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.
-$$f ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 9 x - 18$$
A) $- 3 , - 2,3 ; f ( x ) = ( x + 3 ) ( x + 2 ) ( x - 3 )$
B) $- 3,2,3 ; f ( x ) = ( x + 3 ) ( x - 2 ) ( x - 3 )$
C) $- 2 ; f ( x ) = ( x + 2 ) \left( x ^ { 2 } + x - 9 \right)$
D) $- 3 ; f ( x ) = ( x + 3 ) \left( x ^ { 2 } - x - 6 \right)$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs for $f(x)$ shows 3 or 1 positive real zeros and 1 negative real zero. For $f(-x)$, it shows 3 or 1 positive real zeros (indicating 3 or 1 negative real zeros of $f(x)$). The Rational Zeros Theorem suggests possible rational zeros as $\pm1, \pm2, \pm3, \pm6, \pm9, \pm18$. Testing these, $-3, -2,$ and $3$ are zeros of $f(x)$, leading to the factorization $f(x) = (x + 3)(x + 2)(x - 3)$.

Use Descartes' Rule of Signs and the Rational Zeros Theorem f(x)=x3+2x2−9x−18f ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 9 x - 18f(x)=x3+2x2−9x−18

Use Descartes' Rule of Signs and the Rational Zeros Theorem $f ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 9 x - 18$