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Question Number 67299 by Rio Michael last updated on 25/Aug/19
G(x)= (x+1)(x+3)Q(x) + px +q a) Given that G(x) leaves a remainder of 8 and −24 when divided by (x+1) and (x+3) respectively,find the remainder when G(x) is divided by (x+1)(x+3). b) Given that x+2 is a factor of G(x) and that the graph of G(x) passes through the point with coordinates (0,6) find G(x)
$${G}\left({x}\right)=\:\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right){Q}\left({x}\right)\:+\:{px}\:+{q} \\ $$$$\left.{a}\right)\:{Given}\:{that}\:{G}\left({x}\right)\:{leaves}\:{a}\:{remainder}\:{of}\:\mathrm{8}\:{and}\:−\mathrm{24}\:{when}\:{divided}\:{by}\:\left({x}+\mathrm{1}\right)\:{and}\: \\ $$$$\left({x}+\mathrm{3}\right)\:{respectively},{find}\:{the}\:{remainder}\:{when}\:{G}\left({x}\right)\:{is}\:{divided}\:{by}\:\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right). \\ $$$$\left.{b}\right)\:\:{Given}\:{that}\:{x}+\mathrm{2}\:{is}\:{a}\:{factor}\:{of}\:{G}\left({x}\right)\:{and}\:{that}\:{the}\:{graph}\:{of}\:{G}\left({x}\right)\:{passes}\:{through} \\ $$$${the}\:{point}\:{with}\:{coordinates}\:\left(\mathrm{0},\mathrm{6}\right)\:{find}\:{G}\left({x}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 28/Aug/19
G(x)=(x+1)(x+3)(ax+b)+16x+24 G(−2)=(−2+1)(−2+3)(−2a+b)−32+24=0 2a−b=8 G(0)=(0+1)(0+3)(a(0)+b)+16(0)+24=6 3b+24=6⇒b=−6⇒a=1 ∴Q(x)=ax+b=x−6 G(x)=(x+1)(x+3)(x−6)+16x+24 =(x^2 +4x+3)(x−6)+16x+24 =x^3 +4x^2 +3x−6x^2 −24x−18+16x+24 =x^3 −2x^2 −5x+6
$${G}\left({x}\right)=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\left({ax}+{b}\right)+\mathrm{16}{x}+\mathrm{24} \\ $$$$\:{G}\left(−\mathrm{2}\right)=\left(−\mathrm{2}+\mathrm{1}\right)\left(−\mathrm{2}+\mathrm{3}\right)\left(−\mathrm{2}{a}+{b}\right)−\mathrm{32}+\mathrm{24}=\mathrm{0} \\ $$$$\:\:\:\mathrm{2}{a}−{b}=\mathrm{8} \\ $$$${G}\left(\mathrm{0}\right)=\left(\mathrm{0}+\mathrm{1}\right)\left(\mathrm{0}+\mathrm{3}\right)\left({a}\left(\mathrm{0}\right)+{b}\right)+\mathrm{16}\left(\mathrm{0}\right)+\mathrm{24}=\mathrm{6}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{b}+\mathrm{24}=\mathrm{6}\Rightarrow{b}=−\mathrm{6}\Rightarrow{a}=\mathrm{1} \\ $$$$\therefore{Q}\left({x}\right)={ax}+{b}={x}−\mathrm{6} \\ $$$${G}\left({x}\right)=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{6}\right)+\mathrm{16}{x}+\mathrm{24} \\ $$$$\:\:\:\:\:=\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}\right)\left({x}−\mathrm{6}\right)+\mathrm{16}{x}+\mathrm{24} \\ $$$$\:\:\:\:\:={x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{24}{x}−\mathrm{18}+\mathrm{16}{x}+\mathrm{24} \\ $$$$\:\:\:\:\:={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Aug/19
Sir mr W! I made my answer more compact in the light of your comment. However its only ′unique cubic solution′.
