1. Factorising polynomials in C

$\displaystyle z^4 - 2z^2 - 3$

Help please, this is using imaginary numbers just so you know...

2. Hello,

Originally Posted by iphysics
$\displaystyle z^4 - 2z^2 - 3$

Help please, this is using imaginary numbers just so you know...
Let $\displaystyle t=z^2$

The equation is now $\displaystyle t^2-2t-3$

using the discriminant method :

$\displaystyle \Delta=4+12=16=4^2$

--> $\displaystyle t_1 \& t_2=\frac{2 \pm 4}{2}=3 \text{ or } -1$ are the roots of the polynomial.

--> $\displaystyle t^2-2t-3=(t+1)(t-3)$

Therefore :
\displaystyle \begin{aligned} t^2-2t-3&=(z^2+1)(z^2-3) \\ &=(z^2-(i)^2)(z^2-(\sqrt{3})^2) \\ &=(z-i)(z+i)(z+\sqrt{3})(z-\sqrt{3}) & \leftarrow \text{ because } a^2-b^2=(a-b)(a+b) \end{aligned}

3. Hello, iphysics!

Factor: .$\displaystyle z^4 - 2z^2 - 3$

Factor: .$\displaystyle (z^2 - 3)\,(z^2+1) \;=\;(z^2-3)\,(z^2 - [\text{-}1])$

Factor: .$\displaystyle (z - \sqrt{3})\,(z + \sqrt{3})\,(z - i)\,(z + 1)$

4. Hi,

z^4-2z^2-3

=z^4-3z^2+z^2-3

=z^2(Z^2-3)+1(z^2-3)

=(z^2+1)(z^2-3)

to calculate z value

z^2+1=0

z^2=-1

and z^2-3=0
z^2=3
z=+ or - sqrt3

z=+ or - i