Use Descartes' Rule of Signs and the Rational Zeros Theorem to find all the real zeros of the polynomial function. Use the
zeros to factor f over the real numbers.
-$$f ( x ) = 5 x ^ { 4 } - 8 x ^ { 3 } + 23 x ^ { 2 } - 32 x + 12$$
A) $- 4 , - 1,1 , \frac { 3 } { 5 } ; f ( x ) = ( x - 1 ) ( 5 x - 3 ) ( x + 1 ) ( x + 4 )$
B) $1 , \frac { 3 } { 5 } ; f ( x ) = ( x - 1 ) ( 5 x - 3 ) \left( x ^ { 2 } + 4 \right)$
C) $4 , \frac { 3 } { 5 } ; f ( x ) = ( x - 4 ) ( 5 x - 3 ) \left( x ^ { 2 } + 1 \right)$
D) $- 4 , - 1,1 , - \frac { 3 } { 5 } ; f ( x ) = ( x - 1 ) ( 5 x + 3 ) ( x + 1 ) ( x + 4 )$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates the number of positive real zeros by counting the sign changes in $f(x)$ and for negative real zeros by applying the rule to $f(-x)$. For $f(x) = 5x^4 - 8x^3 + 23x^2 - 32x + 12$, there are two sign changes, suggesting up to 2 positive real zeros. For $f(-x) = 5x^4 + 8x^3 + 23x^2 + 32x + 12$, there are no sign changes, indicating no negative real zeros. The Rational Zeros Theorem suggests possible rational zeros as factors of the constant term over factors of the leading coefficient, which are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ and their variants over $5$. Testing or further analysis would reveal the actual zeros, and among the given options, only choice B correctly identifies $1$ and $\frac{3}{5}$ as zeros and correctly factors $f(x)$ into a product of linear terms for these zeros and a quadratic term that does not factor further over the real numbers.

Use Descartes' Rule of Signs and the Rational Zeros Theorem f(x)=5x4−8x3+23x2−32x+12f ( x ) = 5 x ^ { 4 } - 8 x ^ { 3 } + 23 x ^ { 2 } - 32 x + 12f(x)=5x4−8x3+23x2−32x+12

Use Descartes' Rule of Signs and the Rational Zeros Theorem $f ( x ) = 5 x ^ { 4 } - 8 x ^ { 3 } + 23 x ^ { 2 } - 32 x + 12$