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Prove that every quadratic polynomial of the form $p(z) = az^2 + bz + c$ with $a,b,c\in\mathbb{C}$ and $a\neq0$ can also be written as $p(z) = a(z-w_0)(z-w_1)$ for certain $w_0, w_1 \in \mathbb{C}$.


Now I know that every complex numbers has two possible square roots and my intuition tells me that I should probably apply the quadratic formula $p(z) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ here and go from there, but I'm quite blind to how. Is this just basic algebra that I'm not seeing?

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  • $\begingroup$ The quadratic formula is enough for quadratic polynomials, but the theorem can be generalized to n-degree polynomials where we might not have a formula for the roots. BTW, $p(z)$ is not equal to the expression in your second paragraph: that just gives you the two roots of $p(z)$. $\endgroup$
    – NickD
    Dec 11 '17 at 22:00
  • $\begingroup$ Define $w_0=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $w_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$. Surely they are the complex roots of $p(z)=az^2+bz+c$. Now apply polynomial division to get $p(z)=a(z-w_0)(z-w_1)$. $\endgroup$ Dec 28 '17 at 16:09
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Hint compute $(z-w_0)(z-w_1)$ where $w_i = \frac{-b\pm \delta}{2 a}$ and $\pm \delta$ are the two complex square roots of $b^2-4 a c$.

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Consider $a(z² + (b/a)z + c/a)$ and apply your formula to $z² + (b/a)z + c/a$, so your get $w_0=(b/2a)+\sqrt{(b/2a)^2-c/a}$ and $w_1=(b/2a)-\sqrt{(b/2a)^2-c/a}$.

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