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Algebra

sneha kulkarni's Avatar
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5 Mar 2007 22:05:07 IST
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a prob on bionomial theorem...........pls solve it quickly
None

find the cubic equation whose roots are 1 , -2 , 3


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Cool goIITian

Joined: 3 Mar 2007
Posts: 38
5 Mar 2007 22:38:10 IST
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since roots are1,-2,3
=(x-1)(x+2)(x-3)
=(x^2+x-2)(x-3)
=(x^3+x^2-2x-3x^2-3x+6)
=(x^3-2x^2-5x+6)
here is the answer
plz........... rate me.
edison's Avatar

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Joined: 19 Oct 2006
Posts: 7537
5 Mar 2007 22:44:42 IST
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That is well done vineet
sneha kulkarni's Avatar

Scorching goIITian

Joined: 1 Feb 2007
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5 Mar 2007 22:46:23 IST
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then what about these roots........-1 , 1+i , 1-i

Cool goIITian

Joined: 3 Mar 2007
Posts: 38
5 Mar 2007 23:26:50 IST
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they are
=(x+1)[(x-1)-i][(x-1)+i]
=(x+1)[(x-1)^2-i^]
=(x+1)[(x-1)^+1]
=(x+1)[x^2-2x+2]
=x^3-2x^2+2x+x^2-2x+2
=x^3-x^2+2
that is the answer
if it is right plz.rate me
irfan ahmad's Avatar

New kid on the Block

Joined: 23 Jan 2007
Posts: 6
6 Mar 2007 09:57:33 IST
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they are
=(x+1)[(x-1)-i][(x-1)+i]
=(x+1)[(x-1)^2-i^]
=(x+1)[(x-1)^+1]
=(x+1)[x^2-2x+2]
=x^3-2x^2+2x+x^2-2x+2
=x^3-x^2+2
that is the answer
if it is right plz.rate me

Cool goIITian

Joined: 3 Mar 2007
Posts: 38
7 Mar 2007 10:43:01 IST
0 people liked this

they are
=(x+1)[(x-1)-i][(x-1)+i]
=(x+1)[(x-1)^2-i^]
=(x+1)[(x-1)^+1]
=(x+1)[x^2-2x+2]
=x^3-2x^2+2x+x^2-2x+2
=x^3-x^2+2
that is the answer
if it is right plz.rate me
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
14 Mar 2007 18:10:21 IST
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whatever roots say a,b,c are given to u, form the equation (x-a)(x-b)(x-c) = 0



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