solve the inequality and express in interval notation x^4+4x^3-21x^2 is > or equal to 0?

Answer 1

x^4+4x^3-21x^2
= x^2(x^2+4x-21)
= x^2(x-3)(x+7)
>= 0
then x^2>=0 for all x
so x=0 is a solution
and
(x-3)(x+7) >= 0
says x-3>= 0 and x+7>=0
ie x>=3> AND x>= -7
which says x>=3
since it incorporates both statements
OR
says x-3<= 0 and x+7<=0
ie x<=3> AND x<= -7
which says x<= -7
since it incorporates both statements
so all together
x element (-inf, -7]U[0,0]U[3, inf) in interval notation

the end

|repost|-the above answer is factored wrong

Answer 2

(x+5)(x-5)<=0 x+5=0 x=-5 x-5=0 x=5 you create a signum table x -5 5 x+5 - - - - 0 ++++++ x-5 - - - - - - - - - - - - - 0 +++++ (x+5)(x-5) + + +0 - - - -0 +++ the inequality could be much less or equivalent to 0 so which you will opt for the parte of the table the place the inequality is destructive , So x is between -5;5

Answer 3

x^4+4x^3-21x^2
= x^2(x^2+4x-21)
= x^2(x+7)(x-3) >= 0

x^2 >= 0 for all x

x=0 is a solution

now (x+7)(x-3) > =0 means x+ 7 >=0 and x-3 >=0 or x >= 3

or x+ 7 <= 0 and x-3 <= 0 or x <= -7

so solution x = 0 or ( - inf ,-7] or [3, inf)