How do you solve #(x-4)(x+3)<0# using a sign chart?

1 Answer
Jan 2, 2017

Answer:

#-3 < x < 4#.

Explanation:

Find the critical points. Equate the "factors" to 0 to get the critical points. Use 0 as a critical point also.

#x - 4 = 0" "=>" "x = 4#

#x + 3 = 0" "=>" "x = "-"3#

The critical points are #{"-"3, 0, 4}#. It is only at these points that the sign of #(x-3)(x+4)# may change.

Now, pick a number that lies in each region (in-between/on either side of these critical points), plug it into each factor, and find the sign of each factor in each region. You'll make a table like this one.

#ul("                         "x < "-"3"      -"3 < x < 0"       "0 < x < 4"       "x > 4"   ")#
#x + 3"                  "-"                 "+"                    "+"                "+#
#ul(x - 4"                  "-"                 "-"                    "-"                "+"     ")#
#(x - 4)(x + 3)"  "+"                 "-"                    "-"                "+#

The last row is just the product of the two rows above it.

Also, since we manually added 0 as a critical point, we calculate #(x-4)(x+3)# when #x=0#:

#(0-4)(0+3)" "=" "("-"4)(3)" "=" ""-"12" "<" "0#

Based on this and the chart above,

#(x+3)(x-4)<0# when #-3 < x < 4#.