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
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.
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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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