I'm not understanding something (everything) in this example

P

predondo

New member
Joined
Jul 27, 2021
Messages
7
Capture.PNG
It took me longer than I'm comfortable admitting to understand how it works simplifying x^2 - 1 into (x - 1)(x + 1), (does that only work when the constant is a 1 or negative one?) and I'm immediately stuck on something else!
As I understand it, the entirety of (x - 1)(x +1) is the denominator, so wouldn't all that multiply by the numerator and denominator of 1/(x - 1) resulting in (x - 1)(x +1)/(x - 1)(x +1)(x - 1)?
And, following the formula for adding fractions that preceded this part of the video, the x - 1 denominator should multiply across to the numerator of the other fraction, so the end result would be something like [(3x-1)(x-1)+(x - 1)(x +1)] / (x-1)(x+1)(x-1)
This is obviously not what he'd doing, what information/knowledge am I missing or misunderstanding?
 
predondo said:
View attachment 28409
It took me longer than I'm comfortable admitting to understand how it works simplifying x^2 - 1 into (x - 1)(x + 1), (does that only work when the constant is a 1 or negative one?) and I'm immediately stuck on something else!
As I understand it, the entirety of (x - 1)(x +1) is the denominator, so wouldn't all that multiply by the numerator and denominator of 1/(x - 1) resulting in (x - 1)(x +1)/(x - 1)(x +1)(x - 1)?
And, following the formula for adding fractions that preceded this part of the video, the x - 1 denominator should multiply across to the numerator of the other fraction, so the end result would be something like [(3x-1)(x-1)+(x - 1)(x +1)] / (x-1)(x+1)(x-1)
This is obviously not what he'd doing, what information/knowledge am I missing or misunderstanding?
Click to expand...
Please multiply (a+b)(a-b). What do you get?

When adding fractions we want to get the same denominator. In this case to get the same denominator on the right we only need to multiply the right fraction by (x+1).
 
Last edited:
Do you know how to calculate (a- b)(a+ b)?

By the "distributive law", which says that x(y+ z)= xy+ xz, (a- b)(a+ b)= (a- b)a+ +(a-b)b and applying the distributive law again (a- b)a= a^2- ab and (a- b)b= ab- b^2. So (a- b)(a+ b)= a^2- ab+ ab- b^2= a^2- b^2 since -ab+ ab= 0.

Knowing that, it follows that we can factor a^2- b^2 as (a- b)(a+ b).

It is also good to know that (a+ b)^2= (a+b)(a+ b)= (a+ b)a+ (a+ b)b= a^2+ ab+ ab+ b^2. Now ab and ab do not cancel but add: ab+ ab= 2ab so that (a+ b)^2= a^2+ 2ab+ b^2.

Knowing that it follows that we can factor a^2+ 2ab+ b^2= (a+ b)^2.
 
You must log in or register to reply here.
Top