anonymous
  • anonymous
Can someone teach me how to factor a 5 termed polynomial? here is the polynomial function: 4x^4 + 5x^3 - 14x^2 - 4x +9
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
if it is designed well enough, the rational roots thrm would apply to be able to knock it doen a degree of two
amistre64
  • amistre64
if you solve the setup for "x" you can run some recursions to get approximations of a root.
amistre64
  • amistre64
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I really don't get it. Can you explain it straight forward?
anonymous
  • anonymous
and simple :) pretty please?
amistre64
  • amistre64
the "simple" method relates to the rational roots thrm
amistre64
  • amistre64
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
amistre64
  • amistre64
9 factors to: 1 3 9 4 factors to: 1 2 4 possible rational roots are therefore composed of some combination of: \[\pm\frac{1,3,9}{1,2,4}\]
amistre64
  • amistre64
since the simplest option to test is "1" .... give it a shot when x=1, do we get a zero?
anonymous
  • anonymous
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
amistre64
  • amistre64
by design, 1 is also a double root, which would simplify this down to the usual quadratic stuff
amistre64
  • amistre64
... my eyeballing the graph may be in error tho ... looks lke a double lol
anonymous
  • anonymous
i would use technology yeah, 1 is not a double root, just plays one on tv
amistre64
  • amistre64
ugh, this thing was not "well designed" at all now was it
anonymous
  • anonymous
no there is no good way to factor this in general there isn't so at least the question is honest
amistre64
  • amistre64
1 Attachment
anonymous
  • anonymous
yeah those were the numbers that sprang in to my head first thing
amistre64
  • amistre64
:)
anonymous
  • anonymous
i was just checking by substitution that my guesses were right
amistre64
  • amistre64
so, i spose this is a linear and a cubic prime?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=+4x^4+%2B+5x^3+-+14x^2+-+4x+%2B9
amistre64
  • amistre64
how can a cubic be prime? arent they spose to have at least one real root at all times ....
anonymous
  • anonymous
\[(x-1) (4 x^3+9 x^2-5 x-9)\]
anonymous
  • anonymous
this one has 4 roots usually "prime" means you can't factor using integers
anonymous
  • anonymous
at least i think that is what "prime" means so for example \[x^2+4x-2\] is prime even though it has two zeros
anonymous
  • anonymous
Hmm. I see, thank yooou for explaining. Now, i understand ^_^v Arigato gozaimas.
anonymous
  • anonymous
Another one: how to factor this? x^3-7x+6 I factored it, i got (x^2-1) (x-6), but it's wrong.
amistre64
  • amistre64
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
anonymous
  • anonymous
Hmmm. I see, okaaay. Thanks again @amistre64 v^_^v

Looking for something else?

Not the answer you are looking for? Search for more explanations.