Question

Find the number of possible negative real zeros for f(x)=x^3-4x^2-3x-9.
Please help:)

Answer 1

\(\color{#000000}{ \displaystyle f(x)=3x^3+4x^2-5x +24 }\)

would have possible zeros of,
\(\color{#000000}{ \displaystyle \pm 24/3 }\)
\(\color{#000000}{ \displaystyle \pm 8/3 }\)
\(\color{#000000}{ \displaystyle \pm 6/3 }\)
\(\color{#000000}{ \displaystyle \pm 4/3 }\)
\(\color{#000000}{ \displaystyle \pm 3/3 }\)
\(\color{#000000}{ \displaystyle \pm 2/3 }\)
\(\color{#000000}{ \displaystyle \pm 1/3 }\)

Answer 2

basically the constant term (and all of it's divisors) divided by the first coefficient.