Question

plse any one can factorise ittttt

Answer 1

3x^3+x^2 -22x -8
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it has one sign change as is, so there is at best; 1 positive root

since this thing is a cubic (^3) that leaves 2 or 0 negative roots
2 or 0 complex roots

Answer 2

Cardono developed a formula similar to the quadratic formula for cubics of the form:

x^3 + px + q = 0

Answer 3

so what are its factors

Answer 4

i have given you 3 hints to its possible factorizations; i would appreciate it if you would show some sort of initiative at a solution instead of asking for the results.

Answer 5

u mean i should put x= 1

Answer 6

the cardano can be a bit convoluted, but once the cubic is reduced; its just a plug and play like the quadratic

Answer 7

you can try x = 1; but i doubt it would work

IF this thing has rational roots, they will be of the form related to the ratio of the last to first terms, factors

Answer 8

\[\pm\frac{1,2,4,8}{1,3}\]
so 1 might be an option

Answer 9

Factor theorem may also work.

Answer 10

no, it doesnot work , i have put all the values
@mathslover
plse how factor theorem work
i tried..but did get out from this question

Answer 11

May I know the values you have put on the place of x there?

Answer 12

1,-1,1/2,-1/2...
2,-2,3,-3.....&so on

Answer 13

im thinking of a more calculus approach if we have to make guesses\[g_n=\frac{p-f(g_{n-1})}{3(f'(g_{n-1}))^2}+g_{n-1}\]

Answer 14

@amistre64 I love your calculus approaches.

Answer 15

well, the 3 and ^2 are part of the f' i spose, and p = 0 for a root

\[g_n=\frac{-f(g_{n-1})}{f'(g_{n-1})}+g_{n-1}\]

Answer 16

:)

Answer 17

Well it is sure that factor theorem will NOT work here.

Answer 18

cardanos should work, but i dont use it enough to remember it as well as the quadratic

Answer 19

let x=u - 1/3 and simplify

Answer 20

but @amistre64
i heard cardanos method first time

Answer 21

Amistre always come with innovative ideas.

Answer 22

i ran this on the wolf to see what we are dealing with ... it aint pretty

Answer 23

I was fast at that, I already ran to wolf when I saw you getting confused :)

Answer 24

Cardano's method :

www.macmahon.plus.com/cardano.pdf‎


I never learnt it!

Answer 25

yes @mathslover i also see wolfram
but didn't get anything

Answer 26

im running the calculus trial and error :) one root looks to be getting close to: 2.721...

Answer 27

Well as it works only for the cubic equations in the form : \(\large x^3 + px + q = 0\) , So , we have to transform it to in the above form first.

Answer 28

but if you need exact stuff, cardanos will most likely be the best way

Answer 29

x = u - 1/9 is the transform .. not 1.3 :)

Answer 30

In the pdf , one problem was done like this :

\(\large x^3 + 6x^2 - 3x + 6 =0 \)

Let x = y - 2

and then it solved further.

How we get to know, whether we shall put x = y-2 or something else?

Answer 31

@amistre64 , how u got x = u - 1/9?

Answer 32

descartes found a reduction substitution; the negative coeff of the second term divided by the power of the first term; as long as the first term coeff is 1

in this case; 1/3/3 = 1/9

Answer 33

in the case you presented above: x = u (-6/3) = u-2

Answer 34

\( \large 3x^3+x^2 -22x -8 \\
\large x^3 + \cfrac{x^2}{3} - \cfrac{22x}{3} - \cfrac{8}{3} \)

\(\large \cfrac{\cfrac{1}{3}}{\cfrac{3}{1}} = \cfrac{1}{9} \)

Answer 35

Ok, getting it.

Answer 36

we end up with:\[3u^3-\frac{199}{9}u-\frac{1348}{243}=0\]

Answer 37

So we shall put now x = u -1/9

Answer 38

yes :)

Answer 39

I will do that too, that calculation, I like :)

Answer 40

-1346 or -1348 ?

Answer 41

-1348

Answer 42

My fault I think.

Answer 43

Yes, a little blunder mistake, sorry

Answer 44

arithmetic happens :)

Answer 45

my calculator didn't warn me that I am doing wrong mistake , its' its mistake :)

Answer 46

\(\large { 3(u - \cfrac{1}{9})^3 + (u - \cfrac{1}{9} )^2 - 22( u - \cfrac{1}{9} ) - 8 = 0 \\ 3( u ^3 - \cfrac{1}{729} -\cfrac{u}{3} ( u - \cfrac{1}{9} ) ) + (u^2 + \cfrac{1}{81} - \cfrac{2}{9} u ) - 22u + \cfrac{22}{9} - 8 = 0 \\ 3u^3 - \cfrac{1}{243} - u^2 + \cfrac{u}{9} + u^2 + \cfrac{1}{81} - \cfrac{2}{9} u - 22u + \cfrac{22}{9} - 8 = 0 \\ 3u^3 - \cfrac{1}{243} + \cfrac{u}{9} + \cfrac{1}{81} - \cfrac{2}{9} u - 22u + \cfrac{22}{9} - 8 = 0 \\ 3u^3 - \cfrac{199 u}{9} + \cfrac{22}{9} - 8 - \cfrac{1}{243} + \cfrac{1}{81} = 0 \\ 3u^3 - \cfrac{199 u }{ 9} + \cfrac{-1 + 3 + 22*27 + -8*243}{243} = 0 \\ 3u^3 - \cfrac{199 u }{9} - \cfrac{1348}{243} = 0 \\ }\)

Answer 47

Above is my work , @msingh try to understand what I and amistre were doing till now.

Answer 48

@amistre64 what to do next?

Answer 49

hmm
i got a bit idea and understood till the now the solution

Answer 50

Good msingh. Amistre will guide us now what to do next.

Answer 51

okay