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A Secular Philosopher of Administration


Two polynomials are equal if and only if their corresponding coefficients are equal, so we can conclude{\displaystyle a_{3}=1,\quad a_{2}+a_{3}=0,\quad a_{1}+a_{2}+a_{3}=-1.\,}

a_{3}=1,\quad a_{2}+a_{3}=0,\quad a_{1}+a_{2}+a_{3}=-1.\,

This system of linear equations can easily be solved. First, the first equation simply says that a3 is 1. Knowing that, we can solve the second equation for a2, which comes out to −1. Finally, the last equation tells us that a1 is also −1. Therefore, the only possible way to get a linear combination is with these coefficients. Indeed,{\displaystyle x^{2}-1=-1-(x+1)+(x^{2}+x+1)=-p_{1}-p_{2}+p_{3}\,}


so x2 − 1 is a linear combination of p1p2, and p3.

On the other hand, what about the polynomial x3 − 1? If we try to make this vector a linear combination of p1p2, and p3, then following the same process as before, we’ll get the equation{\displaystyle 0x^{3}+a_{3}x^{2}+(a_{2}+a_{3})x+(a_{1}+a_{2}+a_{3})\,}


{\displaystyle =1x^{3}+0x^{2}+0x+(-1).\,}


However, when we set corresponding coefficients equal in this case, the equation for x3 is{\displaystyle 0=1\,}


which is always false. Therefore, there is no way for this to work, and x3 − 1 is not a linear combination of p1p2, and p3.