How do you solve $(x^{2} + 3) {(x^{2} + 3 x - 4)}^{3} {(x - 1)}^{3} > 0$ $(x^2+3)(x^2+3x-4)^3(x-1)^3 > 0$ ?

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My answer: $(- 4; - 3) \cup (3; + \infty)$ $(-4;-3)uu(3;+oo)$

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Answer 1 · 2018-07-17T20:47:31

It is $- 4 < x < 1$ $-4< x <1$ or $x > 1$ $x > 1$

$x^{2} + 2 > 0$ $x^2+2>0$ holds for all real $x$ $x$ . So we have only consider the factors $x^{2} + 3 x - 4$ $x^2+3x-4$ and $x - 1$ $x-1$

Note that $x^{2} + 3 x - 4 = (x - 1) (x + 4)$ $x^2+3x-4=(x-1)(x+4)$ so both factors are positive if $x > 1$ $x>1$

Both factors are also positive if both factors are negative.This is the case for $- 4 < x < 1$ $-4 < x <1$ .