This is a follow up to this question /advanced-math-answers-5-647546-cpi0.aspx N

Question

This is a follow up to this question /advanced-math-answers-5-647546-cpi0.aspx

Now show that for all polynomials S(z) that are not constantpolynomials & for any complexvalue , S(z)-S() is divisible byz- , so that if S() = 0 then you can sayz- divides S(z). This is a follow up to this question /advanced-math-answers-5-647546-cpi0.aspx

Now show that for all polynomials S(z) that are not constantpolynomials & for any complexvalue , S(z)-S() is divisible byz- , so that if S() = 0 then you can sayz- divides S(z).

Explanation / Answer

When looking at S(z) - S(a), pair up the terms with the commonpowers and notice the b_0 terms cancel, i.e.
S(z) - S(a) = b_n(z^n - a^n) + ... + b_1(z - a)
By the previous problem, z-a divides each of the z^k - a^kterms, so z - a divides S(z) - S(a). If S(a) = 0, then S(z) - S(a)= S(z) and z-a divides S(z) since it divides S(z) - S(a).

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