# How do you write a polynomial in standard form given zeros -4, 1 - i, and 1 + i?

Apr 4, 2016

${x}^{3} + 2 {x}^{2} - 6 x + 8 = 0$.
${x}^{3} - \left(a + b + c\right) {x}^{2} + \left(b c + c a + a b\right) x - a b c = 0$ for the equation whose roots are a, b and c. Here, a+b+c=-2, bc +ca+ab=-2 and abc=-8.
Also, if the roots are $a \mathmr{and} b \pm i c$, the equation is
$\left(x - a\right) \left({\left(x - b\right)}^{2} + {c}^{2}\right)$.
Here, it is $\left(x + 4\right) \left({\left(x - 1\right)}^{2} + {1}^{2}\right) = 0$,