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)
(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, ∞)
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
To solve the inequality (x + 2)(x - 1)(x - 10) > 0, we need to find the intervals where the expression is greater than zero.
Let's find the critical points first:
1. Set each factor to zero: x + 2 = 0, x - 1 = 0, x - 10 = 0
Solving these equations yields: x = -2, x = 1, x = 10
Now, let's create a number line and mark these critical points on it:
-----(-2)-----1-------10------- (where the dashes represent the number line)
Next, we need to test each interval to determine whether the expression is positive or negative in that interval. We can choose any test point from within each interval and substitute it into the expression (x + 2)(x - 1)(x - 10). Let's choose -3, 0, 5, and 11 as test points:
1. For the interval (-∞, -2):
Substitute -3 into the expression: (-3 + 2)(-3 - 1)(-3 - 10) = (-1)(-4)(-13) = -52
Since the result (-52) is negative, the expression is negative in this interval.
2. For the interval (-2, 1):
Substitute 0 into the expression: (0 + 2)(0 - 1)(0 - 10) = (2)(-1)(-10) = 20
Since the result (20) is positive, the expression is positive in this interval.
3. For the interval (1, 10):
Substitute 5 into the expression: (5 + 2)(5 - 1)(5 - 10) = (7)(4)(-5) = -140
Since the result (-140) is negative, the expression is negative in this interval.
4. For the interval (10, ∞):
Substitute 11 into the expression: (11 + 2)(11 - 1)(11 - 10) = (13)(10)(1) = 130
Since the result (130) is positive, the expression is positive in this interval.
Finally, we can write the solution in interval notation:
(0, -2) U (10, ∞)