volume, in cubic centimeters, of a square based box is given 9x^3 + 24x^2 - 44x + 16. Determine possible dimensions of the box if the area of the base, in square centimeters, is 9x^2 - 12x + 4. How do I find the answer?
You're meant to realise that $\displaystyle 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(x + 4)$.

3. Originally Posted by mr fantastic
You're meant to realise that $\displaystyle 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(x + 4)$.
how did you do that? Can you show me step by step? I do not understand how to get that.

4. Originally Posted by Barthayn
how did you do that? Can you show me step by step? I do not understand how to get that.
For a regular solid, Volume = (Area of base) (Height). So it was only natural to try and factorise the volume function using the given area function as one of the factors.

By the way, you should note that the area function also factorises - perfectly ..... You need to factorise it so that you can get an expression for all dimensions (length, width and height).

5. thanks, however, I am still lost on how you got $\displaystyle 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(x + 4)$

6. Originally Posted by Barthayn
thanks, however, I am still lost on how you got $\displaystyle 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(x + 4)$
Think about it ...... If 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(linear factor) and the product of these two factors has to give you a 9x^3 and also a 16, what MUST the linear factor be .....

7. thank you for trying to help me. I got the $\displaystyle 9x^2-12x+4$ fully factored to get me the zeros. However, I have no idea how did you get (x+4) from. Can you show me this?

8. Originally Posted by Barthayn
thank you for trying to help me. I got the $\displaystyle 9x^2-12x+4$ fully factored to get me the zeros. However, I have no idea how did you get (x+4) from. Can you show me this?
Originally Posted by Mr Fantastic
Think about it ...... If 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(linear factor) and the product of these two factors has to give you a 9x^3 and also a 16, what MUST the linear factor be .....
$\displaystyle 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(ax + b)$.

When you expand the right hand side you have to get (among other things) 9x^3 and 16. What does that mean that the value of a and the value of b HAS to be?!

9. thanks, I understand now how you got it. However, is there an easier way to get the (x+4) answer an easier mathematic way?

10. Originally Posted by mr fantastic
$\displaystyle 9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(ax + b)$.

When you expand the right hand side you have to get (among other things) 9x^3 and 16. What does that mean that the value of a and the value of b HAS to be?!
I found out the answer the mathematical way. Would dividing the a value in the first one by the second one will always give me the correct answer for the Ax value for the new binomial? And would dividing the D value (the one without the x) give by the c value give me the answer for the second value? For example:

$\displaystyle 9x^3 / 9x^2 = x. 16 / 4 = 4.$
Therefore the binomial is (x+4).