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Question Number 144060 by Khalmohmmad last updated on 21/Jun/21

p(x)=(x−1)^2 Q(x)+3x+8 p(x)=(x−2)Q(x)+R R=?

$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {Q}\left({x}\right)+\mathrm{3}{x}+\mathrm{8} \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{2}\right){Q}\left({x}\right)+{R} \\ $$$${R}=? \\ $$

Commented by Rasheed.Sindhi last updated on 21/Jun/21

Q(x) is same in both cases!!!?

$${Q}\left({x}\right)\:{is}\:{same}\:{in}\:{both}\:{cases}!!!? \\ $$

Answered by Rasheed.Sindhi last updated on 21/Jun/21

p(x)=(x−1)^2 Q_1 (x)+3x+8.....(i) p(x)=(x−2)Q_2 (x)+R.....(ii) ; R=? ^• Let Q_1 (x)=q(constant p(x)=(x−1)^2 q+3x+8 =qx^2 −2xq+q+3x+8 =qx^2 +(3−2q)x+q+8 When divided by x−2 R=q(2)^2 +(3−2q)(2)+q+8 R=4q+6−4q+q+8 =q+14 ^(• ) Q_1 (x)=ax+b p(x)=(x−1)^2 Q_1 (x)+3x+8 p(x)=(x−1)^2 (ax+b)+3x+8 =(x^2 −2x+1)(ax+b)+3x+8 =ax^3 +bx^2 −2ax^2 −2bx+ax+b+3x+8 =ax^3 +(b−2a)x^2 +(a+3−2b)x+b+8 When divided by x−2 R=p(2)=a(2)^3 +(b−2a)(2)^2 +(a+3−2b)(2)+b+8 =8a+4b−8a+2a+6−4b+b+8 =2a+b+14 No unique answer. R depends upon Q_1 (x).

$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {Q}_{\mathrm{1}} \left({x}\right)+\mathrm{3}{x}+\mathrm{8}.....\left({i}\right) \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{2}\right){Q}_{\mathrm{2}} \left({x}\right)+{R}.....\left({ii}\right)\:\:\:;\:\:{R}=? \\ $$$$\:^{\bullet} {Let}\:{Q}_{\mathrm{1}} \left({x}\right)={q}\left({constant}\right. \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {q}+\mathrm{3}{x}+\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:={qx}^{\mathrm{2}} −\mathrm{2}{xq}+{q}+\mathrm{3}{x}+\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:={qx}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{2}{q}\right){x}+{q}+\mathrm{8} \\ $$$${When}\:{divided}\:{by}\:{x}−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{R}={q}\left(\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{2}{q}\right)\left(\mathrm{2}\right)+{q}+\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:{R}=\mathrm{4}{q}+\mathrm{6}−\mathrm{4}{q}+{q}+\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={q}+\mathrm{14} \\ $$$$\:^{\bullet\:} {Q}_{\mathrm{1}} \left({x}\right)={ax}+{b}\:\:\:\: \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {Q}_{\mathrm{1}} \left({x}\right)+\mathrm{3}{x}+\mathrm{8} \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({ax}+{b}\right)+\mathrm{3}{x}+\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:=\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)\left({ax}+{b}\right)+\mathrm{3}{x}+\mathrm{8} \\ $$$$\:\:\:={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} −\mathrm{2}{ax}^{\mathrm{2}} −\mathrm{2}{bx}+{ax}+{b}+\mathrm{3}{x}+\mathrm{8} \\ $$$$\:\:={ax}^{\mathrm{3}} +\left({b}−\mathrm{2}{a}\right){x}^{\mathrm{2}} +\left({a}+\mathrm{3}−\mathrm{2}{b}\right){x}+{b}+\mathrm{8} \\ $$$${When}\:{divided}\:{by}\:{x}−\mathrm{2} \\ $$$${R}={p}\left(\mathrm{2}\right)={a}\left(\mathrm{2}\right)^{\mathrm{3}} +\left({b}−\mathrm{2}{a}\right)\left(\mathrm{2}\right)^{\mathrm{2}} +\left({a}+\mathrm{3}−\mathrm{2}{b}\right)\left(\mathrm{2}\right)+{b}+\mathrm{8} \\ $$$$=\cancel{\mathrm{8}{a}}+\cancel{\mathrm{4}{b}}−\cancel{\mathrm{8}{a}}+\mathrm{2}{a}+\mathrm{6}−\cancel{\mathrm{4}{b}}+{b}+\mathrm{8} \\ $$$$=\mathrm{2}{a}+{b}+\mathrm{14} \\ $$$${No}\:{unique}\:{answer}. \\ $$$${R}\:{depends}\:{upon}\:{Q}_{\mathrm{1}} \left({x}\right). \\ $$

