IntMath Home » Forum home » Inequalities » Inequalities

IntMath forum | Inequalities

Inequalities [Solved!]

My question

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

3. Solving Non-Linear Inequalties

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.

X

On this site under Algebra: Inequalities - Lesson 3 - Example 2:

Can you break down the factoring of x^3 - 4x^2 + x + 6 &lt; 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.

Re: Inequalities

Phinah

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):

Factor and Remainder Theorems

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.

Re: Inequalities

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.

Re: Inequalities

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`.)

Reply

You need to be logged in to reply.

Search IntMath, blog and Forum