# How do you write a polynomial in standard form given zeros 8, -14, and 3 + 9i?

Apr 23, 2016

${x}^{4} - 22 {x}^{2} - 10080 = 0$

#### Explanation:

Complex roots occur in conjugate pairs. So, the fourth root is $3 - 9 i$.

If ${s}_{1} , {s}_{2} , {s}_{3} \mathmr{and} {s}_{4}$ are the sums of the products of the roots, taken 1, 2, 3 and 4, at a time, respectively, the biquadratic equation has the form
${x}^{4} - {s}_{1} {x}^{3} + {s}_{2} {x}^{2} - {s}_{3} x + {s}_{4} = 0$.

Here, ${s}_{1} = 0 , {s}_{2} = - 22 , {s}_{3} = 0 \mathmr{and} {s}_{4} = - 10080$.

So, the answer is ${x}^{4} - 22 {x}^{2} - 10080 = 0$.

Note that the sum $3 \pm 9 i$=0,
wherever it occurs in the terms of s-sums.