Question

Can someone teach me how to factor a 5 termed polynomial? here is the polynomial function: 4x^4 + 5x^3 - 14x^2 - 4x +9

Answer 1

if it is designed well enough, the rational roots thrm would apply to be able to knock it doen a degree of two

Answer 2

if you solve the setup for "x" you can run some recursions to get approximations of a root.

Answer 3

4x^4 + 5x^3 - 14x^2 - 4x +9 = 0

4x^4 + 5x^3 - 14x^2 +9 = 4x

(4x^4 + 5x^3 - 14x^2 +9)/4 = x

make a guess: g

(4g^4 + 5g^3 - 14g^2 +9)/4 = x'

then use x' as the new guess ... and it should start to converge to one of the roots

Answer 4

I really don't get it. Can you explain it straight forward?

Answer 5

and simple :) pretty please?

Answer 6

the "simple" method relates to the rational roots thrm

Answer 7

define the factors of the last term; and the first term coefficient

divide the last by the first to develop a pool of possible roots to test out

Answer 8

9 factors to: 1 3 9
4 factors to: 1 2 4

possible rational roots are therefore composed of some combination of:
\[\pm\frac{1,3,9}{1,2,4}\]

Answer 9

since the simplest option to test is "1" .... give it a shot

when x=1, do we get a zero?

Answer 10

there is no easy way to do this
you have to find a zero first
fortunately for you, you can check real fast that \(f(1)=0\) and then divide the polynomial by \(x-1\) to factor it

Answer 11

by design, 1 is also a double root, which would simplify this down to the usual quadratic stuff

Answer 12

... my eyeballing the graph may be in error tho ... looks lke a double lol

Answer 13

i would use technology
yeah, 1 is not a double root, just plays one on tv

Answer 14

ugh, this thing was not "well designed" at all now was it

Answer 15

no there is no good way to factor this
in general there isn't so at least the question is honest

Answer 16

yeah those were the numbers that sprang in to my head first thing

Answer 17

:)

Answer 18

i was just checking by substitution that my guesses were right

Answer 19

so, i spose this is a linear and a cubic prime?

Answer 20

http://www.wolframalpha.com/input/?i=+4x^4+%2B+5x^3+-+14x^2+-+4x+%2B9

Answer 21

how can a cubic be prime? arent they spose to have at least one real root at all times ....

Answer 22

\[(x-1) (4 x^3+9 x^2-5 x-9)\]

Answer 23

this one has 4 roots
usually "prime" means you can't factor using integers

Answer 24

at least i think that is what "prime" means so for example
\[x^2+4x-2\] is prime even though it has two zeros

Answer 25

Hmm. I see, thank yooou for explaining. Now, i understand ^_^v Arigato gozaimas.

Answer 26

Another one: how to factor this? x^3-7x+6

I factored it, i got (x^2-1) (x-6), but it's wrong.

Answer 27

trial and error, try using x=1, or x=-1 as a simple test.

there is also a opposite sign rule:

on a continuous function; if f(a) is positive; and f(b) is negative ... then there has to be some value between a and b that produces a root

Answer 28

Hmmm. I see, okaaay. Thanks again @amistre64 v^_^v