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 indicates there can be 3 or 1 positive real zeros and 1 negative real zero. The Rational Zeros Theorem suggests possible rational zeros are ±1, ±2, ±3, ±6, ±9, ±18. Testing these, -3, -2, and 3 are zeros, 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 $f ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 9 x - 18$