# Does the inequality 2*x^3+3*x^2-8*x+3 >0 have any solutions.

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The inequality 2*x^3+3*x^2-8*x+3 >0 has to be solved.

First factorize 2*x^3+3*x^2-8*x+3.

2*x^3+3*x^2-8*x+3

= 2x^3 - 2x^2 + 5x^2 - 5x - 3x + 3

= 2x^2(x - 1) + 5x(x - 1) - 3(x - 1)

= (x - 1)(2x^2 + 5x - 3)

=> (x - 1)(2x^2 + 6x - x - 3)

=> (x - 1)(2x(x + 3) - 1(x + 3))

=> (x - 1)(2x - 1)(x + 3)

(x - 1)(2x - 1)(x + 3) > 0 when:

• (x - 1) > 0, (2x - 1) > 0 and (x + 3) > 0

=> x > 1, x > 1/2 and x > -3

This is true for x `in` `(1, oo)`

• x - 1 > 0, 2x - 1 < 0 and x + 3 < 0

=> x > 1, x < 1/2 and x < -3

This is not possible for any value of x

• x - 1 < 0, 2x - 1 > 0 and x + 3 < 0

=> x < 1, x > 1/2 and x < -3

This is not possible for any value of x

• x - 1 < 0, 2x - 1 < 0 and x + 3 > 0

=> x < 1, x < 1/2 and x > -3

This is true for x `in` (-3, 1/2)

The solution of the given inequality is `(-3, 1/2)U(1, oo)`

Posted on

You have given

`2x^3+3x^2-8x+3>0` . If we call given function as f(x),then we have f(x)>0 .

`f(x)=2x^3+3x^2-8x+3` ,

Apply remainder theorem ,to factorise f(x).

possible factors may x+1,x-1,x+3,x-3

f(1)=2+3-8+3=0

So (x-1) is one factor of f(x)

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

`=(x-1)(2x^2+5x-3)`

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

`=(x-1)(2x-1)(x+3)`

`f(x)>0 if`

`(x-1)(2x-1)(x+3)>0`

So either

`(x-1)>0 and (2x-1)(x+3)>0`

or

`(x-1)<0 and (2x-1)(x+3)<0`

1. if first case

`(x-1)>0 and (2x-1)(x+3)>0`

`so`

`(x-1)>0 , (2x-1) ,(x+3)>0`

`x>1,x>1/2, x> -3`

smallest set is `x>1`

or

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

`x>1 and x<1/2 ,x<-3 `

No possible solution these inequalities have.

2.

`(x-1)<0 and (2x-1)(x+3)<0`

so

x-1 <0  and (2x-1)>0 and (x+3) <0

x<1  and x>(1/2)  and  x<-3

These inequalities have no solution

or

x-1<0 and  (2x-1)<0 ,(x+3)>0

x<1  and  x<(1/2) ,x>-3

These inequalities have solution if

-3<x<(1/2)

Finally

f(x)>0 if `x in(-3,1/2)U(1,oo)`