If $\displaystyle P(x) = x^5 + ax^2 + b$ has multiple roots. Prove that $\displaystyle 3125b^5 = 108 a^3$
Umm... I just cant prove that..... is the question wrong or am i doing something wrong =S?
thanks in advance.
Hello,
I'll try. I did it on a paper and just noticed a tiny mistake, so I'll go freely while typing lol ! But you gotta check it because I'm sure I won't get the result you're given
If P(x) has multiple roots (one or more or all roots are multiple ?), then there exists m such that $\displaystyle P(m)=P'(m)=0$
$\displaystyle 0=m^5+am^2+b$ (1) and $\displaystyle 0=5m^4+2am$ (2)
From (2), we get $\displaystyle m=0$ or $\displaystyle 5m^3+2a=0$
If m=0, then b=0 by (1).
Let's say $\displaystyle m \neq 0$ otherwise that's not funny
So $\displaystyle 5m^3+2a=0 \implies m=-\left(\frac 25 a\right)^{\tfrac 13}$
Substituting in (1) :
$\displaystyle -\left(\frac 25\right)^{\tfrac 53} a^{\tfrac 53}+a*\left(\frac 25\right)^{\tfrac 23} a^{\tfrac 23}+b=0$
$\displaystyle -\left(\frac 25\right)^{\tfrac 53} a^{\tfrac 53}+\left(\frac 25\right)^{\tfrac 23} a^{\tfrac 53}+b=0$
$\displaystyle b=a^{\tfrac 53} \left(\left(\tfrac 25\right)^{\tfrac 53}-\left(\tfrac 25\right)^{\tfrac 23}\right)$
Take the 3rd power of both sides :
$\displaystyle b^3=a^5 \left(\left(\tfrac 25\right)^{\tfrac 53}-\left(\tfrac 25\right)^{\tfrac 23}\right)^3$
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Remember that $\displaystyle (x-y)^3=x^3-y^3+3(xy^2-x^2y)$
We have (I write it in the same order so that you can see) :
$\displaystyle x^3=\frac{2^5}{5^5}$ and $\displaystyle y^3=\frac{2^2}{5^2}$
$\displaystyle xy^2=\left(\frac 25\right)^{\tfrac 53} \cdot \left(\frac 25\right)^{\tfrac 43}=\left(\frac 25\right)^3=\frac{2^3}{5^3}$
and similarly $\displaystyle x^2y=\dots=\frac{2^4}{5^4}$
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$\displaystyle b^3=a^5 \left(\frac{2^5}{5^5}-\frac{2^2}{5^2}+3 \cdot \frac{2^3}{5^3}-3 \cdot \frac{2^4}{5^4}\right)$
$\displaystyle b^3=a^5 \cdot \left(-\frac{108}{3125}\right)$
$\displaystyle \boxed{-3125b^3=108a^5}$
Which is too similar to be a big mistake