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.
-$$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - 4 x - 12$$
A) $- 3 ; f ( x ) = ( x + 3 ) \left( x ^ { 2 } - x - 4 \right)$
B) $- 3 , - 2,2 ; f ( x ) = ( x + 3 ) ( x + 2 ) ( x - 2 )$
C) $- 2 ; f ( x ) = ( x + 2 ) \left( x ^ { 2 } + x - 6 \right)$
D) $- 2,2,3 ; f ( x ) = ( x + 2 ) ( x - 2 ) ( x - 3 )$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates one positive real zero and two negative real zeros. Using the Rational Zeros Theorem, the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12. Testing these, -3, -2, and 2 are zeros. The polynomial factors as $(x + 3)(x + 2)(x - 2)$.

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

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