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.
-$$f ( x ) = x ^ { 4 } - 5 x ^ { 2 } - 36$$
A) $3 ; f ( x ) = ( x - 3 ) ^ { 2 } \left( x ^ { 2 } + 4 \right)$
B) $- 3,3 ; f ( x ) = ( x - 3 ) ( x + 3 ) \left( x ^ { 2 } + 4 \right)$
C) $- 3 , - 2,3,2 ; f ( x ) = ( x - 3 ) ( x + 3 ) ( x - 2 ) ( x + 2 )$
D) $- 2,2 ; f ( x ) = ( x - 2 ) ( x + 2 ) \left( x ^ { 2 } + 9 \right)$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates there can be 2 or 0 positive real zeros (since there are 2 sign changes in $f(x)$) and 0 negative real zeros (no sign changes in $f(-x)$). The Rational Zeros Theorem suggests possible rational zeros could be factors of 36. Testing or using further analysis, we find that $x = 3$ and $x = -3$ are zeros. Factoring $f(x)$ using these zeros gives $f(x) = (x - 3)(x + 3)(x^2 + 4)$, where $x^2 + 4$ has no real zeros.

Use Descartes' Rule of Signs and the Rational Zeros Theorem f(x)=x4−5x2−36f ( x ) = x ^ { 4 } - 5 x ^ { 2 } - 36f(x)=x4−5x2−36

Use Descartes' Rule of Signs and the Rational Zeros Theorem $f ( x ) = x ^ { 4 } - 5 x ^ { 2 } - 36$