Dear Sona
1 . put only odd power factors in numerator and denominator   equal to zero separately as function changes sign only for odd power
2.  Plot these points  on the number line in increasing order  .
3. Now check the coefficients of X and make them positive.
4. expression will be +ve to the right of the highest point allotted on number line.
5.  alternately assign + and - signs in rest of intervals ( from right to left )
example

. 
     To start plotting number line, see coefficient of x of all terms is +ve. otherwise make it +ve
    Since  is an even power. So no number line(don't take 0) for it.
            Plot no. line only for x+1=0,2x+1=0,x-2=0 i.e. x=-1,x=-1/2 and x=2.
          Check   the sign for x>2. It will come +ve. Then put alternate signs from   right to left i.e. b/w -1/2 to 2 as negative. Then b/w -1 to -1/2 as   +ve. Then for x<-1 as -ve. We have to find x for which the expression   is >0. So x belongs to +ve region i.e. x belongs to   (-1,-1/2)U(2,infinity)
 
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
 
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Factorize the polynomial or rational expression into LINEAR factors.

Mark all the critical points (points where any of the factors is zero) on the number line..... so that the no. line is divided into a no. of segments.

Take any convenient value (say x = 0) in any of the divisions, and determine weather the expression is -ve or +ve for that x.

The trick is that the expression changes values only at the critical pts. so if it is positive for one x in a division, it will be positive for the whole division... and it will be negative for the next division.. then again positve... so on... alternate positive and negative divisions..

But there are two traps in this method...

First, if you have linear factors in denominator too... some of the critical points will be out of the domain.

And second, if one of more of the linear factors is repeated


						

							

for > or = 0 we have to choose the areas with "+" sign and for< or = 0 we have to choose the areas with "-" sign