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.
-$$f ( x ) = 5 x ^ { 3 } - 2 x ^ { 2 } + 25 x - 10$$
A) $- 5 , - 1 , \frac { 2 } { 5 } ; f ( x ) = ( 5 x - 2 ) ( x + 1 ) ( x + 5 )$
B) $\frac { 2 } { 5 } ; f ( x ) = ( 5 x - 2 ) \left( x ^ { 2 } + 5 \right)$
C) $10 ; f ( x ) = ( x - 10 ) \left( 5 x ^ { 2 } + 1 \right)$
D) $5 , \frac { 2 } { 5 } , 1 ; f ( x ) = ( 5 x - 2 ) ( x - 1 ) ( x - 5 )$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates there is 1 positive real zero since there is 1 sign change in $f(x)$. For $f(-x)$, there are no sign changes, indicating no negative real zeros. The Rational Zeros Theorem suggests possible rational zeros as factors of $\frac{10}{5} = \frac{2}{5}$, which is a zero. Factoring $f(x)$ by dividing by $5x - 2$ yields a quadratic with no real zeros, hence the factorization into $(5x - 2)(x^2 + 5)$.

Use Descartes' Rule of Signs and the Rational Zeros Theorem f(x)=5x3−2x2+25x−10f ( x ) = 5 x ^ { 3 } - 2 x ^ { 2 } + 25 x - 10f(x)=5x3−2x2+25x−10

Use Descartes' Rule of Signs and the Rational Zeros Theorem $f ( x ) = 5 x ^ { 3 } - 2 x ^ { 2 } + 25 x - 10$