1. ## Factoring Problem

How does

$1 - 9x^{2} + 24x^{4} - 16x^{6}$

become

$(1 - x^{2})(16x^{4} - 8x^{2} + 1)$ ?

2. ## Re: Factoring Problem

Originally Posted by Jason76
How does
$1 - 9x^{2} + 24x^{4} - 16x^{6}$
become
$(1 - x^{2})(16x^{4} - 8x^{2} + 1)$ ?
@Jason76, you are basically lazy are you not?

You demand to know "How does...become"?

Are you to lazy to multiply $(1 - x^{2})(16x^{4} - 8x^{2} + 1)$ out to see how it all works?

3. ## Re: Factoring Problem

Hey Jason76.

You might want to look at this:

Rational root theorem - Wikipedia, the free encyclopedia

4. ## Re: Factoring Problem

$1 - 9x^{2} + 24x^{4} - 16x^{6}$

Pulling out x^{2}

$1 - [x^{2}(9 + 24x^{2} - 16x^{4})]$

$(1 - x^{2})(9 + 24x^{2} - 16x^{4})$

After that, nothing in the right linear has any common factor, so you can't do anymore. But that's not the final answer.

5. ## Re: Factoring Problem

Originally Posted by Jason76
$1 - 9x^{2} + 24x^{4} - 16x^{6}$

Pulling out x^{2}

$1 - [x^{2}(9 + 24x^{2} - 16x^{4})]$

$(1 - x^{2})(9 + 24x^{2} - 16x^{4})$

After that, nothing in the right linear has any common factor, so you can't do anymore. But that's not the final answer.
You did nothing about replay #2.
Are you really that lazy?

6. ## Re: Factoring Problem

Originally Posted by Plato
You did nothing about replay #2.
Are you really that lazy?
You can't do anymore cause there are no more common factors from what I see. 9, 24, and 16 have no common factor.

7. ## Re: Factoring Problem

Originally Posted by Jason76
You can't do anymore cause there are no more common factors from what I see. 9, 24, and 16 have no common factor.
If you do not have a complete understanding of how multiplication works, then you cannot understand factoring.

8. ## Re: Factoring Problem

Originally Posted by Jason76
How does

$1 - 9x^{2} + 24x^{4} - 16x^{6}$

become

$(1 - x^{2})(16x^{4} - 8x^{2} + 1)$ ?
Hello Jason!
First you use rational root theorem to find roots then you will use long polynom division to divide the root out.