If $x = \sqrt { 7 } \text { is a root of } x ^ { 3 } - 3 x ^ { 2 } - 7 x + 21 = 0$ , use synthetic division to factor the polynomial completely and list all real solutions of the equation.
A) $( x - 3 ) ( x - \sqrt { 7 } ) ( x + \sqrt { 7 } ) ; 3 , \sqrt { 7 } , - \sqrt { 7 }$
B) $( x + 3 ) ( x + \sqrt { 7 } ) ^ { 2 } ; - 3 , \sqrt { 7 }$
C) $( x + 3 ) ( x - \sqrt { 7 } ) ^ { 2 } ; - 3 , \sqrt { 7 }$
D)$( x + 3 ) ( x + \sqrt { 7 } ) ^ { 2 } ; 3 , - \sqrt { 7 }$
E) $( x + 3 ) ^ { 2 } ( x - \sqrt { 7 } ) ; - 3 , \sqrt { 7 }$

Question

If $x = \sqrt { 7 } \text { is a root of } x ^ { 3 } - 3 x ^ { 2 } - 7 x + 21 = 0$ , use synthetic division to factor the polynomial completely and list all real solutions of the equation. 
A) $( x - 3 ) ( x - \sqrt { 7 } ) ( x + \sqrt { 7 } ) ; 3 , \sqrt { 7 } , - \sqrt { 7 }$
B) $( x + 3 ) ( x + \sqrt { 7 } ) ^ { 2 } ; - 3 , \sqrt { 7 }$
C) $( x + 3 ) ( x - \sqrt { 7 } ) ^ { 2 } ; - 3 , \sqrt { 7 }$
D)$( x + 3 ) ( x + \sqrt { 7 } ) ^ { 2 } ; 3 , - \sqrt { 7 }$
E) $( x + 3 ) ^ { 2 } ( x - \sqrt { 7 } ) ; - 3 , \sqrt { 7 }$

Quizplus · Accepted Answer

Given $x = \sqrt{7}$ is a root, synthetic division can be used to factor the polynomial. The factored form $(x - 3)(x - \sqrt{7})(x + \sqrt{7})$ indicates the roots are $3, \sqrt{7}, -\sqrt{7}$, which matches option A.

If x=7 is a root of x3−3x2−7x+21=0x = \sqrt { 7 } \text { is a root of } x ^ { 3 } - 3 x ^ { 2 } - 7 x + 21 = 0x=7​ is a root of x3−3x2−7x+21=0

If $x = \sqrt { 7 } \text { is a root of } x ^ { 3 } - 3 x ^ { 2 } - 7 x + 21 = 0$