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# Help

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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)

Oct 2, 2018

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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  = ±√ i     subtract 1 from both sides

x  =  ±√ i  - 1

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

So...the complete factorization is

A. f(x)=(x+2)(x−(−1+i√2))(x−(−1−i√2))   Oct 2, 2018