Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

$${thank}\:{you}\:{all} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com