BF2S Forums Math Help

well, for 3, use the distance formula, but use general formulas for each x and y you plug in
for 1 (x(x-1)(x-2)(x-4)) and then see what you have to do to that to go thru the point

these are just quick guesses, i got a lot of work of my own

maybe you should post what you've done so far with specific questions,like my teachers always say

Last edited by argo4 (2008-12-08 16:14:52)

For number one, since your roots are 0,1,2,4 start off with p(x) = (x-0)(x-1)(x-2)(x-4), then sub in the conditions.


If you want, I could open up maple on the other computer and plug question 2 in...

I still can't get question four...

I don't even know how to draw it minty. I typed it word for word from my review book.

Grade 11.

Quick side question.
How do I simplify this? Can it be simplified?

Is this higher then AP Calculus in the states? lol?

Ryan wrote:

I know you guys are math guru's, so if you don't mind helping me out...
Try to explain as best as possible.

1. Determine the equation of the function in factored form that has the following characteristics: degree 4, zeroes (or roots if thats what you call them) 0,1,2,4; and goes through the point (3,9).

2. Determine the remainder when the polynomial is divided by x + square root of 2,
a) 3x³-2x²+7x-6  b) x^4-2x³+3x²+4x+5

3. Determine the shortest distance from the circle (x-5)²+(y+1)²=5 to the line x+3y=18

4. (I just to see a drawing of what this looks like, maybe in paint). A tangent is drawn from point R to each of two concentric circles with center O. The tangents are on the same side of center. Points P and Q are the points of tangency. Sketch the conditions described and prove that angle PRQ equals angle POQ.

Thank you guys, any help will be greatly appreciated.

you're lucky i'm bored out of my mind and my girlfriend is 200 miles away

1) fourth degree, or four x's in the factored form, so start by making an equation with the proper zeroes, in this case x(x-1)(x-2)(x-4).  Now, when x=3, y=9, so manipulate one of the terms to satisfy this while still retaining the correct zeroes (this is done by scalar multiplication) the easiest choice is to multiply the x term by -3/2, giving you (-3/2)x(x-1)(x-2)(x-4), and you can unfactor it yourself (hopefully)

2) polynomial division, this gets really ugly long hand, there is an equation for the remainder alone, see if you can locate it.

3) this can be solved by calculus, let me know if you really want to see that solution...

4)

edit:  If you connect P-O-Q, there should be a small angle between them.  remember that tangent lines make right angles with radii, and you should be able to work out the geometry  (if nothing else, pick an arbitrary value for PRQ, and prove the two equal, use a small angle in this case)

Last edited by S.Lythberg (2008-12-09 20:12:21)

BS grade Eleven, im taking a 20 Pure course and i have only seen 2, 3, parts of 1, and a simplified version of 4.

yeh, lets see here

#2, use synthetic division, its a lot quicker for this one.


a) 3x³-2x²+7x-6  b) x^4-2x³+3x²+4x+5


X=/= -4

a)

-4|  3     -2      7      -6
   |
   |
   |_________________
      3     -14    63     -258           

3X²-14X+63  Remainder 258

b)

-4|  1    -2    3    4    5
   |
   |
   |_________________
      1    -6    27 -104 -411

X³-6X²+27X-104 Remainder -411                             
         
...at least thats how i remember shit from 3 months ago. 

To do this shit:

Find the NPV, then align coeffecients in descending order as i did.  If a gap is present (ie X^3 then just X) place a zero in the place where the missing one would be.  Drag down the first coeffecient, multiply by NPV, and add to next coeff.  Repeat process.  Every resultant is one veriable root size down from the number above it (for [b], 3X^2 -> 27X.  4X -> -104.  The resultant w/o any root >0, that is remainder.

I hope im not missing something here.

Last edited by N00bkilla55404 (2008-12-09 22:55:47)

Flaming_Maniac wrote:

=NHB=Shadow wrote:

Is this higher then AP Calculus in the states? lol?

lord no

looking through it now

Ryan wrote:

Grade 11.

Quick side question. http://i23.photobucket.com/albums/b386/ … 1228872343
How do I simplify this? Can it be simplified?

I would hazard a guess at a common denominator. I'm not really sure why you would want to, but you certainly could and I think it would be considered more simplified.

I don't see how the proof works at all, or even that they are even, but I am 99% sure I drew the picture correctly.

http://i53.photobucket.com/albums/g44/F … c/math.jpg

1% :p
Spoiler (highlight to read):
SAME side of the center...
No this is like Algebra 2 in the states... Like the PITA stuff they ask you to do at the end.
Most of it isn't that hard, just bulky.

Looks like you got all of them?

Ryan wrote:

Grade 11.

Quick side question. http://i23.photobucket.com/albums/b386/ … 1228872343
How do I simplify this? Can it be simplified?

(8 * √10 - 25)
-----------------
        5

√5 * (8√2 - 5√5)
=-------------------
             5

Well it's simplified, but doesn't look like it.

EDIT:

OK Ryan, I think I may have worked out question 3.

I first took an implicit differential of the formula of the circle to get:

2(x-5) + 2(y+1)(dy/dx) = 0

Rearranging this gives:

dy/dx = (5-x)/(1+y)

Equating this to the gradient of the line (-1/3) I get an expression y = 3x - 16.

I sub this into the equation of the circle and solve the quadratic to get x = 5 +/- (1/√2)

I sketched the circle and the line and found that the point closest to the line is on the right of the centre (5,-1) so I took x = 5 + (1/√2) and y = (3/√2) - 1. This is the point closest to the line.

To find the distance we use the point line distance formula (or whatever it's called) which you can find here.

Plug this in and you get 2.824 (3 dp).

I hope this is right.

Last edited by Vub (2008-12-10 00:20:31)

VicktorVauhn wrote:

Flaming_Maniac wrote:

=NHB=Shadow wrote:

Is this higher then AP Calculus in the states? lol?

lord no

looking through it now

Ryan wrote:

Grade 11.

Quick side question. http://i23.photobucket.com/albums/b386/ … 1228872343
How do I simplify this? Can it be simplified?

I would hazard a guess at a common denominator. I'm not really sure why you would want to, but you certainly could and I think it would be considered more simplified.

I don't see how the proof works at all, or even that they are even, but I am 99% sure I drew the picture correctly.

http://i53.photobucket.com/albums/g44/F … c/math.jpg

1% :p
Spoiler (highlight to read):
SAME side of the center...
No this is like Algebra 2 in the states... Like the PITA stuff they ask you to do at the end.
Most of it isn't that hard, just bulky.

Looks like you got all of them?

They are both on the right side of center.