College Algebra Exam Review 40

College Algebra Exam Review 40 - . The formal proof goes by...

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50 1. ALGEBRAIC THEMES and r D 39 125 x 2 C 636 125 x C 558 125 : Lemma 1.8.12. Let p and d be elements of KŒxŁ , with deg .p/ ± deg .d/ ± 0 . Then there is a monomial m D bx k 2 KŒxŁ and a poly- nomial p 0 2 KŒxŁ such that p D md C p 0 , and deg .p 0 / < deg .p/ . Proof. Write p D a n x n C a n ± 1 x n ± 1 C ²²² C a 0 and d D b s x s C b s ± 1 x s ± 1 C²²²C b 0 , where n D deg .p/ and s D deg .d/ , and s ³ n . (Note that d ¤ 0 , because we required deg .d/ ± 0 .) Put m D .a n =b s /x n ± s and p 0 D p ´ md . Then both p and md have leading term equal to a n x n , so deg .p 0 / < deg .p/ . n Proposition 1.8.13. Let p and d be elements of KŒxŁ , with deg .d/ ± 0 . Then there exist polynomials q and r in KŒxŁ such that p D dq C r and deg .r/ < deg .d/ . Proof. The idea of the proof is illustrated by the preceding example: We divide p by d , obtaining a monomial quotient and a remainder p 0 of degree strictly less than the degree of p . We then divide p 0 by d , obtaining a remainder p 00 of still smaller degree. We continue in this fashion until we finally get a remainder of degree less than deg .d/
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Unformatted text preview: . The formal proof goes by induction on the degree of p . If deg .p/ &lt; deg .d/ , then put q D and r D p . So assume now that deg .p/ deg .d/ , and that the result is true when p is replaced by any polynomial of lower degree. According to the lemma, we can write p D md C p , where deg .p / &lt; deg .p/ . By the induction hypothesis, there exist polynomials q and r with deg .r/ &lt; deg .d/ such that p D q d C r . Putting q D q C m , we have p D qd C r . n Denition 1.8.14. A polynomial f 2 Kx is a greatest common divisor of nonzero polynomials p;q 2 Kx if (a) f divides p and q in Kx and (b) whenever g 2 Kx divides p and q , then g also divides f ....
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