precalculus
posted by mz deedee .
solve the inequality express your solution using interval notation
x^47x^3+12x^2<0

y = x^2 * (x^2  7x + 12)
= x^2 (x3)(x4)
x^2 is always positive, so (x3)(x4) must be negative. That is true only for
3 < x < 4 or x in (3,4) 
x^2(x^2  7x + 12) < 0
x^2(x3)(x4) < 0
critical values are x = 0, 3, 4
test for x< 0 , say x = 1
all we care about is the sign, so
(+)()() < 0 false
test for x between 0 and 3, says x = 1
(+)()() < 0 false
test for x between 3 and 4, say x = 3.5
(+)(+)() < 0 > TRUE
test for x > 4 , say x = 5
(+)(+)(+) < 0 false
so the solution is
3 < x < 4
the graph shown here, confirms this
http://www.wolframalpha.com/input/?i=x%5E47x%5E3%2B12x%5E2
http://www.wolframalpha.com/input/?i=x%5E47x%5E3%2B12x%5E2
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