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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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