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
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).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F454268ad-e730-4c47-902a-aa45b165611b%2F2159cf72-f6b8-496d-977a-ab2187296f2b%2F82yj8fj_processed.png&w=3840&q=75)
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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