Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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

$$\left(\mathrm{1}\right) \\ $$$$\mathrm{let}\:\:\mathrm{x}\:−\:\mathrm{1}\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1} \\ $$$$\mathrm{Hence},\:\:\:\mathrm{remainder}\:\:=\:\:\left(\mathrm{1}\right)^{\mathrm{8}} \:+\:\mathrm{2}\left(\mathrm{1}\right)\:−\:\mathrm{5} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{1}\:+\:\mathrm{2}\:−\:\mathrm{5} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:−\:\mathrm{2} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{let}\:\:\mathrm{x}\:−\:\mathrm{3}\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{3} \\ $$$$\mathrm{Hence},\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}} \:+\:\mathrm{13}\left(\mathrm{3}\right)\:+\:\mathrm{10}\:\:\:\:\:\:\:\left(\mathrm{i}\:\mathrm{assume}\:\:+\:\mathrm{13x}\right) \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{2}\left(\mathrm{9}\right)\:+\:\:\mathrm{39}\:\:+\:\mathrm{10} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{18}\:+\:\:\mathrm{39}\:\:+\:\mathrm{10} \\ $$$$\therefore\:\:\:\mathrm{remainder}\:\:=\:\:\mathrm{67} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com