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algebra

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Find all real values of  x such that
(x^2 + 2x + 5 - x - 4)(3x^2 - x - 4 + 2x + 8) >= 0

Sep 2, 2022

#1
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Rewritten in latex, this is $$(x^2 + 2x + 5 - x - 4)(3x^2 - x - 4 + 2x + 8) \ge 0$$.

Let's simplify the things in the parenthases. Doing so, we get $$(x^2+x+1)(3x^2+x+4)\ge0.$$ These cannot be simplified further.

However, all the signs in this are positive. So for any value of $$x$$ inputted in here, it's always $$\ge 0$$. So the solutions to this are $$-\infty , or in interval, \(x\in(-\infty,+\infty)$$.

Sep 3, 2022