document.write( "Question 28310: I am having difficulty with the following problem -- any help you can give would be greatly appriciated.\r
\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #15406 by venugopalramana(3286) $\"\"$ $\"About$

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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.
\n" ); document.write( "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
\n" ); document.write( "so let us analyse (x+1)(x+2)...product of 2 factors is positive if both are positive or both are negative.
\n" ); document.write( "so x+1 and x+2 are both positive...that is x>-1 and x>-2...or x>-1
\n" ); document.write( "or...x+1 and x+2 are both negative...that is x<-1 and x<-2...or x<-2
\n" ); document.write( "so p(x) will be positive if
\n" ); document.write( "1.x is not equal to 1 and x>-1 or x<-2\r
\n" ); document.write( "\n" ); document.write( "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
\n" ); document.write( "or.........
\n" ); document.write( "-2 is less than x is less than -1 \n" ); document.write( "

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