Question 309805
How do I write the solution set of {x| } for {{{(x+20)(x-10)(x+11)>0}}}
<pre><font size = 4 color = "indigo"><b>
Find all the critical values by setting each of the factors on the
left = 0

x + 20 = 0 gives critical value x = -20

x - 10 = 0 gives critical value x = 10

x + 11 = 0 gives critical value x = -11

Mark them on a number line:
<font size = 1>
----------o-----------------------------------o-----------------------------------------------------------------------------------o---------
-22 -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10 11 12 13
</font>
Choose a value left of -20, the left-most critical value,
say x = -21.  Test it by substituting in the inequality:

{{{(x+20)(x-10)(x+11)>0}}}
{{{(-21+20)(-21-10)(-21+11)>0}}}
{{{(-1)(-21)(-10)>0}}}
{{{-210>0}}}

That is false so we DO NOT shade the part of the number line
left of -20.  So we still have the unshaded number line
<font size = 1>
----------o-----------------------------------o-----------------------------------------------------------------------------------o---------
-22 -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10 11 12 13
</font>
Choose a value between -20 and -11, say x = -12.  Test it by substituting in the inequality:

{{{(x+20)(x-10)(x+11)>0}}}
{{{(-12+20)(-12-10)(-12+11)>0}}}
{{{(8)(-22)(-1)>0}}}
{{{176>0}}}

That is true so we DO shade the part of the number line between -20 and
-11.  So we have:
<font size = 1>
----------o===================================o-----------------------------------------------------------------------------------o--------->
-22 -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10 11 12 13
</font>
Choose a value between -11 and 10, say x = 0.  Test it by substituting in the inequality:

{{{(x+20)(x-10)(x+11)>0}}}
{{{(0+20)(0-10)(0+11)>0}}}
{{{(20)(-10)(11)>0}}}
{{{-2200>0}}}

That is false so we DO NOT shade the part of the number line between -20 and
-11.  So we still have:
<font size = 1>
----------o===================================o-----------------------------------------------------------------------------------o--------->
-22 -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10 11 12 13
</font>
Choose a value right of 10, the right-most critical value,
say x = 11.  Test it by substituting in the inequality:

{{{(x+20)(x-10)(x+11)>0}}}
{{{(11+20)(11-10)(11+11)>0}}}
{{{(21)(1)(22)>0}}}
{{{462>0}}}

That is true so we DO shade the part of the number line to the right of
10.  So we have:
<font size = 1>
----------o===================================o-----------------------------------------------------------------------------------o=========>
-22 -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10 11 12 13
</font>

In set-builder notation the solution set is written:

{x | -20 < x < -11 OR x > 10}

In interval notation the solution set is written:

(-20,-11) U (10,{{{infinity}}})

</pre>

Edwin</pre></font></b>