12. Find a polynomial function of least degree having only real coefficients, a leading coefficient of 1 , and zeros of $1-\sqrt{3}, 1+\sqrt{3}$ and $5-i, \quad 3,3,71$ The polynomial function is $f(x)=$ (Simplify your answer.)
Solution
Step 1 ：The zeros of a polynomial function are the values of x for which the function equals zero. If a polynomial has real coefficients, then any complex zeros must occur in conjugate pairs. In this case, the zeros are \(1-\sqrt{3}\), \(1+\sqrt{3}\), \(5-i\), \(5+i\), \(3\), \(3\), and \(71\).
Step 2 ：We can construct the polynomial by multiplying factors associated with each zero. The factor associated with a zero \(a\) is \((x-a)\).
Step 3 ：By multiplying these factors together, we obtain the polynomial function \(f(x) = x^7 - 89.0x^6 + 1403.0x^5 - 9279.0x^4 + 29220.0x^3 - 38032.0x^2 - 2172.0x + 33228.0\).
Step 4 ：\(\boxed{f(x) = x^7 - 89.0x^6 + 1403.0x^5 - 9279.0x^4 + 29220.0x^3 - 38032.0x^2 - 2172.0x + 33228.0}\) is the polynomial function of least degree having only real coefficients, a leading coefficient of 1, and zeros of \(1-\sqrt{3}\), \(1+\sqrt{3}\), \(5-i\), \(5+i\), \(3\), \(3\), and \(71\).
