Prove the proposition \" Let p(x) be a polynomial. The number z is a root of p(x

Question

Prove the proposition " Let p(x) be a polynomial. The number z is a root of p(x) if and only if there exists a polynomial q(x) such that p(x) = (x-z)q(z)" using the previous proposition "Let n(x) be a polynomial that is not zero. For every polynomial m(x), there exists polynomials q(x) and r(x) such that m(x) q(x)n(x) + r(x) and either r(x) is zero or the degree of r(x) is smaller than the degree of n(x)." Additional information: A root of the polynomial p(x)=(a_d)(x^d)+(a_(d-1))(x^(d-1))+...+(a_1)(x)+a_0 is a number z such that the polynomial evaluated at z is zero, that is, p(z)=(a_d)(z^d)+(a_(d-1))(z^(d-1))+...+(a_1)z+a_0=0. Note: when I use the underscore sign, I imply subscript. So the beginning statement of p(x) reads " a sub d times x to the d power + a sub d minus 1 times times x to the d minus 1....."

Explanation / Answer

Proof. (of Lemma) First we note that X /Y can be equipped with the norm x + Y = inf |x - y | y ?Y for x + Y ? X/Y . For any y1, y2 ? Y , we have |x1 + x2 - (y1 + y2)| = |x1 - y1| + |x2 - y2| inf y1 + y2 ?Y taking inf over y1 and y2 gives subadditivity. Suppose inf y ?Y |x - y | = 0. Then there exists a sequence yn such that |x - yn | ? 0, and since Y is closed, we have that x ? Y so that x + Y = 0 + Y (the zero vector in X/Y ). (Thus, if Y is not closed, · is just a seminorm). Also, if X is complete, so is X/Y . This is because if xn + Y is Cauchy, then xn - xm + Y = inf y ?Y |xn - xm - y | ? 0, and can ?nd ym ? Y such that |(xn - yn) - (xm - ym)| ? 0, so that xn - yn is Cauchy, and by completeness xn - yn ? z, and so xn + Y ? z + Y . Now there exists x0 + Y ? X/Y such that x0 + Y = 1. For e > 0, there exists y such that 1 = |x0 + y | < 1 + e Then let x = |x0 + y | , so that |x| = 1, and furthermore, inf |x - y | = x + Y = y ?Y 1 x = |x0 + y | 1 + e

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