$${Sir}\:{mr}\:{W}!\:{I}\:{made}\:{my}\:{answer}\:{more} \\ $$$${compact}\:{in}\:{the}\:{light}\:{of}\:{your}\:{comment}. \\ $$$${However}\:{its}\:{only}\:'{unique}\:{cubic}\:{solution}'. \\ $$
Commented by Rasheed.Sindhi last updated on 29/Aug/19
mr W sir, Can we translate the problem as follows: (I′ve seen great resemblance between this problem & Q#66832 (with ans 1731) G(x)≡8(mod (x+1) ) G(x)≡−24(mod (x+3) ) G(x)≡0(mod (x+2) ) G(x)≡6(mod x )
$$\:{mr}\:{W}\:{sir},\:{Can}\:{we}\:{translate}\:{the}\:{problem}\:\:\:\: \\ $$$${as}\:{follows}:\:\left({I}'{ve}\:{seen}\:{great}\:{resemblance}\:{between}\right. \\ $$$${this}\:{problem}\:\&\:{Q}#\mathrm{66832}\:\left({with}\:{ans}\:\mathrm{1731}\right) \\ $$$$\:\:\:\:\:\:{G}\left({x}\right)\equiv\mathrm{8}\left({mod}\:\left({x}+\mathrm{1}\right)\:\right) \\ $$$$\:\:\:\:\:\:{G}\left({x}\right)\equiv−\mathrm{24}\left({mod}\:\left({x}+\mathrm{3}\right)\:\right) \\ $$$$\:\:\:\:\:\:{G}\left({x}\right)\equiv\mathrm{0}\left({mod}\:\left({x}+\mathrm{2}\right)\:\:\right) \\ $$$$\:\:\:\:\:\:{G}\left({x}\right)\equiv\mathrm{6}\left({mod}\:{x}\:\:\right) \\ $$
Commented by mr W last updated on 29/Aug/19
indeed an interesting comparision sir!
$${indeed}\:{an}\:{interesting}\:{comparision}\:{sir}! \\ $$
Commented by Rasheed.Sindhi last updated on 02/Sep/19
Sir mr W, I've tried to solve the above problem using Chinese Remainder Theorm.Pl see Q#67697.
Answered by Rasheed.Sindhi last updated on 26/Aug/19
a If any polynomial P(x) is divided by an other polynomial D(x) giving quotient Q(x) and remainder R(x),then P(x)=D(x)×Q(x)+R(x) where R(x) is lower in degree than D(x). Now in G(x)= (x+1)(x+3)Q(x) + px +q if (x+1)(x+3) is divisor having degree 2, then surely px+q having degree 1 (lower than that of divisor) is remainder. Now the work remained is only to determine values of p & q if we could determine. ^• G(x) is divided by x+1 giving remainder 8: ∴G(−1)=8 p(−1)+q=8⇒p−q=−8.........(i) ^• G(x) is divided by x+3 giving remainder −24: ∴ p(−3)+q=−24⇒3p−q=24.......(ii) (ii)−(i):2p=32⇒p=16 p−q=−8⇒16−q=−8⇒q=24 ∴ The required remainder: px+q=16x+24
$${a} \\ $$$${If}\:{any}\:{polynomial}\:{P}\left({x}\right)\:{is}\:{divided}\:{by} \\ $$$${an}\:{other}\:{polynomial}\:{D}\left({x}\right)\:{giving}\:{quotient} \\ $$$${Q}\left({x}\right)\:{and}\:{remainder}\:{R}\left({x}\right),{then} \\ $$$${P}\left({x}\right)={D}\left({x}\right)×{Q}\left({x}\right)+{R}\left({x}\right)\:{where}\:{R}\left({x}\right) \\ $$$${is}\:{lower}\:{in}\:{degree}\:{than}\:{D}\left({x}\right). \\ $$$${Now}\:{in} \\ $$$${G}\left({x}\right)=\:\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right){Q}\left({x}\right)\:+\:{px}\:+{q} \\ $$$${if}\:\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\:{is}\:{divisor}\:{having}\:{degree} \\ $$$$\mathrm{2},\:{then}\:{surely}\:{px}+{q}\:{having}\:{degree}\:\mathrm{1} \\ $$$$\left({lower}\:{than}\:{that}\:{of}\:{divisor}\right)\:{is}\:{remainder}. \\ $$$${Now}\:{the}\:{work}\:{remained}\:{is}\:{only}\:{to} \\ $$$${determine}\:{values}\:{of}\:{p}\:\&\:{q}\:{if}\:{we}\:{could} \\ $$$${determine}. \\ $$$$\:^{\bullet} {G}\left({x}\right)\:{is}\:{divided}\:{by}\:{x}+\mathrm{1}\:{giving}\: \\ $$$${remainder}\:\mathrm{8}: \\ $$$$\:\therefore{G}\left(−\mathrm{1}\right)=\mathrm{8} \\ $$$${p}\left(−\mathrm{1}\right)+{q}=\mathrm{8}\Rightarrow{p}−{q}=−\mathrm{8}………\left({i}\right) \\ $$$$\:^{\bullet} {G}\left({x}\right)\:{is}\:{divided}\:{by}\:{x}+\mathrm{3}\:{giving}\: \\ $$$${remainder}\:−\mathrm{24}: \\ $$$$\therefore\:{p}\left(−\mathrm{3}\right)+{q}=−\mathrm{24}\Rightarrow\mathrm{3}{p}−{q}=\mathrm{24}…….\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right):\mathrm{2}{p}=\mathrm{32}\Rightarrow{p}=\mathrm{16} \\ $$$${p}−{q}=−\mathrm{8}\Rightarrow\mathrm{16}−{q}=−\mathrm{8}\Rightarrow{q}=\mathrm{24} \\ $$$$\therefore\:\:{The}\:{required}\:{remainder}: \\ $$$$\:\:\:\:\:\:\:{px}+{q}=\mathrm{16}{x}+\mathrm{24} \\ $$
Commented by Rio Michael last updated on 27/Aug/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 27/Aug/19
b G(−1)=8 G(−3)=−24 G(−2)=0 G(0)=6 G(x) at least quadratic. G(x)=? •Let G(x)=(x+2)(ax+b) =ax^2 +(b+2a)x+2b G(0)=6⇒2b=6⇒b=3 G(x)=ax^2 +(2a+3)x+6 (((−1)),a,(2a+3),( 6)),(,,(−a),( −a−3)),(,a,(a+3),(−a+3=8)) ) −a+3=8⇒a=−5 G(x)=−5x^2 −7x+6 (((−3)),(−5),(−7),( 6)),(,,(15),( −24)),(,(−5),( 8),( −18)) ) G(−3)=−18 This is against to the given(G(−3)=−24 ∴G(x)≠(x+2)(ax+b) •Let G(x)=(x+2)(ax^2 +bx+c) =ax^3 +bx^2 +cx+2ax^2 +2bx+2c =ax^3 +2ax^2 +bx^2 +2bx+cx+2c =ax^3 +(2a+b)x^2 +(2b+c)x+2c G(0)=6⇒2c=6⇒c=3 G(x)=ax^3 +(2a+b)x^2 +(2b+3)x+6 (((−1)),a,(2a+b),( 2b+3),( 6)),(,,(−a),( −a−b),(a−b−3)),(,a,(a+b),(−a+b+3),(a−b+3=8)) ) a−b=5⇒a=b+5 (((−3)),a,(2a+b),( 2b+3),( 6)),(,,(−3a),( 3a−3b),( −9a+3b−9)),(,a,(−a+b),(3a−b+3),(−9a+3b−3=−24)) ) −9a+3b=−21 −3a+b=−7 −3(b+5)+b=−7 ........ −2b=−7+15=8 ........ b=−4,a=1 G(x)=(x+2)(x^2 −4x+3) ∴ G(x)=x^3 −2x^2 −5x+6 Verification: (((−1)),1,(−2),(−5),6),(,,(−1),( 3),2),(,1,(−3),( −2),8) ) ^• ∴ G(−1)=8 (((−3)),1,(−2),(−5),( 6)),(,,(−3),( 15),(−30)),(,1,(−5),( 10),(−24)) ) ^• ∴G(−3)=−24 ^• G(0)=(0)^3 −2(0)^2 −5(0)+6=6 ^• G(x)=x^3 −2x^2 −5x+6=(x+2)(x^2 −4x+3) All conditions satisfied. ^• Also by dividing x^3 −2x^2 −5x+6 by (x+1)(x+3)=x^2 +4x+3 you can see that the remainder is 16x+24 This attests (a) also.
