# how do i found the the height

• Jun 25th 2012, 10:14 AM
waleedrabbani
how do i found the the height
The volume of a rectangular box is (x^3+6x^2+11x6). The
box is (x+3) cm long and (x+2) cm wide. How high is the box?

May I also add that this is in the long division section of the text book so i am assuming i will have to apply long division in it somewhere, i actually do not understand where to start, I am not looking for an answer, just ways that i can tackle the problem. thank you.
• Jun 25th 2012, 11:12 AM
Plato
Re: how do i found the the height
Quote:

Originally Posted by waleedrabbani
The volume of a rectangular box is (x^3+6x^2+11x6). The
box is (x+3) cm long and (x+2) cm wide. How high is the box?

May I also add that this is in the long division section of the text book so i am assuming i will have to apply long division in it somewhere, i actually do not understand where to start, I am not looking for an answer, just ways that i can tackle the problem. thank you.

Divide the volume by (x+3).
You should get a reminder of zero.
Take the quotient and divide by (x+2) .
You should get a reminder of zero.
That quotient is the height.
• Jun 25th 2012, 09:11 PM
Prove It
Re: how do i found the the height
Quote:

Originally Posted by waleedrabbani
The volume of a rectangular box is (x^3+6x^2+11x6). The
box is (x+3) cm long and (x+2) cm wide. How high is the box?

May I also add that this is in the long division section of the text book so i am assuming i will have to apply long division in it somewhere, i actually do not understand where to start, I am not looking for an answer, just ways that i can tackle the problem. thank you.

As an alternative, for any prism I always remember that the volume is equal to the cross-sectional area multiplied by the height.

You are told the length is \displaystyle \displaystyle \begin{align*} x + 3 \end{align*} and the width is \displaystyle \displaystyle \begin{align*} x + 2 \end{align*}, so the cross sectional area is \displaystyle \displaystyle \begin{align*} (x + 3)(x + 2) = x^2 + 5x + 6 \end{align*}.

Therefore, you have

\displaystyle \displaystyle \begin{align*} V &= AH \\ x^3 + 6x^2 + 11x + 6 &= \left(x^2 + 5x + 6\right)H \\ H &= \frac{x^3 + 6x^2 + 11x + 6}{x^2 + 5x + 6} \end{align*}

Now perform this long division to simplify the height :)