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

Given that $x = \sqrt{6}$ is a root, we can use synthetic division to divide the polynomial by $x - \sqrt{6}$. The result of this division will give us a quadratic polynomial. Since the polynomial is cubic and has real coefficients, if $\sqrt{6}$ is a root, then its conjugate $-\sqrt{6}$ must also be a root. After dividing by $x - \sqrt{6}$ and $x + \sqrt{6}$, we are left with a linear factor, which gives the root $x = -5$. Therefore, the polynomial can be factored as $(x + 5)(x - \sqrt{6})(x + \sqrt{6})$, and the real solutions are $x = -5$, $x = \sqrt{6}$, and $x = -\sqrt{6}$.

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 $x ^ { 3 } + 5 x ^ { 2 } - 6 x - 30 = 0$