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

rational root theorem is your friend here

the 1 term is multiplied by -20
the x^5 is multiplied by 1

factors of 20 are ±1, ±2, ±4, ±5, ±10, ±20
factors of 1 are ±1
so 12 possibilities are ±1, ±2, ±4, ±5, ±10, ±20

trying these, only x=1 (that is, x - 1 = 0) works - as we know an x^5 must have 5 roots, the other 4 must be complex (unless the root is repeated - but read on)

divide the thing by (x-1) and you get x^4 + 9x^2 + 20
this can be factored much like an x^2 - think "what adds to 9 and multiplies to 20" - 4 and 9.

so (x^4 + 9x^2 + 20) = (x² + 4)(x² + 5)
and we know (x-1) is a root

so your equations are
x-1 = 0 => x = 1
x² + 4 = 0 so x² = -4 => x = 2i, -2i
x² + 5 = 0 so x² = -5 => x = i√5, -i√5

so your roots are 1, -2i, 2i, -i√5, i√5

R(x)=x^5-x^4 + 9x^3 - 9x^2 + 20x-20
=x^4(x-1) + 9x^2(x-1) + 20(x-1)
=(x-1) (x^4 + 9x^2 +20)
=(x-1)(x^4 + 5x^2 + 4x^2 + 20)
=(x-1){x^2(x^2 +5 ) + 4(x^2+5)}
=(x-1)(x^2+4)(x^2+5)
make R(x)=0 gives
either x-1=0 => x=1
or x^2+4=0 => x = sqrt(-4) => +2i , -2i
or x^2=5=0 => x= sqrt(-5) =>+-sqrt(-5)

You can see by inspection that x=1 will be a zero of R(x).

Now divide R(x) by (x-1) and you get a 4th order polynomial. Look for another zero and do the smae thing.

i would probably factor it if possible and set it to 0 and see what comes out