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Chapter-2-Remainder-Theorem-and-Factor-Theorem-Exercises-2-1-Remainder-Throrem-1-Find-the-remainder-when-x-8-2x-5-is-divided-by-x-1-2-Find-the-remainder-when-2x-2-13x-10-is-div




Question Number 79866 by 09264910412 last updated on 28/Jan/20
Chapter (2)  Remainder Theorem               and  Factor Theorem  Exercises(2.1)  Remainder Throrem  1.Find the remainder when x^(8 ) +2x−5  is divided by (x−1).  2.Find the remainder when 2x^2  13x+10  is divided by (x−3).
$${Chapter}\:\left(\mathrm{2}\right) \\ $$$${Remainder}\:{Theorem} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{and} \\ $$$${Factor}\:{Theorem} \\ $$$${Exercises}\left(\mathrm{2}.\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{Remainder}}\:\mathrm{T}\boldsymbol{\mathrm{hrorem}} \\ $$$$\mathrm{1}.\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\boldsymbol{\mathrm{x}}^{\mathrm{8}\:} +\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{5} \\ $$$$\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{1}\right). \\ $$$$\mathrm{2}.\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{2x}^{\mathrm{2}} \:\mathrm{13x}+\mathrm{10} \\ $$$$\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{3}\right). \\ $$
Commented by TawaTawa last updated on 28/Jan/20
(1)  let  x − 1  =  0  ∴    x  =  1  Hence,   remainder  =  (1)^8  + 2(1) − 5  ∴   remainder  =  1 + 2 − 5  ∴   remainder  =  − 2    (2)  let  x − 3  =  0  ∴    x  =  3  Hence,   remainder  =  2(3)^2  + 13(3) + 10       (i assume  + 13x)  ∴   remainder  =  2(9) +  39  + 10  ∴   remainder  =  18 +  39  + 10  ∴   remainder  =  67
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{let}\:\:\mathrm{x}\:−\:\mathrm{1}\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1} \\ $$$$\mathrm{Hence},\:\:\:\mathrm{remainder}\:\:=\:\:\left(\mathrm{1}\right)^{\mathrm{8}} \:+\:\mathrm{2}\left(\mathrm{1}\right)\:−\:\mathrm{5} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{1}\:+\:\mathrm{2}\:−\:\mathrm{5} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:−\:\mathrm{2} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{let}\:\:\mathrm{x}\:−\:\mathrm{3}\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{3} \\ $$$$\mathrm{Hence},\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}} \:+\:\mathrm{13}\left(\mathrm{3}\right)\:+\:\mathrm{10}\:\:\:\:\:\:\:\left(\mathrm{i}\:\mathrm{assume}\:\:+\:\mathrm{13x}\right) \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{2}\left(\mathrm{9}\right)\:+\:\:\mathrm{39}\:\:+\:\mathrm{10} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{18}\:+\:\:\mathrm{39}\:\:+\:\mathrm{10} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{67} \\ $$$$ \\ $$

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