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Question Number 19811 by Joel577 last updated on 16/Aug/17
If f(x) = (x +1)g(x) − 2 and g(3) = 4 Find the remainder if f(x) divided by (x + 1)(x − 3)
$$\mathrm{If}\:{f}\left({x}\right)\:=\:\left({x}\:+\mathrm{1}\right){g}\left({x}\right)\:−\:\mathrm{2}\:\mathrm{and}\:{g}\left(\mathrm{3}\right)\:=\:\mathrm{4} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{if}\:{f}\left({x}\right)\:\mathrm{divided}\:\mathrm{by}\: \\ $$$$\left({x}\:+\:\mathrm{1}\right)\left({x}\:−\:\mathrm{3}\right) \\ $$
Commented by myintkhaing last updated on 16/Aug/17
Since, f(x) = (x+1)g(x) −2 and g(3) = 4 when f(x) is divided by (x+1), the remainder, f(−1) = −2 and f(3) = 4g(3) −2 = 14 Let the quotient be Q(x) and the remainder be ax+b when f(x) is divided by (x+1)(x−3) So, f(x) = (x+1)(x−3)Q(x)+ax+b thus f(−1) = −a+b = −2 .........(1) f(3) = 3a+b = 14 .........(2) Solving equations (1) and (2) a = 4 and b = 2 ∴ The required remainder = 4x+2 #
$$\mathrm{Since},\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\left(\mathrm{x}+\mathrm{1}\right)\mathrm{g}\left(\mathrm{x}\right)\:−\mathrm{2}\:\mathrm{and}\:\mathrm{g}\left(\mathrm{3}\right)\:=\:\mathrm{4} \\ $$$$\mathrm{when}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}+\mathrm{1}\right), \\ $$$$\mathrm{the}\:\mathrm{remainder},\:\mathrm{f}\left(−\mathrm{1}\right)\:=\:−\mathrm{2}\:\mathrm{and} \\ $$$$\mathrm{f}\left(\mathrm{3}\right)\:=\:\mathrm{4g}\left(\mathrm{3}\right)\:−\mathrm{2}\:=\:\mathrm{14} \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{quotient}\:\mathrm{be}\:\mathrm{Q}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{remainder} \\ $$$$\mathrm{be}\:\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{when}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right) \\ $$$$\mathrm{So},\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)\mathrm{Q}\left(\mathrm{x}\right)+\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{thus}\:\mathrm{f}\left(−\mathrm{1}\right)\:=\:−\mathrm{a}+\mathrm{b}\:=\:−\mathrm{2}\:………\left(\mathrm{1}\right) \\ $$$$\mathrm{f}\left(\mathrm{3}\right)\:=\:\mathrm{3a}+\mathrm{b}\:=\:\mathrm{14}\:………\left(\mathrm{2}\right) \\ $$$$\mathrm{Solving}\:\mathrm{equations}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{a}\:=\:\mathrm{4}\:\mathrm{and}\:\mathrm{b}\:=\:\mathrm{2} \\ $$$$\therefore\:\mathrm{The}\:\mathrm{required}\:\mathrm{remainder}\:=\:\mathrm{4x}+\mathrm{2}\:# \\ $$
Commented by Joel577 last updated on 16/Aug/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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