Answered by Rasheed.Sindhi last updated on 21/Jun/21

p(x)=(x−1)^2 Q_1 (x)+3x+8 p(x)=(x−2)Q_2 (x)+R ; R=? p(x) when is divided by x−2 giving remainder R (x−1)^2 Q_1 (x)+3x+8 is divided by x−2 giving remainder R R=p(2)=(2−1)^2 Q_1 (2)+3(2)+8 R=p(2)=Q_1 (2)+14 ^• Let Q_1 (x)=q (constant) R=q+14 ^• Let Q_1 (x)=ax+b R=Q_1 (2)+14=a(2)+b+14 R=2a+b+14 ^• Let Q_1 (x)=ax^2 +bx+c R=a(2)^2 +b(2)+c+14 R=4a+2b+c+14 R depends on Q_1 (x) (or Q_2 (x) ). It′s not unique.

$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {Q}_{\mathrm{1}} \left({x}\right)+\mathrm{3}{x}+\mathrm{8} \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{2}\right){Q}_{\mathrm{2}} \left({x}\right)+{R}\:\:;\:\:{R}=? \\ $$$${p}\left({x}\right)\:{when}\:{is}\:{divided}\:{by}\:{x}−\mathrm{2}\:{giving}\: \\ $$$${remainder}\:{R} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} {Q}_{\mathrm{1}} \left({x}\right)+\mathrm{3}{x}+\mathrm{8}\:\:{is}\:{divided}\:{by} \\ $$$${x}−\mathrm{2}\:{giving}\:{remainder}\:{R} \\ $$$${R}={p}\left(\mathrm{2}\right)=\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} {Q}_{\mathrm{1}} \left(\mathrm{2}\right)+\mathrm{3}\left(\mathrm{2}\right)+\mathrm{8} \\ $$$${R}={p}\left(\mathrm{2}\right)={Q}_{\mathrm{1}} \left(\mathrm{2}\right)+\mathrm{14} \\ $$$$\:^{\bullet} {Let}\:\:{Q}_{\mathrm{1}} \left({x}\right)={q}\:\left({constant}\right) \\ $$$${R}={q}+\mathrm{14} \\ $$$$\:^{\bullet} {Let}\:{Q}_{\mathrm{1}} \left({x}\right)={ax}+{b} \\ $$$${R}={Q}_{\mathrm{1}} \left(\mathrm{2}\right)+\mathrm{14}={a}\left(\mathrm{2}\right)+{b}+\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\:\:\:{R}=\mathrm{2}{a}+{b}+\mathrm{14} \\ $$$$\:^{\bullet} {Let}\:{Q}_{\mathrm{1}} \left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\:\:\:\:{R}={a}\left(\mathrm{2}\right)^{\mathrm{2}} +{b}\left(\mathrm{2}\right)+{c}+\mathrm{14} \\ $$$$\:\:\:\:\:\:\:{R}=\mathrm{4}{a}+\mathrm{2}{b}+{c}+\mathrm{14} \\ $$$${R}\:{depends}\:{on}\:{Q}_{\mathrm{1}} \left({x}\right)\:\left({or}\:{Q}_{\mathrm{2}} \left({x}\right)\:\right). \\ $$$${It}'{s}\:{not}\:{unique}. \\ $$$$ \\ $$

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