find-such-a-polynomial-if-is-divided-by-x-2-then-the-remainder-is-5-if-it-isdivided-by-x-3-the-remainder-is-9-if-it-is-divided-by-x-4-the-remainder-is-13-if-divide-by-x-10-and-the-remaid

Question Number 94818 by student work last updated on 21/May/20

$$\:\mathrm{find}\:\mathrm{such}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{if}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{2}\right)\:\mathrm{then}\:\mathrm{the}\: \\ $$$$\mathrm{remainder}\:\mathrm{is}\:\mathrm{5},\:\mathrm{if}\:\mathrm{it}\:\mathrm{isdivided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{3}\right)\:\mathrm{the}\:\mathrm{remainder} \\ $$$$\mathrm{is}\:\mathrm{9},\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{4}\right)\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{13},\:\mathrm{if}\: \\ $$$$\mathrm{divide}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{10}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaider}\:\mathrm{becomes}\:\mathrm{37}\:\mathrm{and} \\ $$$$\mathrm{if}\:\left(\mathrm{x}−\frac{\mathrm{3}}{\mathrm{4}}\right)\:\mathrm{divided}\:\mathrm{by}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{becomes}\:\mathrm{zero}? \\ $$

Commented by Rasheed.Sindhi last updated on 21/May/20

$${The}\:{required}\:{polynomial}\:{is}\:\mathrm{4}{x}−\mathrm{3} \\ $$

Commented by student work last updated on 21/May/20

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{practice}\:\mathrm{der}? \\ $$

Commented by Rasheed.Sindhi last updated on 21/May/20

$${I}\:{shall}\:{write}\:{answer}\:{in}\:{detail}\:{soon}. \\ $$

Commented by student work last updated on 21/May/20

$$\mathrm{you}\:\mathrm{show}\:\mathrm{practice}\:\mathrm{to}\:\mathrm{i}\:\mathrm{learn}\: \\ $$

Commented by mr W last updated on 21/May/20

4x−3 is correct, but not the only one solution, i think. i′ll post my solution later.

$$\mathrm{4}{x}−\mathrm{3}\:{is}\:{correct},\:{but}\:{not}\:{the}\:{only}\:{one} \\ $$$${solution},\:{i}\:{think}. \\ $$$${i}'{ll}\:{post}\:{my}\:{solution}\:{later}. \\ $$

Commented by Rasheed.Sindhi last updated on 21/May/20

$${Yes}\:{sir}\:{it}'{s}\:{linear}\:{answer},\:{and} \\ $$$${if}\:{I}'{ve}\:{done}\:{correct}\:{calculation} \\ $$$${quadratic}\:{is}\:{not}\:{possible}.{Now}\:{I}'{m} \\ $$$${tryingfor}\:{cubic}\:{solution}… \\ $$

Commented by student work last updated on 21/May/20

$$\mathrm{please}\:\mathrm{post}\:\mathrm{me}\:\mathrm{solution}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 21/May/20

say the polynomial searched is f(x). f(x) can be divided by (x−(3/4)) without remainder, therefore f(x)=g(x)(x−(3/4)) with g(x) being an other polynomial. when f(x) is divided by (x−2) we get remainder 5, that means f(x)=p(x)(x−2)+5, i.e. f(2)=5, f(2)=g(2)(2−(3/4))=5 ⇒g(2)=4 similarly f(3)=g(3)(3−(3/4))=9 ⇒g(3)=4 f(4)=g(4)(4−(3/4))=13 ⇒g(4)=4 f(10)=g(10)(10−(3/4))=37 ⇒g(10)=4 that means g(x)=k(x)(x−2)(x−3)(x−4)(x−10)+4 with k(x) being any polynomial. ⇒f(x)=g(x)(x−(3/4)) ⇒f(x)=[k(x)(x−2)(x−3)(x−4)(x−10)+4](x−(3/4)) or ⇒f(x)=[c(x)(x−2)(x−3)(x−4)(x−10)+1](4x−3) with c(x) being any polynomial (including any constant)

$${say}\:{the}\:{polynomial}\:{searched}\:{is}\:{f}\left({x}\right). \\ $$$${f}\left({x}\right)\:{can}\:{be}\:{divided}\:{by}\:\left({x}−\frac{\mathrm{3}}{\mathrm{4}}\right)\:{without} \\ $$$${remainder},\:{therefore} \\ $$$${f}\left({x}\right)={g}\left({x}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$${with}\:{g}\left({x}\right)\:{being}\:{an}\:{other}\:{polynomial}. \\ $$$$ \\ $$$${when}\:{f}\left({x}\right)\:{is}\:{divided}\:{by}\:\left({x}−\mathrm{2}\right)\:{we}\:{get} \\ $$$${remainder}\:\mathrm{5},\:{that}\:{means} \\ $$$${f}\left({x}\right)={p}\left({x}\right)\left({x}−\mathrm{2}\right)+\mathrm{5},\:{i}.{e}.\:{f}\left(\mathrm{2}\right)=\mathrm{5}, \\ $$$${f}\left(\mathrm{2}\right)={g}\left(\mathrm{2}\right)\left(\mathrm{2}−\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{5}\:\Rightarrow{g}\left(\mathrm{2}\right)=\mathrm{4} \\ $$$${similarly} \\ $$$${f}\left(\mathrm{3}\right)={g}\left(\mathrm{3}\right)\left(\mathrm{3}−\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{9}\:\Rightarrow{g}\left(\mathrm{3}\right)=\mathrm{4} \\ $$$${f}\left(\mathrm{4}\right)={g}\left(\mathrm{4}\right)\left(\mathrm{4}−\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{13}\:\Rightarrow{g}\left(\mathrm{4}\right)=\mathrm{4} \\ $$$${f}\left(\mathrm{10}\right)={g}\left(\mathrm{10}\right)\left(\mathrm{10}−\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{37}\:\Rightarrow{g}\left(\mathrm{10}\right)=\mathrm{4} \\ $$$${that}\:{means} \\ $$$${g}\left({x}\right)={k}\left({x}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{10}\right)+\mathrm{4} \\ $$$${with}\:{k}\left({x}\right)\:{being}\:{any}\:{polynomial}. \\ $$$$\Rightarrow{f}\left({x}\right)={g}\left({x}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\left[{k}\left({x}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{10}\right)+\mathrm{4}\right]\left({x}−\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$${or} \\ $$$$\Rightarrow{f}\left({x}\right)=\left[{c}\left({x}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{10}\right)+\mathrm{1}\right]\left(\mathrm{4}{x}−\mathrm{3}\right) \\ $$$${with}\:{c}\left({x}\right)\:{being}\:{any}\:{polynomial} \\ $$$$\left({including}\:{any}\:{constant}\right) \\ $$

Commented by student work last updated on 21/May/20