(0,10) is incorrect.
I would use a sign test as it makes things very easy.
Since you got -2, 1, 10 as the roots it means that only those values would make the equation equal to 0. This means the sign would only change three times.
So between the interval of infinity to -2, the graph is negative. Just plug in a random number from infinity to -2 and it will be negative. From -2 to 1 it is positive, 1 to 10 negative and 10 to infinity positive.
So the intervals at where it is > 0 is
(-2, 1) and (10, ∞)
solve and answer in interval notation
(x+2)(x-1)(x-10)>0
x+2=0 x-1=0 x-10=0
x=-2 x=1 x=10
answer:
(0,-2)(1,0)(10,0)
4 answers
By what you wrote, all of my answer is wrong. My answer should be: (-2,1) and (10,oo)?
Oh yeah sorry (0,2) and (1,0) does not include 0 so unfortunately all your answers were wrong >.<. A good way to check on your answers is to graph them on a graphing calculator.
These are quite easy if they are already factored
(x+2)(x-1)(x-10)>0
so "critical" values are -2, 1,10
so draw open circles at x= -2, x=1 and x=10
you now have 4 different segments of your line, pick any value in each segment and test it in the original relation.
I picked -5, 0, 5, and 20
-5: (-)(-)(-) < 0 wrong
0: (+)(-)(-) > 0 works!
5: (+)(+)(-) < 0 wrong
20> (+)(+)(+) > 0 works!
so -2 < x < 1 OR x > 10
(x+2)(x-1)(x-10)>0
so "critical" values are -2, 1,10
so draw open circles at x= -2, x=1 and x=10
you now have 4 different segments of your line, pick any value in each segment and test it in the original relation.
I picked -5, 0, 5, and 20
-5: (-)(-)(-) < 0 wrong
0: (+)(-)(-) > 0 works!
5: (+)(+)(-) < 0 wrong
20> (+)(+)(+) > 0 works!
so -2 < x < 1 OR x > 10