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.
-$$f ( x ) = 4 x ^ { 4 } - 8 x ^ { 3 } + 5 x ^ { 2 } - 2 x + 1$$
A) $- 1,1 ; f ( x ) = ( x - 1 ) ( x + 1 ) \left( 4 x ^ { 2 } + 1 \right)$
B) no real roots; $\mathrm { f } ( \mathrm { x } ) = \left( \mathrm { x } ^ { 2 } + 1 \right) \left( 4 \mathrm { x } ^ { 2 } + 1 \right)$
C) $- 1$, multiplicity 2; $f ( x ) = ( x + 1 ) ^ { 2 } \left( 4 x ^ { 2 } + 1 \right)$
D) 1 , multiplicity 2; $\mathrm { f } ( \mathrm { x } ) = ( \mathrm { x } - 1 ) ^ { 2 } \left( 4 \mathrm { x } ^ { 2 } + 1 \right)$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates there are 2 or 0 positive real zeros (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 are $\pm1$. Testing $f(1) = 0$ confirms $x = 1$ is a zero. Given the polynomial's degree (4), and the absence of negative real zeros, $x = 1$ with multiplicity 2 and two non-real complex zeros (implied by the factor $4x^2 + 1$) is consistent with the polynomial's behavior.

Use Descartes' Rule of Signs and the Rational Zeros Theorem f(x)=4x4−8x3+5x2−2x+1f ( x ) = 4 x ^ { 4 } - 8 x ^ { 3 } + 5 x ^ { 2 } - 2 x + 1f(x)=4x4−8x3+5x2−2x+1

Use Descartes' Rule of Signs and the Rational Zeros Theorem $f ( x ) = 4 x ^ { 4 } - 8 x ^ { 3 } + 5 x ^ { 2 } - 2 x + 1$