$${b} \\ $$$${G}\left(−\mathrm{1}\right)=\mathrm{8} \\ $$$${G}\left(−\mathrm{3}\right)=−\mathrm{24} \\ $$$${G}\left(−\mathrm{2}\right)=\mathrm{0} \\ $$$${G}\left(\mathrm{0}\right)=\mathrm{6} \\ $$$${G}\left({x}\right)\:{at}\:{least}\:{quadratic}. \\ $$$${G}\left({x}\right)=? \\ $$$$\bullet{Let}\:{G}\left({x}\right)=\left({x}+\mathrm{2}\right)\left({ax}+{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{2}} +\left({b}+\mathrm{2}{a}\right){x}+\mathrm{2}{b} \\ $$$${G}\left(\mathrm{0}\right)=\mathrm{6}\Rightarrow\mathrm{2}{b}=\mathrm{6}\Rightarrow{b}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:{G}\left({x}\right)={ax}^{\mathrm{2}} +\left(\mathrm{2}{a}+\mathrm{3}\right){x}+\mathrm{6} \\ $$$$\begin{pmatrix}{\left.−\mathrm{1}\right)}&{{a}}&{\mathrm{2}{a}+\mathrm{3}}&{\:\:\:\:\:\:\:\:\:\mathrm{6}}\\{}&{}&{−{a}}&{\:\:\:\:−{a}−\mathrm{3}}\\{}&{{a}}&{{a}+\mathrm{3}}&{−{a}+\mathrm{3}=\mathrm{8}}\end{pmatrix}\: \\ $$$$−{a}+\mathrm{3}=\mathrm{8}\Rightarrow{a}=−\mathrm{5} \\ $$$${G}\left({x}\right)=−\mathrm{5}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{6} \\ $$$$ \\ $$$$\begin{pmatrix}{\left.−\mathrm{3}\right)}&{−\mathrm{5}}&{−\mathrm{7}}&{\:\:\:\:\:\:\:\:\:\mathrm{6}}\\{}&{}&{\mathrm{15}}&{\:\:\:\:−\mathrm{24}}\\{}&{−\mathrm{5}}&{\:\:\mathrm{8}}&{\:\:\:\:\:\:\:−\mathrm{18}}\end{pmatrix}\: \\ $$$${G}\left(−\mathrm{3}\right)=−\mathrm{18} \\ $$$${This}\:{is}\:{against}\:{to}\:{the}\:{given}\left({G}\left(−\mathrm{3}\right)=−\mathrm{24}\right. \\ $$$$\:\therefore{G}\left({x}\right)\neq\left({x}+\mathrm{2}\right)\left({ax}+{b}\right) \\ $$$$ \\ $$$$\bullet{Let}\:{G}\left({x}\right)=\left({x}+\mathrm{2}\right)\left({ax}^{\mathrm{2}} +{bx}+{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+\mathrm{2}{ax}^{\mathrm{2}} +\mathrm{2}{bx}+\mathrm{2}{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} +\mathrm{2}{ax}^{\mathrm{2}} +{bx}^{\mathrm{2}} +\mathrm{2}{bx}+{cx}+\mathrm{2}{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{3}} +\left(\mathrm{2}{a}+{b}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{b}+{c}\right){x}+\mathrm{2}{c} \\ $$$${G}\left(\mathrm{0}\right)=\mathrm{6}\Rightarrow\mathrm{2}{c}=\mathrm{6}\Rightarrow{c}=\mathrm{3} \\ $$$$\:\:{G}\left({x}\right)={ax}^{\mathrm{3}} +\left(\mathrm{2}{a}+{b}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{b}+\mathrm{3}\right){x}+\mathrm{6} \\ $$$$\begin{pmatrix}{\left.