Question

is this correct?
solve the equation in the real number system x^4-2x^3+8x^2-32x-126=0

answer: -16 x^5+x^4-32 x-126 = 0

Answer 1

or is there no solution

Answer 2

how the heck are you supposed to solve the zeros of a fourth degree polynomial without technology?
http://www.wolframalpha.com/input/?i=+x^4-2x^3%2B8x^2-32x-126%3D0+

Answer 3

that shows so many answers how do i know the right one

Answer 4

math is extremely hard for me sorry

Answer 5

there are only two real solutions listed

Answer 6

i understand
frankly it is almost impossible to solve this
you need a computer

Answer 7

wolfram lists two real solutions and two complex solutions
how were you supposed to solve this?

Answer 8

it says my answer should be x= or no solution yeah i wish i knew all of this

Answer 9

are you sure there is not a typo in your question? the answers are very ugly

Answer 10

i have about 15 questions i dont understand and due tonight :(

Answer 11

let me double check

Answer 12

can you link or take a screen shot?

Answer 13

it wont let me for some reason i cant even copy and paste here is the question again it is possible i messed it up.




solve the equation in real number system. x^4-2x^3+8x^2-32x-128=0

Answer 14

http://www.wolframalpha.com/input/?i=x^4-2x^3%2B8x^2-32x-128%3D0

Answer 15

that's more like it, \(-2\) and \(4\)

Answer 16

how many more?

Answer 17

can i just give you a hundred medals lol. how many more ?

Answer 18

lets knock a few out quick

Answer 19

in the spirit of "get it over with, i'll understand it some time later"

Answer 20

questions is that what your talkin bout cause i have to open new ones for medals for you!

Answer 21

i get no frequent flyer miles for medals, but if you want to start a new post, fine

Answer 22

otherwise just ask here, when i get tired i'll let you know

Answer 23

solve the inequality algebraically 4x-4>-3x^2. your awesome thank you

Answer 24

start with
\[3x^2+4x-4>0\]

Answer 25

then factor as
\[(3x-2)(x+2)>0\]

Answer 26

zeros are at \(-2\) and \(\frac{2}{3}\) and since this is a parabola that faces up , it is positive outside the zeros,
namely if \(x<-2\) or \(x>\frac{2}{3}\)

Answer 27

if you need interval notation it is
\[(-\infty,-2)\cup (\frac{2}{3},\infty)\]

Answer 28

12 to go

Answer 29

your so quick ;)

Answer 30

half quick

Answer 31

million times quicker than me....solve the fallowing inequality 56x-15>56/x

Answer 32

i wish i could repay you cause this is the fail or pass moments

Answer 33

it is really
\[56x-15>\frac{56}{x}\]??

Answer 34

yeah

Answer 35

lets try to make sure you pass
is that the question for real?

Answer 36

let m check again

Answer 37

this is a pain, but we can do it. you need the whole method (like if you have to take an exam?) or just the answer?

Answer 38

there is more to it sorry

Answer 39

\[-\frac{7}{8}<x<0\] or \[x>\frac{8}{7}\]

Answer 40

ok

Answer 41

is the point of x=0 included in the solution?
are there other finite points of the interval in the solution set?
what is the solution set or is there no solution?

Answer 42

yes, 0 is excluded because you have a zero in the denominator

Answer 43

0 is included?

Answer 44

EX cluded

Answer 45

oh ok

Answer 46

solution is the one i wrote above

Answer 47

\[-\frac{7}{8}<x<0\] or
\[x>\frac{8}{7}\]

Answer 48

this
are there other finite points of the interval in the solution set?
makes no sense to me , but i assume the answer is "no"

Answer 49

11 to go

Answer 50

use the remainder theorem to find the remainder when f(x) is divided by x-1/5. then use the factor theorem to determine whether x-1/5 is a factor of f(x). f(x)=5x^4-x^3-5x+1

Answer 51

christ you math teacher must hate you

Answer 52

im sure she does have never met her too lol

Answer 53

do you need to show your work, or just answer the question?
the answer is
"the remainder is 0, because \(f(\frac{1}{5})=0\)"

Answer 54

and therefore
"yes, \(x-\frac{1}{5}\) is a factor"

Answer 55

show work :(

Answer 56

ok then compute
\[f(\frac{1}{5})=5(\frac{1}{5})^4-(\frac{1}{5})^3-5(\frac{1}{5})+1=0\]

Answer 57

it is actually not that hard
you get
\[\frac{1}{5^3}-\frac{1}{5^3}-1+1=0\]

Answer 58

wow this is crazy

Answer 59

therefore
the remainder is zero
and \(x-\frac{1}{5}\) is a factor
finished

Answer 60

10 to go

Answer 61

when is this due?

Answer 62

lets see how many we can finish in half an hour
i bet we can do them all

Answer 63

due tonight have to decide if im gonna pass or fail.im down to finish in half hour ;)..

use the rational zeros theorem to list the potential rational zeros of the polynomial function do not attempt to find zeros f(x)= -25x^9-x^8+x+5

Answer 64

numerator has to divide 5, denominator has to divide 25
there are not that many
\[\pm1,\pm5,\pm\frac{1}{5},\pm\frac{1}{25}\]

Answer 65

don't forget the \(\pm\)

Answer 66

9 more, lets knock em out so you don't have to retake it

Answer 67

use thw intermediate value theorem to show that the polynomial function has a zero in the given interval. f(x)=x^5-x^4+5x^3-4x^2-17x+8 ; [1.5, 1.8] find value s of 1.5 and 1.8

Answer 68

\[f(1.5)=-7.09375\]

Answer 69

\[f(1.8)=1.99808\]

Answer 70

since one is positive, and the other is negative, there must be a number between 1.5 and 1.8 for which \(f\) is zero
8 more

Answer 71

information is given about a polynomial f(x) whose coefficients are real numbers. find remaining zeros of f. degree3 ; zeros:1,-5-i

Answer 72

still there?

Answer 73

wow missed it somehow

Answer 74

zeros are
\[3, -5-i, -5+i\]

Answer 75

polynomial is
\[(x-3)(x-(-5+i))(x-(-5-i))\]
\[=x^3+7 x^2-4 x-78\]

Answer 76

7 more

Answer 77

this one is a hard one

Answer 78

did i loose ya?

Answer 79

oh wow my computer did something weird so sorry

Answer 80

actually i only have 1 left

Answer 81

similar to the last :)

Answer 82

I sat watching waiting for a response and i had to refresh everything after a long time

Answer 83

if your still up to it here is the last one i will ask you for :)


form a polynomial f(x) with real coefficients having the given degree and zeros. degree; 5 zeros: -4 ; i ; -2+i