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How can I show that $D((x − a)f (x)) = D(f (x))[f (a)]^2$, using the usual definition from Discriminant

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closed as not a real question by Norbert, draks ..., Thomas, Jason DeVito, J. M. Oct 18 '12 at 8:04

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Uh? What is f? A polynomial, I gather, but...what have you done/tried/ thought about this? –  DonAntonio Oct 15 '12 at 21:43
$f$ is a polynomial; $f=a_nx^n+a_{n-1}x^{n-1}+...+a_o$ and you can find two different definitions from Discriminant, both are useful –  Elmo goya Oct 15 '12 at 21:46
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2 Answers 2

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Let $f$ have roots $\{\alpha_i\}$, degree $n$, and leading coefficient $a_n$. Then $(x-a)f(x)$ has roots $\{a\}\sqcup\{\alpha_i\}$, degree $n+1$, and leading coefficient $a_n$. So its discriminant is given by $$ D((x-a)f(x)) = a_n^{2n} \prod_{i<j}(\alpha_i-\alpha_j)^2\prod_i(a-\alpha_i)^2. $$ Now $f(a) = a_n\prod_i(a-\alpha_i)$, and $D(f) = a_n^{2n-2}\prod_{i<j}(\alpha_i-\alpha_j)^2$, so the right-hand side factors neatly as $D(f)f(a)^2$.

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$(x-a)f(x)$ has the same roots as $f$ plus a root at $a$. Therefore, the product of all root differences gets multiplied with $\prod (a-\alpha_i)^2=f(a)^2$.

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