−\mathrm{1}\right)}&{{a}}&{\mathrm{2}{a}+{b}}&{\:\:\:\:\:\mathrm{2}{b}+\mathrm{3}}&{\:\:\:\:\:\mathrm{6}}\\{}&{}&{−{a}}&{\:\:\:−{a}−{b}}&{{a}−{b}−\mathrm{3}}\\{}&{{a}}&{{a}+{b}}&{−{a}+{b}+\mathrm{3}}&{{a}−{b}+\mathrm{3}=\mathrm{8}}\end{pmatrix}\: \\ $$$${a}−{b}=\mathrm{5}\Rightarrow{a}={b}+\mathrm{5} \\ $$$$\begin{pmatrix}{\left.−\mathrm{3}\right)}&{{a}}&{\mathrm{2}{a}+{b}}&{\:\:\:\:\:\mathrm{2}{b}+\mathrm{3}}&{\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}}\\{}&{}&{−\mathrm{3}{a}}&{\:\:\:\mathrm{3}{a}−\mathrm{3}{b}}&{\:\:\:\:\:−\mathrm{9}{a}+\mathrm{3}{b}−\mathrm{9}}\\{}&{{a}}&{−{a}+{b}}&{\mathrm{3}{a}−{b}+\mathrm{3}}&{−\mathrm{9}{a}+\mathrm{3}{b}−\mathrm{3}=−\mathrm{24}}\end{pmatrix}\: \\ $$$$\:\:\:\:\:−\mathrm{9}{a}+\mathrm{3}{b}=−\mathrm{21} \\ $$$$\:\:\:\:−\mathrm{3}{a}+{b}=−\mathrm{7} \\ $$$$\:\:\:\:\:\:\:−\mathrm{3}\left({b}+\mathrm{5}\right)+{b}=−\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….. \\ $$$$\:\:\:\:\:\:\:−\mathrm{2}{b}=−\mathrm{7}+\mathrm{15}=\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….. \\ $$$$\:\:\:\:\:\:\:{b}=−\mathrm{4},{a}=\mathrm{1} \\ $$$${G}\left({x}\right)=\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\therefore\:\:{G}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6} \\ $$$${Verification}: \\ $$$$\begin{pmatrix}{\left.−\mathrm{1}\right)}&{\mathrm{1}}&{−\mathrm{2}}&{−\mathrm{5}}&{\mathrm{6}}\\{}&{}&{−\mathrm{1}}&{\:\:\:\mathrm{3}}&{\mathrm{2}}\\{}&{\mathrm{1}}&{−\mathrm{3}}&{\:−\mathrm{2}}&{\mathrm{8}}\end{pmatrix}\: \\ $$$$\:^{\bullet} \therefore\:\:{G}\left(−\mathrm{1}\right)=\mathrm{8} \\ $$$$\begin{pmatrix}{\left.−\mathrm{3}\right)}&{\mathrm{1}}&{−\mathrm{2}}&{−\mathrm{5}}&{\:\:\:\mathrm{6}}\\{}&{}&{−\mathrm{3}}&{\:\:\:\mathrm{15}}&{−\mathrm{30}}\\{}&{\mathrm{1}}&{−\mathrm{5}}&{\:\:\:\mathrm{10}}&{−\mathrm{24}}\end{pmatrix}\: \\ $$$$\:^{\bullet} \therefore{G}\left(−\mathrm{3}\right)=−\mathrm{24} \\ $$$$\:^{\bullet} {G}\left(\mathrm{0}\right)=\left(\mathrm{0}\right)^{\mathrm{3}} −\mathrm{2}\left(\mathrm{0}\right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{0}\right)+\mathrm{6}=\mathrm{6} \\ $$$$\:^{\bullet} \:{G}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right) \\ $$$${All}\:{conditions}\:{satisfied}. \\ $$$$\:^{\bullet} {Also}\:{by}\:{dividing}\:{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}\:{by} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}\:{you}\:{can}\:{see} \\ $$$${that}\:{the}\:{remainder}\:{is}\:\mathrm{16}{x}+\mathrm{24} \\ $$$${This}\:{attests}\:\left({a}\right)\:{also}. \\ $$
