# How do you write a polynomial in standard form given zeros 1,5,3-i?

Apr 14, 2016

${x}^{4} - 12 {x}^{3} + 51 {x}^{2} - 70 x + 50 = 0$.

#### Explanation:

Complex roots occur in conjugate pairs. So, as $3 - i$ is a root, $3 + i$ is also a root.

The roots are 1, 5, $3 - i \mathmr{and} 3 + i$.

The biquadratic having these as roots is in the form
${x}^{4} - {S}_{1} {x}^{3} + {S}_{2} {x}^{2} - {S}_{3} x + {S}_{4} = 0$, where
${S}_{1} = \sum$(roots) = 12,
${S}_{2} = \sum$(products of the roots, taken two at a time) = 51,
${S}_{3} = \sum$(product of the roots, taken three at a time) = 70
and ${S}_{4} =$product of all the roots =50.

Alternative method:

The factors of the polynomial are x-1, x - 5, ((x-(3-i)) and (x-(3+i)).
$\left(x - 1\right) \left(x - 3\right) \left({\left(x - 3\right)}^{2} + 1\right) = 0$.