Factoring (u^3-1): Homework Solution

[3.141592654]

Homework Statement

This is a limit problem but what I'm stuck on is algebra: find lim as u approaches 1 of ((u^4)-1)/((u^3)-1).

Homework Equations

The Attempt at a Solution

1 is a root and (u-1) is a common factor, so I want to rearrange the numerator and denominator so I can cancel (u-1) in the hopes that it will leave an equation that does not have a zero denominator when I plug in 1 to find the limit.

so the numerator either = ((u^2)-1)((u^2)+1) or (u-1)((u^3)+1) right?

The denominator I'm not sure how to write so that it has a factor (u-1) and = (u^3)-1. Any help is appreciated!

 

[rock.freak667]

u³-1= (u-1)(au²+bu+c)

so just equate coefficients now and you will be able to get it out.

so the numerator either = ((u^2)-1)((u^2)+1) or (u-1)((u^3)+1) right?

Not or. There is only one.

u⁴-1 =(u²)²-1

the right side looks like a²-b² which can be factored as (a+b)(a-b). So the first one is correct. Can you factor u²-1 again?

 

[zcd]

It can be seen immediately that u=1 is one of the solutions, so you can apply long divison with u-1 to find the other part.

 

[Mark44]

Originally posted by 3.141592654
so the numerator either = ((u^2)-1)((u^2)+1) or (u-1)((u^3)+1) right?

Not or. There is only one.

u⁴-1 =(u²)²-1

There is only one factorization if you are limited to prime factorizations, but without this restriction, there can be multiple factorizations. The prime factorization is unique, up to the order of the factors. Pi's first factorization is correct and his second is not, but another would be (u - 1)(u^3 + u^2 + u + 1).

It works the same way with polynomials as it does with integers. For instance, 40 = 4*10 = 5*8 = 2*20 = 2*2*2*5. All of these are valid factorizations of the number 40, but only one of them is a prime factorization.

the right side looks like a²-b² which can be factored as (a+b)(a-b). So the first one is correct. Can you factor u²-1 again?