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Consider a polynomial pn(x) = ao + a₁x +...+ anx" = [ao a₁ z. Synthetic division gives us an] (x) and a constant [bn(z) = [bk (2) = ak + zbk+1(2) for k= n1, n - 2,...,0. Prove that pn(x) = (x − 2)qn−1(x; 2) + bo(z) where qn-1 (x; 2) = [b₁ (2) b₂(z) = an bn (z)](x).

Consider a polynomial pn(x) = ao + a₁x +...+ anx" = [ao a₁ z. Synthetic division gives us an] (x) and a constant [bn(z) = [bk (2) = ak + zbk+1(2) for k= n1, n - 2,...,0. Prove that pn(x) = (x − 2)qn−1(x; 2) + bo(z) where qn-1 (x; 2) = [b₁ (2) b₂(z) = an bn (z)](x).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider a polynomial pn(x) = ao + a₁x +...+ anxn = [ao a₁ an] (x) and a constant z. Synthetic division gives us [bn(z) = an | bk (2) = ak + zbk+1(2) for kn 1, n - 2,..., 0. Prove that pn(x) = (x − 2)¶n−1(x; 2) + bo(2) where qn-1(x; 2) = [b₁(z) b₂(z) ….. bn(z)](x).
Transcribed Image Text:Consider a polynomial pn(x) = ao + a₁x +...+ anxn = [ao a₁ an] (x) and a constant z. Synthetic division gives us [bn(z) = an | bk (2) = ak + zbk+1(2) for kn 1, n - 2,..., 0. Prove that pn(x) = (x − 2)¶n−1(x; 2) + bo(2) where qn-1(x; 2) = [b₁(z) b₂(z) ….. bn(z)](x).
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