If $x = \sqrt { 6 }$ is a root of $x ^ { 3 } + 5 x ^ { 2 } - 6 x - 30 = 0$ , use synthetic division to factor the polynomial completely and list all real solutions of the equation.
A) $( x + 5 ) ( x - \sqrt { 6 } ) ( x + \sqrt { 6 } ) ; - 5 , \sqrt { 6 } , - \sqrt { 6 }$
B) $( x - 5 ) ( x + \sqrt { 6 } ) ^ { 2 } ; 5 , \sqrt { 6 }$
C) $( x - 5 ) ( x - \sqrt { 6 } ) ^ { 2 } ; 5 , \sqrt { 6 }$
D) $( x - 5 ) ( x + \sqrt { 6 } ) ^ { 2 } ; - 5 , - \sqrt { 6 }$
E) $( x - 5 ) ^ { 2 } ( x - \sqrt { 6 } ) ^ { 2 } ; 5 , \sqrt { 6 }$

Question

Quizplus · Accepted Answer

Using synthetic division with $x = \sqrt{6}$, the polynomial can be factored as $(x - \sqrt{6})(x^2 + (5 + \sqrt{6})x + 5\sqrt{6})$. Further simplification and factoring reveal that the quadratic factor can be written as $(x + 5)(x + \sqrt{6})$, leading to the complete factorization $(x - \sqrt{6})(x + 5)(x + \sqrt{6})$. The real solutions are $x = -5, \sqrt{6}, -\sqrt{6}$.

If x=6x = \sqrt { 6 }x=6​ Is a Root Of x3+5x2−6x−30=0x ^ { 3 } + 5 x ^ { 2 } - 6 x - 30 = 0x3+5x2−6x−30=0

If $x = \sqrt { 6 }$ Is a Root Of $x ^ { 3 } + 5 x ^ { 2 } - 6 x - 30 = 0$