Determine the zeros (if any)of the rational function $$g ( x ) = \frac { x ^ { 3 } - 64 } { x ^ { 2 } + 6 }$$ .
A) $x = - \sqrt { 6 } , x = \sqrt { 6 } , x = 4$
B)no zeros
C)x = -4,x = 4
D)x = 4
E) $x = - \sqrt { 6 } , x = \sqrt { 6 } , x = - 4 , x = 4$

Question

Determine the zeros (if any)of the rational function $$g ( x ) = \frac { x ^ { 3 } - 64 } { x ^ { 2 } + 6 }$$ .
A) $x = - \sqrt { 6 } , x = \sqrt { 6 } , x = 4$
B)no zeros
C)​x = -4,x = 4
D)​x = 4
E)​ $x = - \sqrt { 6 } , x = \sqrt { 6 } , x = - 4 , x = 4$

Quizplus · Accepted Answer

To find the zeros of the given rational function, we set the numerator equal to zero and solve for $x$. The numerator is $x^3 - 64$, which can be factored as $(x - 4)(x^2 + 4x + 16)$. Setting this equal to zero gives us $x - 4 = 0$ or $x^2 + 4x + 16 = 0$. The quadratic equation $x^2 + 4x + 16 = 0$ has no real solutions because its discriminant $b^2 - 4ac = 4^2 - 4(1)(16) = 16 - 64 = -48$ is negative. Therefore, the only real zero of the function is $x = 4$.

Determine the Zeros (If Any)of the Rational Function g(x)=x3−64x2+6g ( x ) = \frac { x ^ { 3 } - 64 } { x ^ { 2 } + 6 }g(x)=x2+6x3−64​

Determine the Zeros (If Any)of the Rational Function $g ( x ) = \frac { x ^ { 3 } - 64 } { x ^ { 2 } + 6 }$