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Christal24 · Oct 2, 2009, 11:30 PM

I am stuck again! I can't factorise this to find stationary points!
y= x^3 + x^2 - 4x - 4
y'= 3x^2 +2x-4

how to factorize??

galactus · Oct 3, 2009, 05:44 AM

I assume you want to factor $3x^{2}+2x-4$ ?

It does not factor. That is why you are having a rough time.

Use the quadratic formula.

To tell rather or not a quadratic will factor or not, check the discriminant. $b^{2}-4ac$

If it is a perfect square, then it will factor. We have $2^{2}-4(3)(-4)=52$

52 is not a perfect square so it is not factorable.

Unknown008 · Oct 3, 2009, 08:33 AM

You can however find the stationary points using the formula that galactus said. He didn't say it directly, but you should already know it:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

You'll have two values, where the stationary points are found.

Christal24 · Oct 3, 2009, 08:14 PM

yeah I did that. It's all good now. Thanks ^_^