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March 6, 2015

March 6, 2015

Posted by **.** on Sunday, September 3, 2006 at 9:55am.

could x^5-1 simplified?

what is the limit of cuberoot(x^2-5x-4) as x approaches 4?

For the first one:

could x^5-1 simplified?

Yes, the expression can be factored as the 5th roots of unity. First, divide it by (x-1) to get (x^5-1)=(x-1)*4th deg poly. I'm not going to do the work, so you'll have to do the division. The 4th degree poly can be factored furthered , but I doubt you'd be expected to know how to do this in high school algebra.

For the second question:

what is the limit of cuberoot(x^2-5x-4) as x approaches 4?

We want the lim x->4 of x^2-5x-4

This is the same as (lim x->4)^2 -5(lim x->4) -4 =

16-20-4 = -8

We can take the limit inside of continuous functions, and all polys are continuous. This should be proven as a theorem in your text somewhere. Sometimes it's in the appendix for an intro text.

For the 2nd one I see I missed the cube root part. The limit can still be taken inside, but you'll need to take the 3rd root of the expression. Thus we have,

((lim x->4)^2 -5(lim x->4) -4 =

16-20-4)^(1/3) = something you can do.

BTW, you should read the expression "lim x->4" as the limit of x as x approaches or tends to 4.

Just to reiterate too, if f(x) is continuous, then the lim x->a of f(x) = f(lim x->a). There should be a statement or result to the effect that polynomials of all degrees are continuous, therefore we can evaluate the limit for x=a. If the function is not continuous at a point 'a' we need to look at left and right hand (or one-sided) limits.

i didn't really understand what you meant by 4th deg poly, or what exactly i have to do.

do i just do synthetic division, (x^5-1)/(x-1)?

Yes. I had forgotten what the expression for this was, I see what I meant was synthetic division. I'm more accustomed to the word 'factor(ing)' to describe this process.

In general, if a poly of degree n is factored as two polys of deg p and q, then n=p+q. The degree of a poly is the degree of it's highest term. In this case, x^5-1 should factor as

(x-1)(x^4 + lower degree terms) I don't know if you're exected to factor the x^4 poly, but it does factor; it requires result about the roots of unity to do it.

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