# Thread: Factorise P(z)=9z^3+(9i-12)z^2+(5-12i)z+5i over C if P(-i)=0

1. ## Factorise P(z)=9z^3+(9i-12)z^2+(5-12i)z+5i over C if P(-i)=0

Factorise over C if P(-i)=0

For all the book's questions so far we were told to find one factor after finding a root, then use long division to find the next factor and so on. But I am not sure where to start here because of the "i"s in the polynomial.

So all I know so far is that (Z-(-i)) = (z+i) is a factor. But I don't think I will be able to divide 9z^3+(9i-12)z^2+(5-12i)z+5i by (z+i)

Thank you

2. Originally Posted by anees7112
Factorise over C if P(-i)=0

For all the book's questions so far we were told to find one factor after finding a root, then use long division to find the next factor and so on. But I am not sure where to start here because of the "i"s in the polynomial.

So all I know so far is that (Z-(-i)) = (z+i) is a factor. But I don't think I will be able to divide 9z^3+(9i-12)z^2+(5-12i)z+5i by (z+i)

Thank you
You should be able to do the long division - it's no different than if it was $\displaystyle \displaystyle z - a$ with $\displaystyle \displaystyle a \in \mathbf{R}$ as a factor...

3. Ah ok . Thank you very much ! I did not realise I could

4. Originally Posted by anees7112
Factorise over C if P(-i)=0

For all the book's questions so far we were told to find one factor after finding a root, then use long division to find the next factor and so on. But I am not sure where to start here because of the "i"s in the polynomial.

So all I know so far is that (Z-(-i)) = (z+i) is a factor. But I don't think I will be able to divide 9z^3+(9i-12)z^2+(5-12i)z+5i by (z+i)

Thank you
Here is an alternative

$\displaystyle P(z)=9z^3+(9i-12)z^2+(5-12i)z+5i$

$\displaystyle P(z)=(z+i)\left(9z^2+kz+5)=9z^3+kz^2+5z+9iz^2+kiz+ 5i$

$\displaystyle P(z)=9z^3+(9i+k)z^2+(5+ki)z+5i$

Both the coefficient of $\displaystyle z^2$ and the coefficient of $\displaystyle z$ reveal $\displaystyle k$.