I have to factor the following polynomial completely. But I don't have the answer so I don't know if I did it or it can be factored further.

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- Nov 13th 2008, 02:22 AM #1

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- Nov 13th 2008, 10:02 AM #2

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- Nov 14th 2008, 09:53 AM #3

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I think the answer has to be a lot easier, the book i'm using hasn't aproached roots so far. I'm just factoring polynomials. Said that, I don't understand the point of findig the roots. I looked at your explanation and...well, maybe the book author meant "factor completly" according to what we have covered so far...Anyway, I'll keep trying.

I notice the second part looks very much like a difference of squares.

...And how do I find roots if that were the only chance? Prease an explanation

- Nov 14th 2008, 03:10 PM #4

- Nov 17th 2008, 02:42 AM #5

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Ok! This works for me but how did you know that would be equivalent to

I mean...as it is not a perfect square trinomial I couldn't try finding two factors adding to the product of the external values as I did with this:

I usually find the numerical product of the external values:

And then try to find two factors of 120 that add to the middle term. i.e.

Finally all I have to do is just replacing:

And solve...

And that's it!

As you see this aproach wouldn't have worked with as it isn't a perfect square. Now my question is how can I know will solve for the above trinomial? Is there any special formula or technique?

- Nov 17th 2008, 03:16 AM #6

- Nov 18th 2008, 03:21 AM #7

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Then here is how it works. You factored out the :

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But I didn't know where you obtained that so I solved...

And replaced...

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So now I continue...

and the outside the parentesis will multiply for and so we now have...

Consider...

And finally we apply the formula for the difference of two squares...

And that's it.

Now...there is just one little thing that I don't understand. Hope you don't think I'm a pain in the... but how did you get that up there on the second stage?. I mean...It seems you just put it there arbitrarily so the whole thing could be worked out. Finally I want to thank you. Sincerely.

- Nov 18th 2008, 04:00 AM #8