How exactly would you solve the equation:
"Solve the equation $10x^3+23x^2+5x−2=0$ given that one root is four times a second root."
How would you go about solving this? Any help would be greatly appreciated.
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1You have an answer to your other question. If you don't understand an answer, say something to the answerer. Do not repost. – Cameron Buie May 14 '13 at 11:38
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We know that a cubic polynomial has three zeroes over the complex numbers (including the case where we have double or triple zeroes). So, we can write any cubic polynomial as $a (x-z_1)(x-z_2)(x-z_3)$, where $z_1, z_2, z_3$ are its zeroes.
In this case, we can assume that $z_2 = 4z_1$ by the side condition. We get $$10x^3+23x^2+5x-2 = a (x-z_1)(x-4z_1)(x-z_3).$$ By equating coefficients, we find: $$\begin{align} a &= 10\\ -a(5z_1 + z_3) &= 23\\ a(4z_1^2+5z_1z_3) &= 5\\ a(-4z_1^2z_3) &= -2 \end{align}$$ Solving this system should give you a quadratic equation for $z_1$, from which the result follows.
Johannes Kloos
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