can someone please help me with these math problems!!..?

Question

Problem 1: Find all roots of the polynomial x^3-x^2+16x-16. & Problem 2: Find the vertical asymptote of the rational function f(x)=3x-12/4x-2. & Problem 3: Find the horizontal asymptote of the rational function f(x)=8x-12/4x-2.

Answer 1

#1.

x^3 - x^2 + 16x - 16 = 0
x^2(x - 1) + 16(x - 1) = 0
(x - 1)(x^2 + 16) = 0
x = 1, x = 4i, x = -4i

4i is an imaginary number where i is equal to the square root of -1. If you square i you get -1.

#2.
Set the denominator equal to 0.

4x - 2 = 0
4x = 2
x = 1/2

#3.
8x - 12
---------
4x - 2

The limit as x approaches infinity equals 2. The -12 and -2 are insignificant as X approaches infinity so all you have left is 8x and 4x. If you divide 8x by 4x you get 2.

Answer 2

#1: x=1 and -4i, 4i (if you've learned irrational)
what i did: if the polynmial is more that four, you group. I grouped the first two and the last two and created two pairs. Then i removed the common factor from each pair. The first common factor is x^2 and the second is 16. so i got: x^2(x-1)+16(x-1)=0. Then, because (x-1) is a common factor, I combined the groups to get: (x^2+16)(x-1)=0. This is factored. Now, set each factor to equal 0: x^2+16=0 and x-1=0. (these are seperate equations now) so, now solve for x and you know the rest.

Answer 3

For the first one, you need to find a way to reduce it to a quadratic. Fortunately, it is not hard to see that x = 1 is one of the roots. So, divide (x - 1) into your cubic, reducing it to x^2 + 16. The quadratic has two imaginary roots, 4i and -4i. Your three roots are 1, 4i, and -4i.

Answer 4

I am not sure if this is what you are supposed to do, but all of those questions are very easily answered by plotting those functions on a graph.