Question

Solve inequality algebraically...

(3-x)^3(2x+1)/(x^3-1)<0

Answer 1

changes sign at the zeros of the factors, namely at \(3,1,-\frac{1}{2}\)

Answer 2

break up the real line in to four parts
\[(-\infty, -\frac{1}{2}),(-\frac{1}{2},1), (1,3),(3,\infty)\] and check on one interval to see if it is positive or negative

Answer 3

x<-1/2 negative
-1/2<x<1 positive
1<x<3 negative
3<x positive

Answer 4

i don't think so

Answer 5

for example, \(0\) is in the interval \((-\frac{1}{2},1)\) and if you put \(x=0\) you get
\(\frac{3\times 1}{-1}\) which is negative, not positive

Answer 6

Oh I see what you meant now

Answer 7

To algebraically solve an inequality F(x) = f(x)*g(x)*h(x) < 0, create a Sign Chart (sign table) that figures all the variations of f(x), g(x), and h(x). The sign chart will give the combined variation of F(x).
f(x) = (3 - x)^3 g(x) = (2x + 1) h(x) = x^3 - 1

x | -infinity -1/2 1 3 +infinity
f(x) | + + + 0 -
g(x) | - 0 + + +
h(x) | - - 0 + +
--------------------------------------------------------
F(x) + 0 - || + 0 -
F(x) < 0 (-) in the intervals: (-1/2, 1) and (3, +infinity).
Note: the sign of F(x) is the resulting sign of f(x)*g(x)/h(x).