## using factor method to solve cubic inequality

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
DG1
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Joined: Sat Oct 10, 2015 5:06 am
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### using factor method to solve cubic inequality

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

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

This solves as
x^2 (x-3) - 9 (x-3) < 0
(x^2 - 9) (x-3) < 0

Zeroes

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

Now I get stuck in terms of constructing the table that helps identify where the earn in +ve or -ve.

Thank you
Last edited by PM_Admin on Sat Oct 10, 2015 7:10 pm, edited 1 time in total.
Reason: Splitting new post from very old thread.

maggiemagnet
Posts: 358
Joined: Mon Dec 08, 2008 12:32 am
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### Re: using factor method to solve cubic inequality

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?

DG1
Posts: 3
Joined: Sat Oct 10, 2015 5:06 am
Contact:

### Re: using factor method to solve cubic inequality

Thank you for your kind assistance.