Question

solve the following polynomial inequality
x^3 + x^2 + 64x + 64 > 0

Answer 1

Can you show us what you've done with the question please?

Answer 2

x^3+x^2+64x+64>0
x^2(x+1)+64(x+1)>0
(x^2+64)(x+1)>0
So,
x^2+64>0 or x+1>0
if x^2+64>0
x^2>-64
so x will be a non real number
if x+1>0
x>-1

Answer 3

@shrinifores
For this to be true:
\[\Large (x^2+64)(x+1)>0\] either both sets of brackets must be positive, or both must be negative.

ie (x^2+64) > 0 AND (x+1)>0

or

(x^2+64) < 0 AND (x+1)<0


Also... x^2>-64 is true for all real numbers - x^2 is greater than -64 for all real numbers. (x^2+64) < 0 on the other hand, x^2 < -64 is not true for any x.

Answer 4

Just solve this one, since the other one is never true

(x^2+64) > 0 AND (x+1)>0
^ always positive ^ solve x+1>0