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Help with math !

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plsoft

Journeyman
I'm not too bright with maths, i wud be really grateful if i could get help with my queries.

1. How to solve (find roots) of cubic equations (ax^3+bx^2+cx+d=0) ? If possible please explain the the whole process without short-cuts.

2. Please explain Rank of a Matrix & how to determine them.

For now i will end my queries here, however i have other queries which i wish to ask later. Thank you all in advance.
 

vizay

Right off the assembly line
vist://topmarks.co.uk
use trial and error method to factorise the given equation, foctor thorem is to be applied. eg. 3x^3-x^2-3x+1=0,[Let f(x)=3x^3-x^2-3x+1. ] factor of 1 are +1,-1. Then put this values in x.Now, since f(1)=0, then x-1 is the factor of f(x), divide f(x) by x-1, then f(x)= (x-1)(3x^2+2x-1)=(x-1)(3x^2+3x-x-1)=(x-1){3x(x+1)-1(x+1)}=(x-1)(x+1)(3x-1)
i.e. (x-1)(x+1)(3x-1)=0
therefore 1, -1, 1/3 are the required roots. vizay
 

mediator

Technomancer
Very easy dood!!
U can apply factor theoram......but u cant just put any value to it! For finding the appropriate value use differentiation and find maxima and minima values for x. This should give u the range in which x resides and then put value! After achieving a simple value factorise the remaining problem !
 

Ramakrishnan

The Researcher
There must be many specialised math forums where you can raise queries like this. I don't think digit forum is a suitable place for discussion on a query like this.
 
OP
plsoft

plsoft

Journeyman
I agree with you ramakrishnan, however math forums n other math related sites tend to explain problems in a general way. I wanted a specific n straight-to-the-point answer to my query, thats why i posted it here.
 

__Virus__

Ambassador of Buzz
nitish_mythology said:
Should this post be 'In General' section?
i thnk so..

no no no... its a question and hence rightly placed ;)

and the answer.. why dont u apply remainder theorom or factor theorom or trial and error method (sux)... its very easy.. i mean i will just divie the whole equeation by x-1 and then blablabla its damn easy I wish u were down in hyd I wudhave explained u :D maths maths maths :D
 
OP
plsoft

plsoft

Journeyman
i don't wanna be annoying but cud someone pls tell me how to divide the equation by (X-1). Just consider it as explaining it to a rookie, the thing is haven't had a brush with math for quite some time. thanks
 

Ishan

In the zone
Take an example: 3x3+4x2+2x+1

3x3+4x2+2x+1
3x3+3x2+x2+x+x+1
(x+1)(3x2+x+1)

Now How I did this:
Step 1
First write the ques as it is.. Now leave one line below it. Ok Now check that substituting which value of x gives the result of the equation as zero(Here it is “-1”)
Step 2
Now in the third line write the factor like as (x+1)…….If you get x=”-2” then write the factor as (x+2) and if u get it as x=”3” then write it as (x-3){. i.e. change the sign of the value of x and write it after x.}
Uptill now ur problem should look like this:
3x3+4x2+2x+1

(x+1)()

Step 3
Now see what to multiply with x {i.e. the first term of the factor} to get the first term of the expression{i.e. 3x3}. Here it is 3x2. So write this in the empty bracket. And write 3x3 in the second line.
3x3+4x2+2x+1
3x3+
(x+1)(3x2+ )

Step 4
Now multiply the second term of the factor {i.e. 1} with the term u wrote in the empty bracket.{i.e. 3x2}.OK So u’ll get the same. 3x2. Now write this in the second line.
3x3+4x2+2x+1
3x3+3x2+
(x+1)(3x2+ )
Step 5
But we need 4x2 instead of 3x2. So add x2 to the second line.
3x3+4x2+2x+1
3x3+3x2+x2+
(x+1)(3x2+ )
Step 6
Now see that what will u need to multiply with x to get x2. It is x. so write x in the third line.
3x3+4x2+2x+1
3x3+3x2+x2+
(x+1)(3x2+x + )
Step 7
Now multiply this x with 1 so that u’ll get x. write this in the second line.
3x3+4x2+2x+1
3x3+3x2+x2+x
(x+1)(3x2+x + )
But we need 2x in the ques. So add x in the second line.
3x3+4x2+2x+1
3x3+3x2+x2+x+x+
(x+1)(3x2+x + )
Step 8
Now see what we need to multiply with x of the factor (x+1) to get the last x we wrote just now(latest) in the second line.It is 1 right? So write 1 in the third line.
3x3+4x2+2x+1
3x3+3x2+x2+x+x+
(x+1)(3x2+x +1 )
Step 9
Now multiply 1 of the factor (x+1) with this last one in the third line so we get 1 again. This 1 u write in the second line.
3x3+4x2+2x+1
3x3+3x2+x2+x+x+1
(x+1)(3x2+x +1 )
Step 10
That’s it. U have done. The two expressions in the last line are the factors of the given ques. So ur factors are (x+1) and (3x2+x +1 ).

Note. U can do the same method with polynomials having power 4. So go on. See few more examples.

**5x3-2x2+4x-7
5x3-5x2+ 3x2-3x+7x-7
(x-1)(5x2+3x+7)--à u can simplify this further.

**6x4-3x3+2x2-4x-15
6x4+6x3-9x3-9x2+11x2+11x-15x-15
(x+1)(6x3-9x2+11x-15)--à u can simplify this further.
 

Ishan

In the zone
One more thing friends....Please this is a QnA forum and we come here to share our knowledge and not to show how smart we are.
If u wish u can give the solution of the problem whatever the problem is.else dont just pass comments and discourage others.
So plz remeber that one day u could also be at plsoft's place. the situation might be different.

Just remeber that if thing i seasy for u it may be hard for others... eg.. playong cricket is hard for me but working on computers is a bit easy...BUt it is vice-versa for Sachin Tendulkar.. So if he'll have some problem and if god wishes he aks me for help...I'll help him rather than saying go to the computor vendor...Ok

Sorry for the lecture.:):eek:
 
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plsoft

plsoft

Journeyman
thanks ishan, now i know how to solve the problem, n thanks 4 understanding my situation. A problem is a problem for the person who doesn't know, no matter how silly n simple it may be for others.
 
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