# How do i solve this?

1. Oct 3, 2004

2. Oct 3, 2004

### Dr-NiKoN

x^2 + 7x > 6 + 2x^3 = -2x^3 + x^2 + 7x - 6 > 0

Try solving for:
-2x^3 + x^2 + 7x - 6 = 0

3. Oct 3, 2004

### Gughanath

I cant solve that sorry...could you help any further?

4. Oct 3, 2004

### Tide

I noticed the sum of the coefficients is 1 so (x-1) is a factor:

$$2x^3-x^2-7x+6 = (x-1)(x+2)(2x-3)$$

That should get you started!

5. Oct 6, 2004

### aekanshchumber

try it,
x^2 + 7x>6x + 2x^3
or, 2x^3 - x^2 -x<0
or, x(2x^2 - x - 1)<0
this implies, either x<0 or (2x^2 - x - 1) <0
as the expression is negative which is only possible if one of the term is negative.

6. Oct 6, 2004

### Integral

Staff Emeritus
$$x^2 + 7x > 6 + 2x^3$$