solve for $p$ & $q$, simplify $2z^2+pz+q$, roots are $-1-2i$ and $-1+2i$

Question

I am currently working on solving polynomials that contain complex numbers, the question posed is as follows:

Find $p$ and $q$. $$2z^2+pz+q$$ with roots at $-1-2i$ and $-1+2i$.

The solution is given as shown below:

I am not sure how this answer has been achieved, I know that the roots must cause the equation to equal to zero however on the first line of work with the 2 at the start of the right hand side I dont know where this two has came from. Please could someone help me with this?

Edit:

given a simple polynomial such as $$x^2+5x+6 = (x+2)(x+3)$$ however $$2x^2+5x+6 \neq 2(x+2)(x+3)$$

This is why I am unsure how this can be applied to an expression containing complex roots.

Answer 1

The $2$ at the right-hand side is the leading coefficient of the polynomial $2z^2+pz+q$ and so has to be pulled out:

$2z^2+pz+q = 2(z-z_1)(z-z_2)$

where $z_1,z_2$ are the conjugate roots.

Answer 2

A monic quadratic's factorization$$(z-a)(z-b)=z^2+cz+d$$over $\Bbb C$ implies (by comparing coefficients) $c=-(a+b),\,d=ab$. In our example$$a,\,b=-1\pm 2i\implies a+b=-2,\,ab=5.$$But the quadratic's coefficients have all been doubled, so $p=2c=4,\,q=2d=10$. In particular,$$z=-1\pm 2i\iff z^2+2z+5=0\iff 2z^2+4z+10=0.$$

Answer 3

\begin{align} \color{blue}{2}z^2+pz+q&=\color{blue}{2}(z^2+\frac{p}2z+\frac{q}2)=\color{blue}{2}(z-\alpha)(z-\beta)\\ \end{align} where $\alpha=-1-2i$ and $\beta=-1+2i$. Continuing, we have \begin{align} &=\color{blue}{2}(z-(-1-2i))(z-(-1+2i))\\ &=\color{blue}{2}(z+1+2i)(z+1-2i)\\ &=\color{blue}{2}((z+1)^2-(2i)^2)\\ &=\color{blue}{2}(z^2+2z+1+4)\\ &=\color{blue}{2}(z^2+2z+5) \end{align}

The coefficients of the terms must be the same, so $$\frac{p}2=2 \Rightarrow p=4$$ and $$\frac{q}2=5\Rightarrow q=10$$ as required.

Answer 4

Note

$$z_1+z_2=-\frac p2,\>\>\>\>\>z_1z_2=\frac q2$$

where $z_1 =-1+2i$ and $z_2 =-1-2i$. Since $z_1+z_2=-2$, and $z_1z_2=5$, then

$$p=4,\>\>\>\>q=10$$