lol thats great but my teach is pregnant and wants us to show work.. which means lots of synthetic division. With the calc I can figure out what roots will work though right? So I wont have to do every one?

LOL, graphic calculator....

6x^3 - x^2 - 13x + 8 = 0

-> 6x^3 - x^2 - 5x - 8x + 8 = 0

-> x ( 6x^2 - x - 5) - 8 (x - 1) = 0

-> x (6x + 5) (x - 1) - 8 (x - 1) = 0

-> (x-1) (6x^2 + 5x - 8) = 0

And I am sure you can solve that now, right?!

x = 1, (-5+-(25+4*8*6)^(0.5))/12

SUCH a long way of doing it [sm=tongue_smile.gif]


just factor it out and then use quad formula to get roots. <3 easy math

Hee hee... I thought we are supposed to show our work?! hee hee..

Maybe you can show how you factor it out... I am sure the original poster would appreciate it!

Well i took applied calc last semster and our teacher did not care if we showed work or used our caculator. It is truly amazing all the stuff the caclulator can do once you know how to use it.

It is truly amazing how you don't know how to do it properly once you know how to use the calculator.. (joke, just messin' w/ you Nick.. ha!)

What calculator do you use?! TI?

oo.. are you supposed to show that? aahh. that equation is so simple i didnt think much showing work was necessary.



TI rocks my world btw.. 5 years in a row

lol... I actaully dont rember somethings too good. [:@] It sucks. But after seeing your examples i think i could run through it the old fashion way. I also rember sythentic division. Which i did not think that was that hard. Just kinda weird.

But i have a TI-84 plus.

How would you factor it out, though.. just want to see your thought process...

TIs are easy to use, right?!

I think I had a TI-81 ages ago, but then I sold it... and got a HP 48G or something in that name at high school... (again, a long time ago, ha!) I still got that calculator today, ha!!

yeah TI's are great. I wuold say 99.9999% of people in school use them. There are very few of the other brands.

sry, i meant factoring by division. i see it as a quicker way of doing it.

How would you factoring by division?

(No worry, I know it is not a hard problem.. but just curious on how your thought process work.. just want to see how you would attack this problem...)

See what you do is divide the multiude dynamic by flux capacitor rate and square that with hexdiagonal portional rate. and your change is a dollar two ninty five.

Hope that helps

I got one of them Flux capacitors in my RC...

Should learn how to do it without the calculator now. In college no calculators are allowed in calculus tests.

I've taken many calculus classes in college and we've always been able to use calculators on tests.

ahhh the days of math. i hated trig at first but once it clicked, it was kinda fun for me. as for calculus, no idea how to do that since i never took the class

Most of the math classes / tests I have taken -> it won't help if you got a calculator... ha!

lol well I figuresd it out today, kinda..

I have another ? tho.

Find all the zero's of the function.

f(x) = x^4 -8x^3 +13x^2 +32x -68

all my maths classes caculators are allowed. Teachers are starting to see the real world where no one who needs this math dose it by hand. I am working on my last math class now Stats. Already taken college algb and applied calc.

Nah, got to be hardcore, man.. no calculators... (joke)

What?! More?! Geez...

I guess teachers nowadays teach whatever synthetic division? (I never heard of that before..) ha!

I guess it just means you try some #s to see if you get it to be 0.. something like that.. someone needs to teach me that.. ha! (Let me Google it.. !)

Anyway, this might be hard to see, kind of like magic.....

x^4 - 8x^3 + 13x^2 + 32x - 68

= x^2 (x^2 -8x + 12) + x^2 + 32x - 68

= x^2 ( x-6) (x-2) + (x+34) (x-2)

= (x-2) (x^3 - 6x^2 + x + 34)

= (x-2) [x^3 +2x^2 - 8x^2 + x + 34]

= (x-2) [x^2(x+2) - (8x-17)(x+2)]

= (x-2) (x+2) (x^2 -8x + 17)

= (x-2)(x+2)(x -(4+i))(x -(4-i)) (u guys learned complex #s yet?)

oh yea.. synthetic division is the name.. few already mentioned it.
http://www.purplemath.com/modules/synthdiv.htm