# I NEED MAJOR HELP!(Question on Polynomials)

• Sep 25th 2007, 07:54 PM
zarlock99
I NEED MAJOR HELP!(Question on Polynomials)
Ok so this is my tips question which is due tomorrow for marks and I havnt been able to think how to do it.

The Volume of a cylindrical can is
4(pi)x^3 + 28(pi)x^2 + 65(pi)x + 50(pi) cm^3. If the can has a height of
(x +2)cm, what is the radius of the cylinder? Now so far I figured the way to figure this out is to take the formula for Volume of a cylinder which is V=pi*r^2 *h and rearange is so your solving for r^2 instead so it would be r^2=(pi)(h)(V). The problem is im pretty sure I need to factor that equation out or expand w/e it is. I know how to expand out an equation where theres just x2 + x + 0 lets say but I dont know how to expand it when you have that extra x so its now x3 + x2 + x + 0. Can someone plz help me right now whoever's on at this time? Thanks!
• Sep 25th 2007, 08:02 PM
Jhevon
Quote:

Originally Posted by zarlock99
Ok so this is my tips question which is due tomorrow for marks and I havnt been able to think how to do it.

The Volume of a cylindrical can is
4(pi)x^3 + 28(pi)x^2 + 65(pi)x + 50(pi) cm^3. If the can has a height of
(x +2)cm, what is the radius of the cylinder? Now so far I figured the way to figure this out is to take the formula for Volume of a cylinder which is V=pi*r^2 *h and rearange is so your solving for r^2 instead so it would be r^2=(pi)(h)(V). The problem is im pretty sure I need to factor that equation out or expand w/e it is. I know how to expand out an equation where theres just x2 + x + 0 lets say but I dont know how to expand it when you have that extra x so its now x3 + x2 + x + 0. Can someone plz help me right now whoever's on at this time? Thanks!

$V = \pi r^2 h$

$\Rightarrow r^2 = \frac V{\pi h} = \frac {4x^3 + 28x^2 + 65x + 50}{x + 2}$

now, do polynomial long division (or synthetic division) to find $r^2$, then square root both sides when done
• Sep 25th 2007, 08:07 PM
zarlock99
But what happened to the pi's within the equation for the volume?
• Sep 25th 2007, 08:11 PM
Jhevon
Quote:

Originally Posted by zarlock99
But what happened to the pi's within the equation for the volume?

i factored them out and canceled them with the $\pi$ in the denominator
• Sep 25th 2007, 08:17 PM
zarlock99
So when I use synthetic devision i get this
x3 x2 x
-2| 4 28 65 50
-8 -40 -50
_______________
4 20 25 0 R=0

So would that be 4x^3 + 20x^2 + 25x + 0? Then how so I take the square root of that?
• Sep 25th 2007, 08:19 PM
zarlock99
By the way, the answer is supposed to be 2x + 5, I just need to know how to get that answer.
• Sep 25th 2007, 08:23 PM
Jhevon
Quote:

Originally Posted by zarlock99
So when I use synthetic devision i get this
x3 x2 x
-2| 4 28 65 50
-8 -40 -50
_______________
4 20 25 0 R=0

So would that be 4x^3 + 20x^2 + 25x + 0? Then how so I take the square root of that?

no, you just divided, how would you end up with the same degree polynomial?

the answer is $4x^2 + 20x + 25$

now take the square root of that and that is r
• Sep 25th 2007, 08:31 PM
zarlock99
ooooooooo I see, iunno but for some reason when I was looking through my notes I thought I saw after using synthetic devision u still stick with the x3, x2, and x but it never made sence, now i does tho. Thanks man a lot, u just saved me from losing marks.
• Sep 25th 2007, 08:42 PM
zarlock99
well actually its not completely resolved unless the answer that was given is wrong. When I square root that I get 2x + 4.47 + 5, I got the 4.47 by taking the square root of 20x, but the answer that was given is (2x + 5)
• Sep 25th 2007, 09:15 PM
Jhevon
Quote:

Originally Posted by zarlock99
well actually its not completely resolved unless the answer that was given is wrong. When I square root that I get 2x + 4.47 + 5, I got the 4.47 by taking the square root of 20x, but the answer that was given is (2x + 5)

you cannot take square roots like that!

anyway, you just foil: $4x^2 + 20x + 25 = (2x + 5)(2x + 5) = (2x + 5)^2$

so $r^2 = (2x + 5)^2$

$\Rightarrow r = 2x + 5$