SOLUTION: Can you explain to me how to find all zeros of the following polynomial in factored form: f(x)=7x^3-x^2+7x-1

f(x) = 7x³-x²+7x-1

Factor the first two terms:

f(x) = x²(7x-1)+7x-1

Factor the last two terms.  [Note: that you can always factor
a 1 out of any expression, even prime polynomials]

So we factor 1 out of the last two terms:

f(x) = x²(7x-1)+1(7x-1)

Now we can factor out (7x-1)

f(x) = (7x-1)(x²+1)

x²+1 is the sum of two squares but we can make it into the
difference of two squares by using the fact that 1 = -i²,
so we substitute -i² for 1 in the second parentheses:

f(x) = (7x-1)(x²-i²)

f(x) = (7x-1)(x-i)(x+i)

That is the factored form of f(x)

The zeros are found by setting f(x) = 0 and solving for x

       (7x-1)(x-i)(x+i) = 0

     7x-1=0; x-i=0; x+i=0
       7x=1;   x=i;   x=-i
        x=

The three zeros are , i, and -i.

Edwin