# How do you write a polynomial in standard form given the zeros x=-9 and -i?

Mar 13, 2016

${P}_{2} = {\left(x + 1\right)}^{2} = {x}^{2} + 2 x + 1$
${P}_{3} = {P}_{2} \left(x + 9\right) = {x}^{3} + 11 {x}^{2} + 19 x + 9$

#### Explanation:

Clearly the $x = - i$ it satisfies the roots $x = \pm i$
Thus we have ${\left(x + 1\right)}^{2}$. Now using the other root (x+9) we can write a 3rd order polynomial...
${P}_{2} = {\left(x + 1\right)}^{2} = {x}^{2} + 2 x + 1$
${P}_{3} = {P}_{2} \left(x + 9\right) = {x}^{3} + 11 {x}^{2} + 19 x + 9$

Remark: Notice that I did not use the root given in the question $x = - I$. The reason is that the complex number in general comes in pair out of even powered polynomial. For example out of the binomial. Recall the quadratic formula:
${x}_{1 , 2} = \frac{- b \pm \sqrt{b - 4 a c}}{2 a}$ Note that the complex number emerges for negative value of $\left(b - 4 a c\right) i . e . \implies b < 4 a c$ and note the true set is a pair of $\pm \text{ complex numbers}$