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Two roots of $4x^3+8x^2+Kx-18=0$ are equal numerically but opposite in value. Find the value of K.

I've tried plugging in two roots $x_0$ and $-x_0$, to get the relationship $k=-4x^2$. Plugging it back in doesn't achieve anything, I'm not sure how to proceed, I have seen people do this with the quadratic, but similarly trying to force this answer is too complicated, this is an SAT II Level 2 question and should be solvable in under a minute!

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If $a$ is one of the two opposite roots, then $$ \begin{align} (x^2-a^2)(4x-r)&=4x^3+8x^2+Kx-18\\ 4x^3-rx^2-4a^2x+a^2r&=4x^3+8x^2+Kx-18 \end{align} $$ The coefficient of $x^2$ reveals the value of $r$. The constant term then reveals the value of $a^2$. And then the linear coefficient reveals the value of $K$.

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  • $\begingroup$ So the answer is $-12$. Nice! $\endgroup$ – MaximusFastidiousIrreverence Apr 22 '18 at 4:35
  • $\begingroup$ The answer is -9! Thanks! $\endgroup$ – Mike Apr 22 '18 at 5:01
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Hint: by Vieta's relations the sum of all three roots is $-8/4=-2\,$. Now, if two of them sum to $0$ $\ldots$

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Let the roots be $a, - a, b$. Then

$$4x^3+8x^2+Kx-18=4(x^2-a^2)(x-b)$$

So, $-4b=8$, $-4a^2=K$ and $4a^2b=-18$.

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