Factorized form with (x-1) and/or (x+1)

A

Andrew Rubin

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Hi everyone

I am having some difficulties understanding a general concept in factorization. I would be glad if anyone could explain it to me or give me a reference. To illustrate, I will use the following factorization:

[MATH]x^3-x^2=x^2(x-1)[/MATH]
I get that [MATH]x^2[/MATH] is a common factor, but I don't understand why [MATH](x-1)[/MATH] is included. I come across several similar factorizations in my course, where the factorized formula include [MATH](x-1) [/MATH] or [MATH](x+1)[/MATH], so I guess there is a general concept that I have not grasped here.
 
If m is a factor of P, what can we say about P/m?
It's a factor: P/(P/m) = m.
Example: 5 is a factor of 15, what about 3? It has to be a factor too.
If x - 1 is the result of division of x3 - x2 by x2, then it's a factor too.
 
Factorization means finding a product of two or more factors that is equal to a given expression. It is typically the opposite of using the distributive property to expand a given product.

In the same way, factorizing a number (e.g. 15) means finding a product that is equal to it (in this case, [MATH]3\times5[/MATH]). Just giving one of the numbers would not answer the question, just as only giving one factor, [MATH]x^2[/MATH], would not answer your question.

In your example, if we gave you [MATH]x^2(x-1)[/MATH] and asked you to expand (distribute) it, would you be able to do that? That is how you would check that your factorization is correct. The second factor is there so that you will get the correct product.
 
Another note: Does x3-x2 really equal x. Remember that x is a place holder for a number and equals means that what is to the left and right of the equal sign, well they must be equal.

Does 33-32 really equal 3? Or does 33-32=32(3-1)??
 
Dr.Peterson said:
Factorization means finding a product of two or more factors that is equal to a given expression. It is typically the opposite of using the distributive property to expand a given product.

In the same way, factorizing a number (e.g. 15) means finding a product that is equal to it (in this case, [MATH]3\times5[/MATH]). Just giving one of the numbers would not answer the question, just as only giving one factor, [MATH]x^2[/MATH], would not answer your question.

In your example, if we gave you [MATH]x^2(x-1)[/MATH] and asked you to expand (distribute) it, would you be able to do that? That is how you would check that your factorization is correct. The second factor is there so that you will get the correct product.
Click to expand...

Thank you for your good explanation! I now see how [MATH]x^2(x−1) = x^3−x^2 [/MATH]. I am working on factorization and I am trying to expand the factorization logic to another problem. I have

[MATH]f'(x) = e^{-2x}-2e^{-2x}x[/MATH]
which in factorized form becomes,

[MATH]f'(x) = e^{-2x}(1-2x)[/MATH]
What I get, by distribution, does not make sense

[MATH]e^{-2x} \cdot 1 - e^{-2x} \cdot (-2x) [/MATH][MATH]= e^{-2x}- e^{-2x}(-2x)[/MATH][MATH]= e^{-2x}+2xe^{-2x}[/MATH]
Hopefully you can provide some more intuition on this. I think most of my course makes sense, but I am struggling with grasping factorization, which is a bit frustrating as it seems like it should be pretty straigthforward.
 
lev888 said:
Where did "-" before e come from?
Click to expand...
Sorry, I guess the appropriate use of the distributive law is

[MATH]e^{-2x} \cdot 1 + e^{-2x} \cdot (-2x)[/MATH]
as we add factors when using this law, so I get

[MATH]e^{-2x} - 2xe^{-2x}[/MATH]
To me this now seem to be equivalent to the expression above before factorization, i.e.

[MATH]e^{-2x} - 2e^{-2x}x[/MATH]
because the order of multiplication should not matter?
 
Exactly.

Negatives can lead to errors in distribution or factoring, so when I see them, I slow down and write more steps! And you're right that the commutative property of multiplication allows us to change the order within a term. I would tend to write this one as
[MATH]e^{-2x} - 2xe^{-2x}[/MATH] rather than the way you initially wrote it, but that's just a matter of taste.
 
Dr.Peterson said:
Exactly.

Negatives can lead to errors in distribution or factoring, so when I see them, I slow down and write more steps! And you're right that the commutative property of multiplication allows us to change the order within a term. I would tend to write this one as
[MATH]e^{-2x} - 2xe^{-2x}[/MATH] rather than the way you initially wrote it, but that's just a matter of taste.
Click to expand...
Thank you, this was really helpful!
 
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