$\text { If } x = \sqrt { 3 } \text { is a root of } x ^ { 3 } - x ^ { 2 } - 3 x + 3 = 0$ , use synthetic division to factor the polynomial completely and list all real solutions of the equation.
A) $( x + 1 ) ( x + \sqrt { 3 } ) ^ { 2 } ; - 1 , \sqrt { 3 }$
B) $( x + 1 ) ^ { 2 } ( x - \sqrt { 3 } ) ; - 1 , \sqrt { 3 }$
C) $( x + 1 ) ( x + \sqrt { 3 } ) ^ { 2 } ; 1 , - \sqrt { 3 }$
D) $( x + 1 ) ( x - \sqrt { 3 } ) ^ { 2 } ; - 1 , \sqrt { 3 }$
E) $( x - 1 ) ( x - \sqrt { 3 } ) ( x + \sqrt { 3 } ) ; 1 , \sqrt { 3 } , - \sqrt { 3 }$

Question

$\text { If } x = \sqrt { 3 } \text { is a root of } x ^ { 3 } - x ^ { 2 } - 3 x + 3 = 0$ , use synthetic division to factor the polynomial completely and list all real solutions of the equation. 
A) $( x + 1 ) ( x + \sqrt { 3 } ) ^ { 2 } ; - 1 , \sqrt { 3 }$
B) $( x + 1 ) ^ { 2 } ( x - \sqrt { 3 } ) ; - 1 , \sqrt { 3 }$
C) $( x + 1 ) ( x + \sqrt { 3 } ) ^ { 2 } ; 1 , - \sqrt { 3 }$
D) $( x + 1 ) ( x - \sqrt { 3 } ) ^ { 2 } ; - 1 , \sqrt { 3 }$
E) $( x - 1 ) ( x - \sqrt { 3 } ) ( x + \sqrt { 3 } ) ; 1 , \sqrt { 3 } , - \sqrt { 3 }$

Quizplus · Accepted Answer

Using synthetic division with $x = \sqrt{3}$, we can factor the polynomial. Since $\sqrt{3}$ is a root, $-\sqrt{3}$ will also be a root due to the polynomial's real coefficients. Factoring completely, we get $(x - 1)(x - \sqrt{3})(x + \sqrt{3})$, which gives the real solutions $1, \sqrt{3}, -\sqrt{3}$.

If x=3 is a root of x3−x2−3x+3=0\text { If } x = \sqrt { 3 } \text { is a root of } x ^ { 3 } - x ^ { 2 } - 3 x + 3 = 0 If x=3​ is a root of x3−x2−3x+3=0

$\text { If } x = \sqrt { 3 } \text { is a root of } x ^ { 3 } - x ^ { 2 } - 3 x + 3 = 0$