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.
-$$f ( x ) = 3 x ^ { 3 } - 2 x ^ { 2 } + 9 x - 6$$
A) $6 ; f ( x ) = ( x - 6 ) \left( 3 x ^ { 2 } + 1 \right)$
B) $3 , \frac { 2 } { 3 } , 1 ; f ( x ) = ( 3 x - 2 ) ( x - 1 ) ( x - 3 )$
C) $\frac { 2 } { 3 } ; f ( x ) = ( 3 x - 2 ) \left( x ^ { 2 } + 3 \right)$
D) $- 3 , - 1 , \frac { 2 } { 3 } ; f ( x ) = ( 3 x - 2 ) ( x + 1 ) ( x + 3 )$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates there is 1 positive real zero because there is 1 sign change in $f(x)$. For $f(-x)$, there are no sign changes, indicating no negative real zeros. The Rational Zeros Theorem suggests possible rational zeros as factors of $\frac{6}{3}$, which are $\pm1, \pm2, \pm3, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{6}{3}$. Testing these, $\frac{2}{3}$ is a zero. Factoring $f(x)$ using $\frac{2}{3}$ leads to $f(x) = (3x - 2)(x^2 + 3)$, where $x^2 + 3$ has no real zeros.

Use Descartes' Rule of Signs and the Rational Zeros Theorem f(x)=3x3−2x2+9x−6f ( x ) = 3 x ^ { 3 } - 2 x ^ { 2 } + 9 x - 6f(x)=3x3−2x2+9x−6

Use Descartes' Rule of Signs and the Rational Zeros Theorem $f ( x ) = 3 x ^ { 3 } - 2 x ^ { 2 } + 9 x - 6$