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what are the solutions of following equation? Ix3-6x2+11x-6I
Through trial an error method we can see that (x-1) is a factor. Then by vanishing method the polynomial can be solved as :f(x) = x^3-6x^2+11x-6= x^3-x^2-5x^2+5x+6x-6= x^2(x-1) -5x(x-1) +6(x-1)= (x-1)(x^2-5x+6)= (x-1)(x-2)(x-3)So under modulus function it becomes |(x-1)(x-2)(x-3)| Now it f(x) > 0 when x > 3 f(x)
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