Solve the following polynomial inequality:  2(x^3 - 2x^2 + 3) < x(x-1)(x+1)  



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Posted on (Answer #1)

We can distribute on both sides, to get:


Subtract `x^3` from both sides, and add `x` to both sides to get:

`x^3-4x^2+x+6 < 0`   

One way to do this is to factor the left hand side.  You might notice that if you plug in `x=-1` the left hand side is 0.  Thus `(x+1)` is a factor:


`x*ax^2=x^3` , so we must have `a=1`  

`(x+1)(x^2+bx+c) = x^3-4x^2+x+6`

`x*bx+1*x^2 = -4x^2` , so we must have `b=-5`


`x*c+1*-5x = x` , also, `1*c=6` , thus `c=6`



But we can factor further:

`x^2-5x+6 = (x-2)(x-3)`



`x=-1` , `x=2` , and `x=3` are places where the left hand side is zero, so they are places where there is a switch from negative to positive, or positive to negative.

So we look at: `x<-1` , `-1<x<2` ,  `2<x<3` , and `x>3`   

One at a time:





`x+1<0` , `x-2 < -3 < 0` . `x-3<-4<0`

So we have a negative times a negative times a negative, which is negative.

So if `x<-1` then `(x+1)(x-2)(x-3)<0` , so `2(x^3-2x^2+3)<x(x-1)(x+1)`  



Next up:



` ``x+1 > 0` , `x-2 < 0` , `x-3<-1<0`  

So we have a positive times a negative times a negative, which is positive.  Thus it isn't true that `(x+1)(x-2)(x-3)<0 `



Now we try `2<x<3`

Then `x+1>0` , `x-2>0` , and `x-3 < 0`

So we have a positive times a positive times a negative, which is negative.  So it is true that ` `  

So if  2<x<3  then `` , so ``  



Finally, suppose x>3

Then x+1, x-2, and x-3 are all positive.  So multiplying them all together gives you a positive, which doesn't satisfy the inequality.


So the inequality works if   x< -1 or if   2<x<3




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