If p(x) is a quadratic polynomial, then 1/p(x) can be put in the form A/(x-a) + B/(x-b) where a,b,A, and B are (real) constants.

Could someone show me why this is true? I put false but got it wrong.

Thanks,

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- Jun 24th 2010, 12:18 PMcentenialTrue/False Question
If p(x) is a quadratic polynomial, then 1/p(x) can be put in the form A/(x-a) + B/(x-b) where a,b,A, and B are (real) constants.

Could someone show me why this is true? I put false but got it wrong.

Thanks, - Jun 24th 2010, 12:22 PMAlso sprach Zarathustra
The fundamental theorem of algebra! (Google it!)

- Jun 24th 2010, 12:27 PMAlso sprach Zarathustra...
not with all quadratic polynomial you can do what you describe.

let,$\displaystyle p(x)=x^2+x+1$, $\displaystyle p(x)$ have no real roots. - Jun 24th 2010, 12:29 PMAckbeet
... and you would need to find real, unrepeated roots in order to do the partial fraction expansion (which is what this is) and end up with real numbers in all the places described.