Question

What are the possible rational zeros of f(x) = x4 + 6x3 - 3x2 + 17x - 15?

Answer 1

Rational root theorem states that if the coefficient of the leading term is 1 (it is here, as \(x^4 = 1x^4\)) that all possible rational zeros must be factors of the coefficient of the constant term (which is \(-15\) here\).

Answer 2

Remember that any polynomial with rational zeros \(r_1,r_2,...r_n\) can be written as a product:
\[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\]\(a\) is a constant to make the curve fit an arbitrary point, if needed. If you expand the factored polynomial, the constant term will end up as \[r_1*r_2*...r_n\]

Unfortunately, it isn't always easy to figure out what the zeros are from looking at the product, if the zeros were not prime numbers! You can quickly get a list of possible zeros that is much longer than the actual list of zeros.

Answer 3

\(\bf f(x) = {\color{red}{ 1}}x^4 + 6x^3 - 3x^2 + 17x - {\color{blue}{ 15}}
\\ \quad \\
\textit{possible zeros}\implies \pm\cfrac{{\color{blue}{ \textit{factors of constant}}}}{{\color{red}{ \textit{factors of leading term coefficient}}}}\)

Answer 4

@jdoe0001 extends it to the (unwelcome) case where the leading term coefficient is something other than 1. You know you're having a bad day when both the constant and the leading term coefficient both have about 19 different factors :-)

Answer 5

heehe

Answer 6

As it turns out, after you go to the trouble to make the list of possible rational zeros of this equation, none of them turn out to be actual zeros.