Commented by mr W last updated on 27/Aug/19
i have doubt that the solution is unique. we know only that Q(0)=−6 and Q(−2)=−8. one solution is Q(x)=x−6 which means G(x)=(x+1)(x+3)(x−6)+16x+24 or G(x)=(x+2)(x−1)(x−3)⇒(x+2)(x^2 −4x+3). but why not Q(x)=(x^2 /2)+2x−6 which means G(x)=(1/2)(x+2)(x^3 +6x^2 −5x+6)?
$${i}\:{have}\:{doubt}\:{that}\:{the}\:{solution}\:{is} \\ $$$${unique}.\:{we}\:{know}\:{only}\:{that} \\ $$$${Q}\left(\mathrm{0}\right)=−\mathrm{6}\:{and}\:{Q}\left(−\mathrm{2}\right)=−\mathrm{8}. \\ $$$${one}\:{solution}\:{is}\:{Q}\left({x}\right)={x}−\mathrm{6}\:{which} \\ $$$${means}\:{G}\left({x}\right)=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{6}\right)+\mathrm{16}{x}+\mathrm{24} \\ $$$${or}\:{G}\left({x}\right)=\left({x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\Rightarrow\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\right). \\ $$$${but}\:{why}\:{not}\:{Q}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{x}−\mathrm{6}\:{which} \\ $$$${means}\:{G}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}\right)? \\ $$
Commented by Rasheed.Sindhi last updated on 28/Aug/19
I agree with you sir! This is only ′unique cubic solution′. I wanted to solve the problem in a tricky way but I failed.This is a lengthy and straivhtforward way. But you know something is better than nothing. I′ll be pleased if some one add a more compact and tricky way! Thanks for your precious comment Sir!
$${I}\:{agree}\:{with}\:{you}\:{sir}!\:{This}\:{is}\:{only} \\ $$$$'{unique}\:{cubic}\:{solution}'. \\ $$$${I}\:{wanted}\:{to}\:{solve}\:{the}\:{problem}\:{in}\:{a} \\ $$$${tricky}\:{way}\:{but}\:{I}\:{failed}.{This}\:{is}\:{a} \\ $$$${lengthy}\:{and}\:{straivhtforward}\:{way}. \\ $$$${But}\:{you}\:{know}\:{something}\:{is}\:{better}\:{than} \\ $$$${nothing}. \\ $$$${I}'{ll}\:{be}\:{pleased}\:{if}\:{some}\:{one}\:{add}\:{a}\:{more} \\ $$$${compact}\:{and}\:{tricky}\:{way}! \\ $$$$\boldsymbol{{Thanks}}\:{for}\:{your}\:{precious}\:{comment} \\ $$$$\boldsymbol{{Sir}}! \\ $$
Commented by Rasheed.Sindhi last updated on 28/Aug/19
Sir your approach:to determine Q(x) when Q(0)=−6 & Q(−2)=−8 (Something like functional equation) might be better but again this is a guessing approach,I think.
