Question 3696: Identify the real zeros to the nearest tenth for the function f(x) = x3 - 6x - 9.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! f(x) = x^3 - 6x - 9
= x^3 - 3x^2 + 3x^2 -
6x - 9
= x^2(x-3) + 3(x^2 -2x -3)
= x^2(x-3) + 3(x-3)(x+1)
= (x-3)(x^2 + 3x +3)
So, x -3=0 or x^2 +3x+3 = 0
By quadratic formula
x = (-3+ sqrt(9-12) )/2 = (-3 + sqrt(3) i)/2
or (-3 - sqrt(3))/2
( The approximated values are -1.5 -0.87 i or
-1.5 +0.87 i ,to the nearest hundreds.)
I don't think to the tenth is good enough.)
Kenny
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