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avatar+130 

The numbers 1,-1 and a+1 are the roots of the equation x^3+6x^2-2bx-3c-12=0. 

Find the values of a,b,c.

 

TY :)

 Mar 3, 2021
 #1
avatar+117 
+1

Easier to read: The numbers \(1,-1\) and \(a+1\) are the roots of the equation \(x^3+6x^2-2bx-3c-12=0\). Find the values of \(a,b,c\).

 Mar 3, 2021
 #2
avatar+118069 
+2

Since  1  is a root,   by the Remainder Theorem :

 

1^3  + 6(1)^2 - 2b(1)  -3c  -12  = 0

 

1  + 6  - 2b - 3c  - 12   =   0

 

-2b - 3c  =  5              (1)

 

Likewise....since  -1  is   a root

 

(-1)^3 + 6(-1)^2  - 2b(-1) - 3c  - 12    =   0

 

-1 + 6  + 2b - 3c  - 12  =  0

 

2b - 3c  =  7      (2)

 

Add ( 1)  and (2)  and we get that

 

-6c = 12

 

c  = -2

 

2b -3(-2)  =    7

 

2b  = 1

 

b = 1/2

 

-3(-2)  - 12  =  -6

 

By Vieta

 

(1)(-1) (a + 1)  =  6

 

-a - 1    = 6

 

-a  =  7

 

a  =  -7

 

 

cool cool cool

 Mar 3, 2021
edited by CPhill  Mar 3, 2021

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