How do I find all zeros of 1. R(x)=x^5-x^4+9x^3-9x^2+20x-20

Please show how i approach this thankyou so much for your time

First, look for rational zeros.

The rational Root Theorem says that if rational roots exist, then they will look like this
p/q
(a fraction involving two integers, p and q)
where p is a divider of the "constant" (here, this is the -20 at the end)
and
where q is a divider of the highest-degree coefficient (here, it is the implicit "1" in front of x^5)

p/q = p/1 = p
we only need to worry about the possible dividers of 20 (and we have to check both + and -)

The suspects are:
1, 2, 4, 5, 10, 20 (plus or minus in each case).
[in this case, because ALL even powers have negative coefficients and ALL odd powers have positive coefficient, then there are no negative roots, only check the positive values)

You try them out, one by one, to see if they cause R(x) = 0

If so, then the value is a zero (a root is simply another name for the same thing).

Once you have found a root, then (x - root) is a factor.

You can divide the quintic (the 5th degree polynomial) by the factor(s) and you get a smaller polynomial, for which you continue to search for roots.

If a root is there more than once (and it can happen) you WILL HAVE to do the division in order to check if the same root works for the lower-degree polynomial.

------------

Factoring can also be good.

Here, because of the + - sequence, you can parse the function this way:

R(x) = (x^5 - x^4) + 9(x^3 - x^2) + 20(x - 1)
R(x) = x^4(x-1) + 9x^2(x-1) + 20(x-1)
(x-1) can be factored out:

R(x) = (x^4 + 9x^2 + 20)(x - 1)

Obviously, since (x-1) is a factor, then x = +1 will be a zero:

R(1) = (whatever)(1 - 1) = (whatever)(zero) = zero

At this point, you are left with
x^4 + 9x^2 + 20
which happens to be a "bi-quadratic"

If you replace the variable with u = x^2