# Repeated factors

• Sep 9th 2010, 11:58 PM
arze
Repeated factors
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!
• Sep 10th 2010, 12:28 AM
Opalg
Quote:

Originally Posted by 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!

That is a good way to start the problem. To continue, divide your equation (2) by 2, then multiply it by $\displaystyle \alpha$ and subtract the resulting equation from equation (1). That will get rid of the terms containing $\displaystyle a$, leaving you with a cubic equation for $\displaystyle \alpha$ (for which you ought to be able to spot one of the roots).