Find a polynomial of least degree with only real coefficients and having the given zeros.
-$3 + \sqrt { 6 } , 3 - \sqrt { 6 }$, and 3
A) $f ( x ) = x ^ { 3 } - 9 x ^ { 2 } + 21 x - 9$
B) $f ( x ) = x ^ { 3 } - 11 x ^ { 2 } + 21 x + 9$
C) $f ( x ) = x ^ { 3 } + 9 x ^ { 2 } + 21 x - 9$
D) $f ( x ) = x ^ { 3 } - 11 x ^ { 2 } + 21 x + 11$

Question

Quizplus · Accepted Answer

The polynomial of least degree with real coefficients that has the given zeros can be found by multiplying the factors corresponding to each zero. The factors for the zeros $3 + \sqrt{6}$ and $3 - \sqrt{6}$ are $x - (3 + \sqrt{6})$ and $x - (3 - \sqrt{6})$, respectively. Multiplying these two factors gives a quadratic polynomial with real coefficients because the product will eliminate the square root terms. The third zero, 3, corresponds to the factor $x - 3$. Multiplying all three factors together gives a cubic polynomial. The correct polynomial, after simplification, is $x^3 - 9x^2 + 21x - 9$, which matches choice A.

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