Question

I was stuck at number four where I have to multiply each terms by a constant that will make all coefficients integers. M y equation is 0=3.14x^3+6.28x^2-50.24 Height is 4inches and radius is 2inches. so I have (x+2) However, when I multiplied them all by 2, it still shows in decimals. IF I have multiplied them my 100, it looks like 0=314x^3+628x^2-5024. when I began to do (P/Q) P:5024 is pretty big number to find out the possibilities.. same as 314.

Question: Apply the formula of a right circular cylinder to find the volume of the object. I have = radius=2inch, height=4inch.. volume=50.42 Basically my polynomial equation is 0=3.14x^3+6.28x^2-50.24 Since its decimals, I have to make it all coefficients integers.. it said in order to find them, multiply all by the constant, which is 2.....

Answer 1

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Answer 2

sharp wat

Answer 3

5024/314 = 16

Answer 4

so, possible roots will be factors of 16

Answer 5

you could have divided everything by 3.14

Answer 6

can't you just use the quadratic equation here? factoring will be annoying

Answer 7

@hartnn will still be decimal

Answer 8

x^3+2x^2-16=0

Answer 9

\(6.26\ne2(3.14)\)

Answer 10

aawww :(

Answer 11

by any chace is it 6.28 ?

Answer 12

more than likely meant to be :-)

Answer 13

mayb I should write all the question so u may know what IM talking about. it s confusing right??

Answer 14

yeah.

Answer 15

ldrin.batak, u r right I better do quadratic eq. instead of factoring but I still have to find out the Fundamental Theorem of Algebra, the Rational Root theorem, Descartes`s rule of sign and the factor theorem......

Answer 16

how you got this :my polynomial equation is 0=3.14x^3+6.26x^2-50.24
?
and 2pi means it has to be 6.28 ...and not 6.26...

Answer 17

@LIZ-0- don't round to decimals yet... also are your powers of \(x\) correct?

Answer 18

well i got this "my eq." by rewriting the equation with one variable woud result in a polynomial eq. that you could slove to find the radius... which it looks like... x=sqrt( (volume)/(piexheight))

Answer 19

volume: v=sqrt( (v)/(piexheight)) so then when I rewrite the formula using the variable x for the radius.. which looks like.. x^2= ( (50.24)/ (pie) (x+2) ) subsitute..... pie(3.14)x^3+6.28x^2=50.24 this is wat I got.

Answer 20

MY BIG MISTAKE it not 6.26 its 6.28 sry... I wrote 6.28 but I typed it in 6.26.... sry.......... hartnn and oldrin.bataku

Answer 21

if h=x+2
\(V=\pi r^2 h \implies 50.24 =\pi r^2(r+2) \implies 50.24/\pi=r^3+2r^2\\ \implies r^3+2r^2-16=0\)
try and solve this ...

Answer 22

so, possible roots are factors of 16......try r=+2 or -2 ....

Answer 23

oh, you already have it in the question :P
r= 2inch, h= 2+2=4 inch

Answer 24

sry Im really slow and don know wat 2 solve....

Answer 25

what do you need to find out ??
the volume is already given to be 50.42 ....

Answer 26

I have to find rational root theorem and decartes rule sign and the factor theorem.. now I am confused what is my polynomial eq. in order to find theses questions.......

Answer 27

"find rational root theorem" ...how can one find a theorem ? :O ...
you already found correct polynomial equation, x^3+2x^2-16=0

Answer 28

It said Find the solution to this eq. algebraically using the Fundamental Theorem of Algebra, the Rational root theorem, decartes sign rule and factor theorem....

Answer 29

fundamental theorem of algebra for this eq. x^3+2x^2-16=0,, there are 3 roots.degree... (is that a answer for this questions??)

Answer 30

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root...

This question... so many words...

Answer 31

yup so many words, still it helps.

Answer 32

so your polynomial, one of the ways it can be factored is:

x^3+2x^2-16=0
so =
(x - 2)*(x^2 + 4x + 8) = 0
so one of the roots is x=2... as
(2 - 2)*(x^2 + 4x + 8) = 0
( 0 ) *(x^2 + 4x + 8) = 0 ( anything multiplied by zero = 0 )

so do you have to work out the other 2 roots?

Answer 33

cos they work out as complex numbers
basically you're finding

(x^2 + 4x + 8) = 0

so do you know how to find the x values that solve that...?

Answer 34

wait Jack, but the constant is 2 and I have (x+2) but y (x-2) to conti. finding solutions??

Answer 35

...what? i didn't read everything above (as you oldrin and hartnan basically wrote a novel)
y(x-2)... where did this come from?

Answer 36

it come from x^3+2x^2-16=0
so =
(x - 2)*(x^2 + 4x + 8) = 0
so one of the roots is x=2... as
(2 - 2)*(x^2 + 4x + 8) = 0
( 0 ) *(x^2 + 4x + 8) = 0 ( anything multiplied by zero = 0 )


it said (x-2) but yyyy?????

Answer 37

oh, ok, sorry i factorised x-2 out of your original polynomial

x^3+2x^2-16 = (x - 2)*(x^2 + 4x + 8)

same as basic factorizing of quadratics:
2x^2 - x -3 = (2x - 3) (x + 1)

Answer 38

and so to continue using this as an example:
2x^2 - x -3 = (2x - 3) (x + 1)

so if 2x^2 - x -3 = 0
(2x - 3) (x + 1) = 0

so x = -1 or x = 1.5... yeah?

Answer 39

u ok with that liz?

Answer 40

when I factorize, I have to do (x-2) instead of (x+2)

Answer 41

sorry is that a question or a statement?

Answer 42

thats a question...sry

Answer 43

i suppose you could factorise x+2 out of it
k so you got:
((x+2) * x^2) -16 = 0...
that dosent factorise it enough to give you one of the roots
u need 2 or 3 multiples, whereas you still have a "-16" in there by itself, so the order of operations wont give you 0 unless (x+2)* x^2 = 16... which isnt feasible or really easy to work out roots from that

Answer 44

ok..

Answer 45

k so stick with x-2 as your first factor... so = 0 when x = +2

then do u need to work out the other 2 complex roots of your polynomial...?

or move on to the other questions?

Answer 46

oh yup... move on the other quest...

Answer 47

so it looks like you've already used descartes' to find the number of real roots... yeah?

Answer 48

oh thats true..... !

Answer 49

basically im finish???

Answer 50

so bam, one down

rational root theorum... dude ive never seen this before, i can solve polynomials ok and vaguely understand them... but this is some serious maths ;)

Answer 51

yeah, if you dont need to find the complex roots of your poly

Answer 52

then ur pretty much done

Answer 53

ok.. but wat abou fundamental algebra ... fundamental theorem of algebra...

Answer 54

that was the first one we were doing... find the roots of your poly... we found the real, the complex ones are found by continuing to solve:

(x - 2)*(x^2 + 4x + 8) = 0
so we've got
x-2 = 0 when x = +2

so all we need is the other half:
(x^2 + 4x + 8) = 0 when... complex numbers are involved

so complete the square
then you end up with
(x+2)^2 = -4

so roots are x+2 = 2i or x+2 = -2i
so x = 2i -2 OR
x = -2i -2

Answer 55

food, shower calling
slaters liz

Answer 56

sry Jack, but thnx!!Actually I understood wat u said!! thats really helps!! thnx! enjoy ur evening!