Math Help!

[o1lucky132005]

Alright, i hope you guys are good with math, cuz i'm not. I've tried working these problems forever, and i just can't figure them out!



The directions say: State the number of complex roots of each equation. Then find the roots and graph the related function.



I can do everything except finding the roots, i can't figure out how. I know you have to factor it somehow, but i can't figure out these two, hopefully someone can help. (note: ^ means to the power of)



1. t^3 + 2t^2 - 4t - 8= 0



2. 6c^3 - 3c^2 - 45c = 0



I also have 2 more problems from a different assignment that i'm having problems with (is it obvious that i'm a little behind in my math homework?) they are:



1. The diameter of a circl has endpoints at (1, -6) (-3, -4). Find the equation for this circle in general form.



2. A circle has a center at (3,5) and passes thru the point (-2, 2). Find the equation.



Thanks ahead of time for any help!



Megan

[junglemonkey]

Right this is convienient since I have my higher maths exam tomorrow and the circle and polynomials are in it, so I'll give it a go. If I'm wrong, please forgive me lol.



For the first question you use synthetic division. It's difficult to show it on the net but if you have any questions or can't follow it just ask.

So first of all you try putting numbers in and find one that is equal to 0. To do this, write out all the factors of the last number. You're working with the idea that if f(a) = 0 then x - a is a factor

so h = + or - 8, + or - 1, + or - 2, + or - 4 (don't have the wee symbol for "plus or minus") To try them you can make it easier if you have a Ti-83 or 82. The number you want to try, say it's one, you press the following



1. Type the function in exactly as you have it up there but substitute the 2. T's for X's using the X,T,theta button, then press enter

1 then STO- then the X, T, theta button (Don't have the theta symbol lol) Press enter.

3. If that number isn't suitable store another number and press second function adn enter until you see your function appear on the screen again and press enter. Otherwise you can just type them all in indiviually, I just find that method less time consuming



So start trying them:-

f(-1) = -3

f(1) = -9

f(-2) = -2

f(2) = 0, so you have your factor!!



So if f(2) = 0 then x - 2 is a factor.



What you now do is draw up a table like this, with the coefficients of your function at the top but you have lines under the lower two and next to it. And what you want to do is take the one down, multiply it by the two, then take it up diagonally to under the two then add as you go down to under the line, then multiply by the two etc (this probably won't trun out right on the net) Also, ignore the underscores, its just so it looks more like it's supposed to

1____2_____-4_____-8







2 _______ 2______8_____8

---------------------------------

____1____ 4______4_____0 <--- That 0 means your factor is right.

____^^^^^^^^^^^

That's the co-efficients of your other term.



So now you have (t-2)(t? + 4t + 4) which can factorise further to:-



(t-2)(t-2)(t-2)

=(t-2)?



-------------------------------------------------------------------



In this one you have to be careful since there is no number at the end. In your synthetic division table you put 0 but you DO still have to count it. I did the question in the same way as last time but got

(c-3)(6c?-15) which factorises even further to

3(c-3)(2c?-5) I think.



-------------------------------------------------------------



To find the equation of a circle you need the centre and the radius length. It didn't specify a specific form to put it in, so I'll just assume.



Using the midpoint formula, you can get the centre.



m = (the average of the two x co-ordinates, the average of the two y co-ordinates)

so that turns out to be (-3+1/2 , -4+(-6)/2)

=(-1, -5)



So now you need the radius length which can be done using the distance formula. You can either find the diameter and half it, or you can use the point you just found. I suggest using the first way just incase the point is wrong.

=√(difference between the x co-ordinates)? + (difference between the y co-ordinates)?



That, for me turns out to be √20/2 which is where something may have gone wrong. But! with a little bit of fraction work, you square that to get 20/4 = 5, so r = 5



So you put that into the generic circle equation and you get

(x+1)? - (x+5)? = 5



------------------------------------------------------------------



If you draw a diagram of the circle, you'll find that you can join up the two points to make a radius. I used the distance formula and got √34 (either I am wrong or you're getting really bizarre figures for your hw)



So put that into the equation and you get:-

(x-3)? + (y-5)? = 34

[Erowid]

aight hope I can help a lil bit... Ummm to find the factors for synthetic division much faster then just plugging them in, (if you have a graphing calc) GRAPH it and see where the polynomial intersects the X axis. and what you do is graph this equation then a line at y=0 and press 2nd-Trace and then 5, then you scroll to the point that you want to find and press enter 3 times. so your intersections are 2 and -2 and I think one of them is a double root but I don't have time to figure it out. hope I was a lil bit of help. Now for circles... I just totally forgot... sry

[Michie]

So now you have (t-2)(t? + 4t + 4) which can factorise further to:-



(t-2)(t-2)(t-2)

=(t-2)?

Sorry junglemonkey...but I did that question (my way-traditional factoring) and you've made an error:



(t^2 + 4t + 4) - does not end up being (t - 2) (t - 2)...because there's a PLUS sign next to 4t.



Here was my solution from where you erred:



(t-2)(t^2 + 4t +4) = 0

(t-2)(t+2)(t+2) = 0



Therefore, the roots should be at

(t-2) : +2



Because : ((2) - 2) = 0

And when you multiply (0) ((2) + 2) ((2) + 2) you get 0



And



(t+2): - 2



Because: ((-2) + 2) = 0

And when you multiply ((-2) -2) (0) (0) you get 0 again.



I believe that's correct. Check your graphic calc. And good luck.

[Michie]

2. 6c^3 - 3c^2 - 45c = 0

Heres how I did this one:

First I factored out the common factors, which is 3c



3c (2c^2 - c - 15)



Then I further worked on the expression:

I factored through to:

3c (2c + 5) (c - 3)



Therefore, the roots are:



3c = 0 : 0 multiplied by anything is zero



(2c + 5): if you take the second number in the binomial and find it's opposite, the opposite is - 5. Firthermore, you have to consider the 2 in 2c. You have to cancel that out too



you end up with (2c + 5)--> c= -5/2

Plug that in- 2 (-5/2) + 5 = = (-10/5) + 5. Which is essentially -5 + 5, which equals zero.



3(-5/2) (0) (-5/2 -3) equals zero.



And the last:

(c - 3) - c= +3



I think thatis it. Check the calc again. I hope this helps. It's hard to explain math through computers communication.

[o1lucky132005]

hey, i just wanted to tell everyone thanks, you all helped a bunch!



Love,

Megan

[Michie]

No prob Megan. If you ever need more help in math, you can post again

[junglemonkey]

thanks Michie I knew I'd made a mistake somewhere There had to be a repeated root for it to just touch the x-axis. I used the discriminant on it to find out.



Erowid, I prefer to do maths algebraically. We have a non-calculator paper so we have to be able to do stuff like that without one.