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**arze** Find the value of *a* for which the function $\displaystyle 2x^3-ax^2-12x-7$ has repeated factors.

If $\displaystyle \alpha$ is the repeated root, $\displaystyle (x-\alpha)$ is the factor

$\displaystyle f(\alpha)\equiv 2\alpha^3-a\alpha^2-12\alpha-7=0$ ___(1)

$\displaystyle f'(\alpha)\eqiuv 6\alpha^2-2a\alpha-12=0$ ___(2)

$\displaystyle \alpha (2)-3\times (1)\longrightarrow (-2a\alpha^2-12\alpha)-(-3a\alpha^2-36\alpha - 7)=0$

$\displaystyle a\alpha^2+24\alpha+7=0$

This is as far as I got.

Thanks!