Question

Sat, Myininya?

Answer 1

Fundamental Theorem of Algebra,
p(x) = 0 has a solution in the complex
plane.
Since P (x) has odd degree it must have one real root since the coefficients are real and
the complex zeros come in conjugate pairs



the Rational Root Theorem,
the possible rational roots must be of the form p/q where p divides 114 and q divides 1.

therefore the possible rational zeros are ±1, ±2, ±3, ±6, ±19,±38, ±57
, ±114,


Descartes' Rule of Signs,
f(x)=x3+9x2+15x−114
+ + + -
there is only 1 change so this means there is one possible real root
f(−x)=−x3+9x2−15x−114
- + - -
2 changes so this means there are 2 possible or 0 possible negative real roots
so there could be 1+ and 2- real roots or there could be 1+ and 0- real roots which means there are 2 complex roots


Factor Theorem.

Answer 2

Jimmy?

Answer 3

just looking - i didnt know descartes rule of signs

Answer 4

I wanted help with Factor Therom:
x^3 + 8x^2 +15x-114=0

Answer 5

oh ok
if a polynomial f(x) is divisible by (x-a) then f(a) = 0
its really a special case of remainder theorem

Answer 6

Now what should i dO?

Answer 7

hello people i ate a very yummy taco

Answer 8

Good.

Answer 9

ok - well its trial and error really
f(2) = 2^3 + 8*(2)^2 + 30 -114 = 8 + 32 + 30 - 114 = -
f(3) = 27 + 72 + 45 - 114 = 144-114 = 30
try f(-2) = -8 + 32 - 30 - no good
f(-3) = -27 + 72 -45 - no good
f(-4) = -64 + 128 - 60
its so tedious - maybe this doesnt factor - try wolfram- lol

Answer 10

LOL.

Answer 11

Alright. ty.

Answer 12

hey who want a medal

Answer 13

seriously :- we can conclude that theres one real root and its not an integer

Answer 14

that is really confusing.
a czn of mine is in FLV, they made the things un-understandable. LOL

Answer 15

I dont' know how helpful this is but in general it can reduce the possibilities from the rat root thrm:

there may not be a rational root, but we can use the boundedness theorem to find the interval for the real root

4 | 1 9 15 -114
4 52 134
===========
1 13 67 20
bottom >= 0 implies all real roots must be strictly less than 4

-9 | 1 9 15 -114
-9 0 -135
=============
1 0 15 -249
bottom row sign alternate (0 can be pos or neg) implies all roots must be strictly greater than -9

all real roots must lie on (-9,4)

Answer 16

The above is no help in this case, wolfram indicates 1 irrational, 2 complex

Answer 17

which method is that?

Answer 18

Fundamental Theorem of Algebra,


the Rational Root Theorem,


Descartes' Rule of Signs,

factor therom?

Answer 19

i've heard it called the boundedness theorem; let me see if i can find it stated online; i'll respond here if i find it.

Answer 20

One of those fours.

Answer 21

that is called synthetic division as well

Answer 22

no, but usually taught in the same time frame as those in a computational algebra course

Answer 23

Oh, okay.

Answer 24

@myininya, did i copied that right?

Answer 25

Scroll down to "tutorial" the WTAM website calls it The Upper and Lower Bound Theorem

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut39_zero2.htm

Answer 26

ty.

Answer 27

i luv you saifoo