I'm struggling with the problem $f(x)= x^3-5x^2-4x-20$. I've found the factors but none of them seem to be working, how exactly can I find all the zeroes by applying the fundamental therom of algebra? The factors I have found are: 1, 2, 4, 5, 10, 20. Making all the possible zeroes negative or positive versions of those numbers.
2026-03-27 15:35:29.1774625729
How to find zeroes of polynomial function?
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I'm not sure, but you seem to misunderstand the term factor when applied to polynomials. The Fundamental Theorem of Algebra says that for a cubic polynomial, like the one you present, will have at most 3 factors. Yet you list six factors, which coincidentally are the six factors of 20.
Now, the 20 in your polynomial is the constant, the least significant part of the polynomial. The FTA looks at the term with the highest degree as the most significant part of the polynomial; in fact, the FTA says that degree gives you the number of roots (including multiple roots and including complex numbers).
The factors of a polynomial are the linear terms which, when multiplied together, give the polynomial. To compare this to integers (it's not a perfect analogy), we would want to list just the prime factors of 20, not all of them. So, $20 = 2*2*5$.
If $f(x)= x^3-5x^2-4x+20$ (note the change of sign; as written in the original question, $f(x)$ had two complex roots), then $f(x)=(x+2)(x-2)(x-5)$. This means the factors are $(x+2), (x-2),$ and $(x-5)$. And the zeroes (roots) are $x=-2, x=2$, and $x=5$.
Hope this helps clarify these things in your understanding.