Please can you help apply the factor method ( reference lesson on this topic) to find solution intervals. I can get the answer using test values but not via the method you illustrate in your lesson on this topic

The lesson

here? If yes, then:

x^3 - 3x^2 - 9x + 27 < 0

This solves as

x^2 (x-3) - 9 (x-3) < 0

(x^2 - 9) (x-3) < 0

This "factors" as. Important difference!

Zeroes

...of the corresponding

**equality**!

x < +3 or -3 for (x^2-9)<0

x < 3 for (x - 3)<0

1st, find the zeroes:

x^2 - 9 = 0 for (x - 3)(x + 3) = 0 so x = -3, x = +3

x - 3 = 0 for x = +3

So you have the three factors, x - 3 (twice) and x + 3, and you have the three zeros, x = 3 (twice) and x = -3. These two zeroes (one with "multiplicity 2") split the number line into intervals: (-infinity, -3), (-3, +3), (+3, +infinity).

Now look at the factors, to find their individual signs on each of these intervals:

factor x - 3:

This factor is positive for x - 3 > 0, so for x > 3. When x < 3 (that is, to the LEFT of x = 3), then the factor is negative.

factor x + 3:

This factor is positive for x + 3 > 0, so for x > -3. When x < -3 (that is, to the LEFT of x = -3), then the factor is negative.

Now draw the table with rows for the three factors. Make column bars for the two zeroes.

Code: Select all

```
table of signs:
-----+----+----+-----
x - 3: | |
-----+----+----+-----
x - 3: | |
-----+----+----+-----
x + 3: | |
-----+----+----+-----
--------|----|-------->
-oo -3 +3 +oo
--------|----|--------
```

There's a number line at the bottom. Now fill in the rows with the signs. For x - 3, remember that it's "minus" to the left of x = 3 and "plus" to the right:

Code: Select all

```
table of signs:
-----+----+----+-----
x - 3: - | - | +
-----+----+----+-----
x - 3: - | - | +
-----+----+----+-----
x + 3: | |
-----+----+----+-----
--------|----|-------->
-oo -3 +3 +oo
--------|----|--------
```

For the factor x + 3, it's "minus" to the left of x = -3 and "plus" to the right:

Code: Select all

```
table of signs:
-----+----+----+-----
x - 3: - | - | +
-----+----+----+-----
x - 3: - | - | +
-----+----+----+-----
x + 3: - | + | +
-----+----+----+-----
--------|----|-------->
-oo -3 +3 +oo
--------|----|--------
```

Now multiply the signs down the columns (that is, on each interval):

Code: Select all

```
table of signs:
-----+----+----+-----
x - 3: - | - | +
-----+----+----+-----
x - 3: - | - | +
-----+----+----+-----
x + 3: - | + | +
-----+----+----+-----
--------|----|-------->
-oo -3 +3 +oo
--------|----|--------
poly: - + +
```

So the polynomial is negative on (-infinity, -3), and positive on the other two intervals. It's on the x-axis at x = -3 and x = +3. It crosses (from negative to positive) at x = -3, but just touches at x = +3.

Does that make sense?