How do you solve #(x + 8) ( 2x + 8) ( 3x - 27) \geq 0#?

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Answer 1 · 2017-11-24T09:28:41

#( -8 ) <= x <= ( -4 )#

OR

# x >= 9#

Interval Notation: #[ -8, -4 ] uu [ 9, oo )#

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We are given the inequality

#( x + 8 ) ( 2z + 8 ) ( 3x - 27 ) >= 0#

We will factorize as much as possible as follows:

# (x + 8) * 2(x + 4) * 3( x- 9 ) >= 0#

We can rewrite the above expression as

# 6(x + 8)(x + 4)( x- 9 ) >= 0#

We observe that

#( x + 8)# is ZERO when #x = ( -8 )#

#( x + 4)# is ZERO when #x = ( -4 )#

#( x - 9)# is ZERO when #x = ( +9 )#

Next, we will take each of these values and place them on a Number Line.

We will also be checking around these values to figure out whether the entire expression is either POSITIVE or NEGATIVE.

Please refer to the attached TABLE for the procedure.

With reference to the TABLE, we must find
when is our polynomial #>= 0#

That is exactly what our original problem is looking for.

We find our required intervals to be

#( -8) <= x <= (-4) #

OR

#x >= +9#