Find a polynomial of least degree with only real coefficients and having the given zeros.
-$- \sqrt { 3 } , \sqrt { 3 }$, and 8 (multiplicity 2 )
A) $f ( x ) = x ^ { 4 } - 16 x ^ { 3 } + 61 x ^ { 2 } + 48 x - 192$
B) $f ( x ) = x ^ { 3 } + 8 x ^ { 2 } - 3 x - 24$
C) $f ( x ) = x ^ { 3 } - 8 x ^ { 2 } - 3 x + 24$
D) $f ( x ) = x ^ { 4 } + 16 x ^ { 3 } - 61 x ^ { 2 } - 48 x + 192$

Question

Quizplus · Accepted Answer

Given zeros are $-\sqrt{3}, \sqrt{3}$, and $8$ with multiplicity 2. The polynomial with these zeros can be constructed as $(x + \sqrt{3})(x - \sqrt{3})(x - 8)^2$. Simplifying, $(x^2 - 3)(x - 8)^2$ leads to a polynomial of degree 4, which matches the form of option A after full expansion and combining like terms.

Find a Polynomial of Least Degree with Only Real Coefficients −3,3- \sqrt { 3 } , \sqrt { 3 }−3​,3​

Find a Polynomial of Least Degree with Only Real Coefficients $- \sqrt { 3 } , \sqrt { 3 }$