Question

Determine the zeros of f(x) = x3 – 12x2 + 28x – 9.

Answer 1

YOU need an answer !? I'm surprised lol ...

Answer 2

LOL.

Answer 3

Method as well.

Answer 4

When you mean zeros do you mean the points of intersection of the graph?

Answer 5

idk, that''s all wht the question says. i guess that's same.

Answer 6

There is a method

Answer 7

y = x^3 – 12x^2 + 28x – 9.
When the graph cuts the x axis y is equal to zero.
0 = x^3 – 12x^2 + 28x – 9.
Solve it using whatever method you like.

x = 9
x = 0.381966
x = 2.61803

I hope what I think your asking is what I'm answering. What is your chapter on?

Answer 8

There's even a formula! (similar to the general solution for quadratic equations), but it's much much much longer :-P

Answer 9

I'll post the method (which I forgot ages ago)

Answer 10

now @chayC.

np man!

Answer 11

how*

Answer 12

?

Answer 13

try completing the cube,

Answer 14

hmm

Answer 15

f(x) = x3 – 12x2 + 28x – 9.
I don't think you can easily complete the cube, so we will use the long method

Answer 16

substitute x for (t - (b/(3a))), which is t + 4

(t+4)^2 - 12(t+4)^2 + 28(t+4) - 9 = 0

Answer 17

oops it's:

(t+4)^3 - 12(t+4)^2 + 28(t+4) - 9 = 0

Answer 18

expand to get

t^3 -20t -25 = 0

Answer 19

my czn showed me this method, it was long as hell.
thanks for your help. :0

Answer 20

it's easier than using the general solution :-P

Answer 21

after all that.. I give up

Answer 22

I think the best way would be to guess the first rational root

Answer 23

lol, nvm.. thanks for your time.. :)

Answer 24

the way to do it is to take the last term, which has factors 2, 3, 9

Answer 25

oh and 1

Answer 26

signed and unsigned

Answer 27

how about solving it by taking commons?

Answer 28

plug in those values, and see which one gives you zero for the function

Answer 29

I got 9, so we can factor out (x-9)

use long division to find out what the rest of the equation looks like after you factor out x-9

Answer 30

that's right..

but how to get x-9 with taking commons?

Answer 31

(x-9)(x^2-3x+1) now it's easy since you can use the general formula for quadratic roots

Answer 32

taking commons..... hmm

Answer 33

i can solve the right part with using quadratic formula.
but how you got x-9?

Answer 34

trial and error :-D

Answer 35

Cool. Thanks.

Answer 36

best to memorize that general method (or the general formula for cubic roots if you're guinness world record genius, but you will be able to derive it if you know the procedure)

Answer 37

Okay, thanks!! :)
i have it in written form, but havn't memorized it.

Answer 38

There is a method that involves complex numbers. I haven't tried it but it might be easier. http://en.wikipedia.org/wiki/Cubic_function#Lagrange.27s_method

Answer 39

let \[\zeta =-\frac{1}{2}+\frac{\sqrt{3}}{2}i\]so that zeta^2 + zeta + 1 = 0

Answer 40

now, set \[s_0 = x_0+x_1+x_2\]
\[s_1 = x_0 + \zeta x_1+\zeta^2x_2\]
\[s_2=x_0 +\zeta^2x_1+\zeta x_2\]

Answer 41

where, x_0, x_1, and x_2 are the possible roots of ax^3+bx^2+cx+d=0

Answer 42

Dang! that's scary. :O

Answer 43

lol. that's Lagrange's ,ethod.

Answer 44

method*

Answer 45

now you can do the reverse Fourier Transform to get the real roots
\[x_0 = \frac{1}{3}(s_0+s_1+s_2)\]
\[x_1 = \frac{1}{3}(s_0+\zeta^2s_1+\zeta s_2)\]
\[x_2 = \frac{1}{3}(s_0 +\zeta s_1+\zeta^2s_2)\]

Answer 46

finding s_0 is straightforward
\[s_0=-\frac{b}{a}=12\]

finding s_1 and s_2 is more involved

Answer 47

trololololol.

he's asking about the zeros of the function, either descartes' or rational theorem.
he's prolly in the fundamental theorem of algebra.