Find all rational zeros and factor f(x).
-$$f ( x ) = 4 x ^ { 3 } - 28 x ^ { 2 } - x + 7$$
A) $1 , - 1,7 ; f ( x ) = ( 2 x - 1 ) ( 2 x + 1 ) ( x - 7 )$
B) $\frac { 1 } { 2 } , - \frac { 1 } { 2 } , 7 ; f ( x ) = ( x - 2 ) ( x + 2 ) ( x - 7 )$
C) $2 , - 2,7 ; f ( x ) = ( x - 2 ) ( x + 2 ) ( x - 7 )$
D) $\frac { 1 } { 2 } , - \frac { 1 } { 2 } , 7 ; f ( x ) = ( 2 x - 1 ) ( 2 x + 1 ) ( x - 7 )$

Question

Quizplus · Accepted Answer

To find the rational zeros of the polynomial, we can use the Rational Root Theorem. The possible rational roots are factors of the constant term divided by factors of the leading coefficient. For $f(x) = 4x^3 - 28x^2 - x + 7$, the constant term is 7 and the leading coefficient is 4. The factors of 7 are ±1, ±7, and the factors of 4 are ±1, ±2, ±4. This gives possible rational roots of ±1/4, ±1/2, ±1, ±7/4, ±7/2, ±7. Testing these, we find that 1/2 and -1/2 are roots, as well as 7. Factoring by synthetic division or other methods, we find the factorization matches option D: $f(x) = (2x - 1)(2x + 1)(x - 7)$.

Find All Rational Zeros and Factor F(x) f(x)=4x3−28x2−x+7f ( x ) = 4 x ^ { 3 } - 28 x ^ { 2 } - x + 7f(x)=4x3−28x2−x+7

Find All Rational Zeros and Factor F(x) $f ( x ) = 4 x ^ { 3 } - 28 x ^ { 2 } - x + 7$