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Find all the zeros of the function f(x) = x^3 -2x^2 +16x -32

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4 votes

Answer:

Real zeros: x = 2

All zeros: x = 2, x = -4i and x = 4i

Explanation:


f(x)=x^3-2x^2+16x-32\\\\\text{The zeros:}\\\\f(x)=0\Rightarrow x^3-2x^2+16x-32=0\\\\x^2(x-2)+16(x-2)=0\\\\(x-2)(x^2+16)=0\iff x-2=0\ \vee\ x^2+16=0\\\\x-2=0\qquad\text{add 2 to both sides}\\\boxed{x=2}\\\\x^2+16=0\qquad\text{subtract 16 from both sides}\\x^2=-16<0\qquad\text{no real solutions}\\\\\text{In the set of the complex numbers}\\x^2=-16\to x=\pm√(-16)\\\boxed{x=-4i\ \vee\ x=4i}

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