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Factorize the expression $$x^3-3x^2-4x+12$$ Hence calculate the ranges of values of $x$ for which $x^3-3x^2>4x-12$.

I factorised it to obtain $(x-2)(x-3)(x+2)$ but I don't how how to get to the next step.

Jimmy R.
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    The next step consists in finding when your polynomial $(x-2)(x-3)(x+2)$ is positive. Can you do that if you know the sign of each factor? Now build a sign chart. – Jean-Claude Arbaut Nov 06 '14 at 16:07

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Hint: to have $(x-2)(x-3)(x+2)=x^3-3x^2-4x+12>0$, you need either $1$ or $3$ factors to be positive, the others being negative...

Milly
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Your second question boils down to when $$x^3-3x^2>4x-12\iff x^3 - 3x^2 - 4x + 12 = (x-2)(x-3)(x +2) > 0$$

You should know the three zeros of your function by reading off what the factors tell you. Make a sign chart INSIDE the intervals:

$$(-\infty, -2),\; (-2, 2),\; (2, 3), \text { and } (3, +\infty)$$ to determine when where $(x-2)(x-3)(x+2)$ is positive. You need only test one value in each interval to check for positivity.

amWhy
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$$ {x}^{3}-3\,{x}^{2}-4\,x+12={x}^{3}-4\,x-3\,{x}^{2}+12=x(x^2-4)-(3x^2-12)= $$ $$ =x(x^2-4)-3(x^2-4)=(x-3)(x^2-4)=(x-3)(x-2)(x+2). $$

Leox
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  • Oh, sorry. Really, you are right – Leox Nov 07 '14 at 15:57
  • No problem. The title of the question can easily trigger an appropriate answer, for that title question. For some reason, though, OPs frequently ask very different questions in the body of the post. – amWhy Nov 07 '14 at 16:00