Try + and - the prime factors of 48:
1, 2, 3, 4, 6, 8, 12 and 16.
When you find one that works, call it 'a' and divide the f(x) polymonial by (x-a) to get the quadratic factor. The other two roots can easily be found with that.
It looks to me like one root is +3. Therefore x-3 is a factor.
Now evaluate [x^3-3x^2+16x-48]/(x-3)
It should give you a quadratic equation with no remainder. Factor that.
find all zeros of the following polynomial. write the polynomial in factored form.
f(x)=x^3-3x^2+16x-48
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