Consider a polynomial p₁(x) = ao+a₁2x +...+ ª₂x² = [ªo a₁ an] (x) and a constantz. Synthetic division gives us...[b₁ (2)=a₂[bk (2) = ak + zbk+1(2) for k=n-1,n-2,...,0.Prove that pn(x) = (x − 2)qn-1(x; 2) + bo(z) where qn-1(x; 2) = [b₁(2) b₂(2)... bn(z)](x).

A shortcut for dividing a polynomial by a linear binomial of the form x - k is synthetic division.…

Consider a polynomial p₁(x) = ao+a₁2+...+ ª₂z² = [ªo a₁ ... an] (x) and a constant z. Synthetic division gives us [bn (2) = an bx (2) = a + 2bx+1(3) for k=n-1,n-2,...,0. Prove that p₁(x) = (x − 2)qn−1(x; 2) + bo(2) where qn-1 (z; 2) = [b₁(2) b₂(2). b₁(z)](x).

Consider a polynomial p₁(x) = ao+a₁2+...+ ª₂z² = [ªo a₁ ... an] (x) and a constant z. Synthetic division gives us [bn (2) = an bx (2) = a + 2bx+1(3) for k=n-1,n-2,...,0. Prove that p₁(x) = (x − 2)qn−1(x; 2) + bo(2) where qn-1 (z; 2) = [b₁(2) b₂(2). b₁(z)](x).

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.2: Properties Of Division

Problem 42E

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Question

Consider a polynomial p₁(x) = ao+a₁2x +...+ ª₂x² = [ªo a₁ an] (x) and a constant
z. Synthetic division gives us
...
[b₁ (2)=a₂
[bk (2) = ak + zbk+1(2) for k=n-1,n-2,...,0.
Prove that pn(x) = (x − 2)qn-1(x; 2) + bo(z) where qn-1(x; 2) = [b₁(2) b₂(2)... bn(z)](x).

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