help?

**Barthayn** · September 25th 2009, 04:12 PM

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?

**mr fantastic** · September 25th 2009, 04:25 PM

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 $9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(x + 4)$ .

**Barthayn** · September 25th 2009, 04:40 PM

Originally Posted by mr fantastic

You're meant to realise that $9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(x + 4)$ .

how did you do that? Can you show me step by step? I do not understand how to get that.

**mr fantastic** · September 25th 2009, 04:46 PM

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).

**Barthayn** · September 25th 2009, 05:00 PM

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

**mr fantastic** · September 25th 2009, 05:10 PM

Originally Posted by Barthayn

thanks, however, I am still lost on how you got $ 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 .....

**Barthayn** · September 26th 2009, 02:45 PM

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

**mr fantastic** · September 26th 2009, 03:24 PM

Originally Posted by Barthayn

thank you for trying to help me. I got the $9x^2-12x+4$ 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 .....

$9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(ax + b)$ .

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?!

**Barthayn** · September 26th 2009, 05:48 PM

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

**Barthayn** · September 26th 2009, 05:57 PM

Originally Posted by mr fantastic

$9x^3 + 24 x^2 - 44x + 16 = (9x^2 - 12x + 4)(ax + b)$ .

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:

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

Thread: help?

LinkBack

Thread Tools

Display

help?