Solve (z+3)(2-z)(1-2z)>0?

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Answer 1 · 2015-07-10T23:14:50

$z in (-3, 1/2) uu (2, oo)$

Let $f(z) = (z+3)(2-z)(1-2z) = (z+3)(2z-1)(z-2)$

Then $f(z) = 0$ when $z = -3$ , $z = 1/2$ and $z = 2$

These three points split the real line into four intervals:

$(-oo, -3)$ , $(-3, 1/2)$ , $(1/2,2)$ and $(2,oo)$

If $z in (-oo, -3)$ then

$(z+3) < 0$ , $(2z-1) < 0$ , $(z-2) < 0$ so $f(z) < 0$

If $color(red)(z in (-3, 1/2))$ then

$(z+3) > 0$ , $(2z-1) < 0$ , $(z-2) < 0$ so $color(red)(f(z) > 0)$

If $z in (1/2, 2)$ then

$(z+3) > 0$ , $(2z-1) > 0$ , $(z-2) < 0$ so $f(z) < 0$

If $color(red)(z in (2, oo))$ then

$(z+3) > 0$ , $(2z-1) > 0$ , $(z-2) > 0$ so $color(red)(f(z) > 0)$

So the solution is $z in (-3, 1/2) uu (2, oo)$

graph{ (x+3)(2-x)(1-2x) [-40, 40, -12.24, 27.76]}