Find all rational zeros and factor f(x).
$f ( x ) = 12 x ^ { 3 } + 61 x ^ { 2 } + 4 x - 5$
A) $- 3,4 , - 5 ; f ( x ) = ( x + 3 ) ( x - 4 ) ( x + 5 )$
B) $- \frac { 1 } { 12 } , 1 , - 5 ; f ( x ) = ( 12 x + 1 ) ( x - 1 ) ( x + 5 )$
C) $- 3,4 , - 5 ; f ( x ) = ( 3 x + 1 ) ( 4 x - 1 ) ( x + 5 )$
D) $- \frac { 1 } { 3 } , \frac { 1 } { 4 } , - 5 ; f ( x ) = ( 3 x + 1 ) ( 4 x - 1 ) ( x + 5 )$

Question

Quizplus · Accepted Answer

The rational root theorem suggests that the rational zeros of a polynomial are of the form ±(factors of the constant term)/(factors of the leading coefficient). For $f(x) = 12x^3 + 61x^2 + 4x - 5$, the constant term is -5, and the leading coefficient is 12. The factors of -5 are ±1, ±5, and the factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12. Testing these, $-\frac{1}{3}$, $\frac{1}{4}$, and -5 are the zeros that satisfy the polynomial, leading to the factorization $(3x + 1)(4x - 1)(x + 5)$.

Find All Rational Zeros and Factor F(x) f(x)=12x3+61x2+4x−5f ( x ) = 12 x ^ { 3 } + 61 x ^ { 2 } + 4 x - 5f(x)=12x3+61x2+4x−5

Find All Rational Zeros and Factor F(x) $f ( x ) = 12 x ^ { 3 } + 61 x ^ { 2 } + 4 x - 5$