# How do you solve and write the following in interval notation: # x^4+4x^3-21x^2>=0#?

##### 2 Answers

#### Answer:

#### Explanation:

but

#### Answer:

#### Explanation:

Factor this to find when the expression is equal to

Factor

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

Factor with

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

Thus the expression equals

Use one test point in every interval between two pairs of zeros, and another test point between a zero and positive/negative infinities.

Before

We can plug in

#x=-8# into#x^2(x+7)(x-3)# .

#(-8)^2# is positive,#(-8+7)# is negative, and#(-8-3)# is negative, so their product will be positive. Thus the entire interval#(-oo,-7]# will be#>=0# .

Between

Plugging in

#x=-1# yields:

#overbrace((-1)^2)^+overbrace((-1+7))^(+)overbrace((-1-3))^(-) <=0# This is negative, so the interval is not a part of the solution set.

Between

#overbrace((1)^2)^+overbrace((1+7))^(+)overbrace((1-3))^(-) <=0# This is also negative, and not a part of the solution set.

From

Plugging in

#x=4# yields:

#overbrace((4)^2)^+overbrace((4+7))^(+)overbrace((4-3))^+ >=0#

Since this is positive,

Thus the entire solution is the union of the two positive intervals:

#(-oo,-7]uu[3,+oo)#