Factoring $qz^2+2pz+q$ via the quadratic equation

Question

I am factoring $qz^2+2pz+q$ via the quadratic equation. So I get $z=\frac{-2p\pm \sqrt{4p^2-4(q)(q)}}{2q}=\frac{-p\pm \sqrt{p^2-q^2}}{q}$ If the two roots are $z_1,z_2$ then I think that $qz^2+2pz+q=(z-z_1)(z-z_2)$ but my instructor is saying that I should get $qz^2+2pz+q=q(z-z_1)(z-z_2)$. I feel really frustrated because I can't see why that would be. Why can't I just use the quadratic equation?

Thanks.

Answer 1

You can use quadratic equation for $z_1$ and $z_2$ but since the original polynomial is not monic you need an extra factor $q$ to have the correct coefficient for $z^2$

$$\color{red}qz^2+2pz+q=\color{red}q(z-z_1)(z-z_2)$$

Answer 2

Hint: think about what your "just use the quadratic equation" argument would say if you tried it for $2x^2 - 2 = 0$ and $3x^2 - 3 = 0$. Then you should be able to see why it doesn't work.