Question

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

Answer 1

`-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`.