first question:
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.
Yes, you can use synthetic division to simplify the expression (x^5-1)/(x-1). Synthetic division is a method used to divide a polynomial by a linear factor.
To perform synthetic division, you need to write the coefficients of the polynomial in the dividend expression. In this case, the dividend is x^5-1, so the coefficients are 1, 0, 0, 0, 0, -1.
Then, write the linear factor (x-1) and its opposite sign (+1) as the divisor. Divide the first term of the dividend (1) by the divisor (1) and write the result above the line. Multiply the divisor by the result (1) and write it below the next coefficient.
```
1
________
1 | 1 0 0 0 0 -1
1
---
0
```
Then, add the next coefficient (0) to the result (1) and write it above the line. Multiply the divisor by the new result (1) and write it below the next coefficient.
```
1 1
________
1 | 1 0 0 0 0 -1
1
---
0 1
```
Repeat this process for each coefficient until you have gone through all the terms of the polynomial.
```
1 1 1 1 1
________
1 | 1 0 0 0 0 -1
1
---
0 1 1 1 1
```
The final result is the quotient of the division, which is the polynomial x^4 + x^3 + x^2 + x + 1.
Therefore, the simplified form of x^5-1 is (x-1)(x^4 + x^3 + x^2 + x + 1).
Regarding the limit of the cube root expression, the correct way to evaluate it is to substitute x=4 into the expression and take the cube root of the resulting value.
The expression is cuberoot(x^2-5x-4). So, when x=4:
cuberoot(4^2 - 5*4 - 4) = cuberoot(16 - 20 - 4) = cuberoot(-8).
Since the cube root of a negative number is a complex number, we cannot evaluate the limit at x=4.
Therefore, the limit of cuberoot(x^2-5x-4) as x approaches 4 does not exist.