SOLUTION: I am having difficulty with the following problem -- any help you can give would be greatly appriciated. Let p(x)=(x-1)to the second power(x+1)(x+2). Where is p(x)>0? Where is p

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: I am having difficulty with the following problem -- any help you can give would be greatly appriciated. Let p(x)=(x-1)to the second power(x+1)(x+2). Where is p(x)>0? Where is p      Log On


   



Question 28310: I am having difficulty with the following problem -- any help you can give would be greatly appriciated.
Let p(x)=(x-1)to the second power(x+1)(x+2). Where is p(x)>0? Where is p(x)<0? Answer in interval notation.

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Let p(x)=(x-1)to the second power(x+1)(x+2). Where is p(x)>0? Where is p(x)<0? Answer in interval notation.
Let p(x)=(x-1)^2*(x+1)(x+2)....(x-1)^2 being a perfect square is always positibe or zero in the least when x=1
so let us analyse (x+1)(x+2)...product of 2 factors is positive if both are positive or both are negative.
so x+1 and x+2 are both positive...that is x>-1 and x>-2...or x>-1
or...x+1 and x+2 are both negative...that is x<-1 and x<-2...or x<-2
so p(x) will be positive if
1.x is not equal to 1 and x>-1 or x<-2
now (x+1)(x+2) will be negative if one of them is negative and another positive. that is x lies between -2 and -1
or.........
-2 is less than x is less than -1