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What is the completely factored form of f (x) = x ^3 +4x ^2+7x+6?

 

A. f(x)=(x+2)(x−(−1+i√2))(x−(−1−i√2))


B. f(x)=(x−2)(x−(−1+i√2))(x−(−1−i√2))

C. f(x)=(x+2)(x−(−3+i√10))(x−(−3−i√10))

 

D. f(x)=(x−2)(x−(−3+i√10))(x−(−3−i√10)

Guest Oct 2, 2018
 #1
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 x ^3 +4x ^2+7x+6

 

-2 is a root because   (-2)^3  + 4(-2)^2  + 7(-2)  + 6  =   -8 + 16  - 14 + 6  = 0

 

So  (x + 2)  is a factor.....using synthetic division, we can find the remaining polynomial as follows

 

 

-2  [ 1    4     7      6  ]

            -2     -4    -6

     _______________

       1    2     3      0

 

The remaining polynomial is

 

x^2  + 2x  + 3   set to 0

 

x^2 + 2x + 3   = 0        subtract 3 from both sides

 

x^2  + 2x    =  -3        take 1/2 of 2  = 1...square it  = 1...add it to both sides

 

 

x^2 + 2x  + 1   = -3 + 1      factor the left side, simplify the right

 

(x + 1)^2   = -2       take both roots

 

x + 1  = ±√[-2]

 

x = 1  = ±√[2] i     subtract 1 from both sides

 

x  =  ±√[2] i  - 1

 

So...the other two factors are    -1 + √[2] i  and   -1 - √[2] i

 

So...the complete factorization is

 

A. f(x)=(x+2)(x−(−1+i√2))(x−(−1−i√2))

 

 

cool cool cool

CPhill  Oct 2, 2018

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