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)
+128475 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))