$${Sir}\:{your}\:{approach}:{to}\:{determine}\:{Q}\left({x}\right) \\ $$$${when}\:{Q}\left(\mathrm{0}\right)=−\mathrm{6}\:\&\:{Q}\left(−\mathrm{2}\right)=−\mathrm{8}\: \\ $$$$\left({Something}\:{like}\:{functional}\:{equation}\right) \\ $$$${might}\:{be}\:\:{better}\:{but}\:{again}\:{this}\:{is}\:\:{a}\:{guessing} \\ $$$${approach},{I}\:{think}. \\ $$
Commented by mr W last updated on 28/Aug/19
since G(x)=(x+1)(x+3)Q(x)+16x+24, we only need to determine Q(x). from G(−2)=0 and G(0)=6 we get Q(−2)=−8 and Q(0)=−6. that means y=Q(x) must pass through (0,−6) and (−2,−8). but there are infinite polynomial functions which satisfy this condiction, as linear function: the only one is Q(x)=x−6 as quadratic function: infinite, e.g. Q(x)=(x^2 /2)+2x−6 ...... that is why i think the solution is not unique.
$${since}\:{G}\left({x}\right)=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right){Q}\left({x}\right)+\mathrm{16}{x}+\mathrm{24}, \\ $$$${we}\:{only}\:{need}\:{to}\:{determine}\:{Q}\left({x}\right). \\ $$$$ \\ $$$${from}\:{G}\left(−\mathrm{2}\right)=\mathrm{0}\:{and}\:{G}\left(\mathrm{0}\right)=\mathrm{6}\:{we} \\ $$$${get}\:{Q}\left(−\mathrm{2}\right)=−\mathrm{8}\:{and}\:{Q}\left(\mathrm{0}\right)=−\mathrm{6}.\:{that} \\ $$$${means}\:{y}={Q}\left({x}\right)\:{must}\:{pass}\:{through} \\ $$$$\left(\mathrm{0},−\mathrm{6}\right)\:{and}\:\left(−\mathrm{2},−\mathrm{8}\right).\:{but}\:{there}\:{are} \\ $$$${infinite}\:{polynomial}\:{functions}\:{which} \\ $$$${satisfy}\:{this}\:{condiction}, \\ $$$${as}\:{linear}\:{function}:\: \\ $$$${the}\:{only}\:{one}\:{is}\:{Q}\left({x}\right)={x}−\mathrm{6} \\ $$$${as}\:{quadratic}\:{function}:\: \\ $$$${infinite},\:{e}.{g}.\:{Q}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{x}−\mathrm{6} \\ $$$$…… \\ $$$${that}\:{is}\:{why}\:{i}\:{think}\:{the}\:{solution}\:{is} \\ $$$${not}\:{unique}. \\ $$
Commented by mr W last updated on 28/Aug/19
Rasheed sir, i appreciate very much that you always try to go deeply behind the question, not only just to give a solution.
$${Rasheed}\:{sir},\:{i}\:{appreciate}\:{very}\:{much} \\ $$$${that}\:{you}\:{always}\:{try}\:{to}\:{go}\:{deeply}\:{behind} \\ $$$${the}\:{question},\:{not}\:{only}\:{just}\:{to}\:{give}\:{a} \\ $$$${solution}. \\ $$
Commented by Rasheed.Sindhi last updated on 28/Aug/19
θαnkSs SI^(⧫) R! I always received encourage from you!
$$\theta\alpha{nk}\mathcal{S}{s}\:\mathbb{S}\overset{\blacklozenge} {\mathbb{I}R}!\:{I}\:{always}\:{received}\:{encourage} \\ $$$${from}\:{you}! \\ $$
Commented by Rio Michael last updated on 29/Aug/19
thank you all
$${thank}\:{you}\:{all} \\ $$

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