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I have this quadratic equation $$2x^2+x-3=0$$ that I wish to reconstruct from its roots. $$D=b^2-4ac=25$$ $$x_1=\frac {-b\pm \sqrt D} {2a} = 1 \text{ and } \frac {-2} 3$$ Now, I've always learned that by taking the product of the roots you get back your original equation. $$2x^2+x-3=(x-1)(x+1,5)$$ But this is incorrect, and if a do a polynomial long division I get $$2x^2+x-3=(x-1)*2(x+1,5)$$ So what I've learned appears to be wrong. But what is the correct way? All my old textbooks and online resources tell me nothing of this.

Is it something like this? $$\text{original quadratic equation}=n(x-x_1)l(x-x_2) \text{ where } n,l \in \mathbb{Z} $$

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You cannot reconstruct a quadratic equation from its roots, only a monic quadratic equation.

More precisely, if $q(x)=0$ is a quadratic equation with roots $x_1$ and $x_2$, then $q(x)=a(x-x_1)(x-x_2)$ for some $a\ne 0$. The roots do not determine $a$.

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