If the polynomial $x^3+3px+q$ has a factor of the form $(x-a)^2$, then prove that $q^2+4p^3=0$.
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See https://math.stackexchange.com/questions/593568/multiple-root-of-a-polynomial-is-also-a-root-of-the-derivative – lab bhattacharjee Aug 21 '18 at 11:55
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2 Answers
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$$(x-a)^2(x-b)=x^3-bx^2-2ax^2+a^2x+2abx-ba^2$$ and by identification
$$b=-2a.$$
Then
$$q^2+4p^3=(2a^3)^2+4(-a^2)^3=0.$$
You can also perform the long division of the given polynomial by $(x-a)^2$ and cancel the remainder.
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$a$ is a double root so that
$$a^3+3pa+q=3a^2+3p=0.$$
Eliminate $a$.