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3n-5-4n-4-7n-3-5n-2-5-n-1-There-can-be-no-residue-a-0-b-2-c-4-d-5-e-9-




Question Number 183381 by Shrinava last updated on 25/Dec/22
((3n^5  + 4n^4  − 7n^3  + 5n^2  − 5)/(n + 1))  There can be no residue:  a)0   b)2   c)4   d)5   e)9
$$\frac{\mathrm{3n}^{\mathrm{5}} \:+\:\mathrm{4n}^{\mathrm{4}} \:−\:\mathrm{7n}^{\mathrm{3}} \:+\:\mathrm{5n}^{\mathrm{2}} \:−\:\mathrm{5}}{\mathrm{n}\:+\:\mathrm{1}} \\ $$$$\mathrm{There}\:\mathrm{can}\:\mathrm{be}\:\mathrm{no}\:\mathrm{residue}: \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{0}\left.\:\left.\:\:\mathrm{b}\right)\mathrm{2}\:\:\:\mathrm{c}\right)\mathrm{4}\:\:\:\mathrm{d}\right)\mathrm{5}\:\:\:\mathrm{e}\right)\mathrm{9} \\ $$
Commented by Shrinava last updated on 25/Dec/22
sorry:  5n^2  − 4
$$\mathrm{sorry}:\:\:\mathrm{5n}^{\mathrm{2}} \:−\:\mathrm{4} \\ $$
Commented by Frix last updated on 25/Dec/22
same answer
$$\mathrm{same}\:\mathrm{answer} \\ $$
Answered by Frix last updated on 25/Dec/22
((3n^5 +4n^4 −7n^3 +5n^2 −4)/(n+1))=  =3n^4 +n^3 −8n^2 +13n−13+(9/(n+1))  (9/(n+1))=r ⇔ n=(9/r)−1 ⇒ r≠0
$$\frac{\mathrm{3}{n}^{\mathrm{5}} +\mathrm{4}{n}^{\mathrm{4}} −\mathrm{7}{n}^{\mathrm{3}} +\mathrm{5}{n}^{\mathrm{2}} −\mathrm{4}}{{n}+\mathrm{1}}= \\ $$$$=\mathrm{3}{n}^{\mathrm{4}} +{n}^{\mathrm{3}} −\mathrm{8}{n}^{\mathrm{2}} +\mathrm{13}{n}−\mathrm{13}+\frac{\mathrm{9}}{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{9}}{{n}+\mathrm{1}}={r}\:\Leftrightarrow\:{n}=\frac{\mathrm{9}}{{r}}−\mathrm{1}\:\Rightarrow\:{r}\neq\mathrm{0} \\ $$

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