Find the zeros and their multiplicities. Consider using Descartes' rule of signs and the upper and lower
bound theorem to limit your search for rational zeros.
-$$f ( x ) = 20 x ^ { 3 } - 47 x ^ { 2 } + 11 x + 6$$
A) $1 , - 3$, and $- 2$ (each with multiplicity 1 )
B) $- \frac { 1 } { 4 } , \frac { 3 } { 5 }$, and 2 (each with multiplicity 1 )
C) $\frac { 1 } { 4 } , - \frac { 3 } { 5 }$, and $- 2$ (each with multiplicity 1 $)$
D) $- 1,3$, and 2 (each with multiplicity 1)

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates the number of positive and negative real roots based on sign changes. For $f(x) = 20x^3 - 47x^2 + 11x + 6$, there are two sign changes, suggesting up to two positive real roots. For $f(-x) = -20x^3 - 47x^2 - 11x + 6$, there is one sign change, indicating exactly one negative real root. The Rational Root Theorem suggests that rational roots, if they exist, are of the form $\pm \frac{p}{q}$, where $p$ is a factor of the constant term (6) and $q$ is a factor of the leading coefficient (20). Testing rational roots within the bounds and considering the given options, option B provides a set of roots that are plausible given the constraints of the polynomial, even though the actual process of finding these specific roots would require polynomial division or another root-finding method to verify.

Find the Zeros and Their Multiplicities f(x)=20x3−47x2+11x+6f ( x ) = 20 x ^ { 3 } - 47 x ^ { 2 } + 11 x + 6f(x)=20x3−47x2+11x+6

Find the Zeros and Their Multiplicities $f ( x ) = 20 x ^ { 3 } - 47 x ^ { 2 } + 11 x + 6$