x^3 - 4x^2 + x + 6 < 0
x^3 - 4X^2) + x + 6 <0
X^2(X-4)+(X+6)<0
(X^2+1)(X-4)(X+6)<0

So the roots given this way are different from the roots given in the solution.

X

On this site under Algebra: Inequalities - Lesson 3 - Example 2:
Can you break down the factoring of x^3 - 4x^2 + x + 6 < 0 or give me the right starting point?
Thanks.

Relevant page
<a href="/inequalities/3-solving-non-linear-inequalities.php">3. Solving Non-Linear Inequalties</a>
What I've done so far
I used grouping.
x^3 - 4x^2 + x + 6 < 0
x^3 - 4X^2) + x + 6 <0
X^2(X-4)+(X+6)<0
(X^2+1)(X-4)(X+6)<0
So the roots given this way are different from the roots given in the solution.

You can do grouping only if the expressions in brackets are the same.

So if our question was like this:

Factor `x^3 - 6x^2 + x - 6`

Then we would proceed as follows:

`x^3 - 6x^2 + x - 6 = x^2(x-6) + (x-6)`

Now we can group them like this:

`(x^2 + 1)(x - 6)`

But in the example you are talking about, `x^3 - 4x^2 + x + 6`, there is no common part that we can extract into brackets.

For this problem (if you have to do it on paper rather than using a computer), you would need to use the techniques given in this section (which is what I linked to in my solution on that page):

By the way, be careful with upper- and lower-case variables.

In mathematics, `X` does not equal `x`. They stand for different numbers.

X

Phinah
You can do grouping only if the expressions in brackets are <b>the same</b>.
So if our question was like this:
Factor `x^3 - 6x^2 + x - 6`
Then we would proceed as follows:
`x^3 - 6x^2 + x - 6 = x^2(x-6) + (x-6)`
Now we can group them like this:
`(x^2 + 1)(x - 6)`
But in the example you are talking about, `x^3 - 4x^2 + x + 6`, there is no common part that we can extract into brackets.
For this problem (if you have to do it on paper rather than using a computer), you would need to use the techniques given in this section (which is what I linked to in my solution on that page):
<a href="http://www.intmath.com/equations-of-higher-degree/2-factor-remainder-theorems.php">Factor and Remainder Theorems</a>
By the way, be careful with upper- and lower-case variables.
In mathematics, `X` does not equal `x`. They stand for different numbers.

But the issue is that in the Remainder and Factor Theorems page, with each example and exercise, we are given a factor from the start along with the polynomial. For instance, Examples 2 and 3:(x-2), Example 4: (x + 4), etc.

With this inequality example that I am trying to factor, we cannot immediately use those methods without first having a starting point. I want to learn to do it on paper. That is why I was asking for a starting point. I see none. Thanks.

X

But the issue is that in the <a href="http://www.intmath.com/equations-of-higher-degree/2-factor-remainder-theorems.php">Remainder and Factor Theorems</a> page, with each example and exercise, we are given a factor from the start along with the polynomial. For instance, Examples 2 and 3:(x-2), Example 4: (x + 4), etc.
With this inequality example that I am trying to factor, we cannot immediately use those methods without first having a starting point. I want to learn to do it on paper. That is why I was asking for a starting point. I see none. Thanks.

Hello Phinah. It's just trial and error with a bit of careful observation.

Look at the final number in your example question. It's 6. We can multiply the following to get 6:

`1 xx 6`
`-1 xx -6`
`2 xx 3`
`-2 xx -3`

So that gives me a big hint that one of the following numbers will be a good place to start:

`-6, -3, -2, -1, 1, 2, 3, 6`

You just keep going until you hit one that "works" (that is, when you substitute it into the polynomial, it will give you `0`.)

X

Hello Phinah. It's just trial and error with a bit of careful observation.
Look at the final number in your example question. It's 6. We can multiply the following to get 6:
`1 xx 6`
`-1 xx -6`
`2 xx 3`
`-2 xx -3`
So that gives me a big hint that one of the following numbers will be a good place to start:
`-6, -3, -2, -1, 1, 2, 3, 6`
You just keep going until you hit one that "works" (that is, when you substitute it into the polynomial, it will give you `0`.)