Question 181856
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If you applied the difference of two squares and grouping you should have obtained:


*[tex \LARGE \text{          }\math \left((p + 1)(p - 1)\right) - 4q(q - 1)]


That's as completely factored as anyone can make this one.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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