Need help on what looks like polynomial factoring
- Thread starter sleepykid
- Start date
Another thing to think of is that I remember seemingly complex problems like this being easy to solve via substitution; i.e. if you can find like, if long, terms, you can use "u" to take the place of a big chunk. Here "u" might be (x[sup:6e8bgas8]2[/sup:6e8bgas8] + 15x)^.5, or to the one half power. I'm still stumped though.
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- Apr 12, 2005
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I might start with u^2 = x^2 + 15x
Actually, I would start with Domain considerations. x^2 + 15x >= 0 -- Why? This leads to x >= 0 or x <= -15. If we get anything in -15 < x < 0, it's no good. Notice, please, that this is what gave me permission to use u^2, rather than "u". u^2 is never negative. We just suggested that x^2 + 15x is never negative!
Yours is fine, though. This leaves u^2 + 2u - 24 = 0. Solve this using any method at your disposal. One word of warning -- there are more Domain issues. Your definition of u, again, is never negative. If you EVER get anything negative for u, simply discard it.
Actually, I would start with Domain considerations. x^2 + 15x >= 0 -- Why? This leads to x >= 0 or x <= -15. If we get anything in -15 < x < 0, it's no good. Notice, please, that this is what gave me permission to use u^2, rather than "u". u^2 is never negative. We just suggested that x^2 + 15x is never negative!
Yours is fine, though. This leaves u^2 + 2u - 24 = 0. Solve this using any method at your disposal. One word of warning -- there are more Domain issues. Your definition of u, again, is never negative. If you EVER get anything negative for u, simply discard it.