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Question-208836

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Question Number 208836 by Adeyemi889 last updated on 24/Jun/24
Answered by Rasheed.Sindhi last updated on 25/Jun/24
Ax^3 −A+Bx^2 +Bx+B+(Cx+D)(x^2 −2x+1) =Ax^3 −A+Bx^2 +Bx+B+Cx^3 +Dx^2 −2Cx^2 −2Dx+Cx+D =(A+C)x^3 +(B+D−2C)x^2 +(B−2D+C)x+(−A+B+D) =x^3 +5x^2 +4x+5 Comparing coefficients: A+C=1...i B+D−2C=5...ii B+C−2D=4...iii −A+B+D=5...iv iv ⇒B+D=A+5 ii ⇒ A+5−2C=5⇒A=2C i ⇒3C=1⇒C=1/3⇒A=2/3 iii ⇒B+C−2D=4 ⇒B+1/3−2D=4 ⇒B−2D=4−1/3=11/3 B+D=2/3+5=17/3 −3D=−6/3⇒D=2/3 B+D=17/3⇒B=17/3−2/3=5 ⇒B=5
$${Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+\left({Cx}+{D}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$={Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+{Cx}^{\mathrm{3}} +{Dx}^{\mathrm{2}} −\mathrm{2}{Cx}^{\mathrm{2}} −\mathrm{2}{Dx}+{Cx}+{D} \\ $$$$=\left({A}+{C}\right){x}^{\mathrm{3}} +\left({B}+{D}−\mathrm{2}{C}\right){x}^{\mathrm{2}} +\left({B}−\mathrm{2}{D}+{C}\right){x}+\left(−{A}+{B}+{D}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5} \\ $$$${Comparing}\:{coefficients}: \\ $$$${A}+{C}=\mathrm{1}…{i} \\ $$$${B}+{D}−\mathrm{2}{C}=\mathrm{5}…{ii} \\ $$$${B}+{C}−\mathrm{2}{D}=\mathrm{4}…{iii} \\ $$$$−{A}+{B}+{D}=\mathrm{5}…{iv} \\ $$$${iv}\:\Rightarrow{B}+{D}={A}+\mathrm{5} \\ $$$${ii}\:\Rightarrow\:{A}+\mathrm{5}−\mathrm{2}{C}=\mathrm{5}\Rightarrow{A}=\mathrm{2}{C} \\ $$$$\:{i}\:\Rightarrow\mathrm{3}{C}=\mathrm{1}\Rightarrow{C}=\mathrm{1}/\mathrm{3}\Rightarrow{A}=\mathrm{2}/\mathrm{3} \\ $$$${iii}\:\Rightarrow{B}+{C}−\mathrm{2}{D}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{B}+\mathrm{1}/\mathrm{3}−\mathrm{2}{D}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{B}−\mathrm{2}{D}=\mathrm{4}−\mathrm{1}/\mathrm{3}=\mathrm{11}/\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{B}+{D}=\mathrm{2}/\mathrm{3}+\mathrm{5}=\mathrm{17}/\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3}{D}=−\mathrm{6}/\mathrm{3}\Rightarrow{D}=\mathrm{2}/\mathrm{3} \\ $$$${B}+{D}=\mathrm{17}/\mathrm{3}\Rightarrow{B}=\mathrm{17}/\mathrm{3}−\mathrm{2}/\mathrm{3}=\mathrm{5} \\ $$$$\Rightarrow{B}=\mathrm{5} \\ $$
Commented by Adeyemi889 last updated on 25/Jun/24
yes sir the last part of the question (Cx+B) is with (X−1)^2 i notice in ur solution sir u only wrote (Cx+B)
$${yes}\:{sir}\:{the}\:{last}\:{part}\:{of}\:{the}\:{question}\:\left({Cx}+{B}\right)\:{is}\:{with}\:\left({X}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${i}\:{notice}\:{in}\:{ur}\:{solution}\:{sir}\:{u}\:{only}\:{wrote}\:\left({Cx}+{B}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 25/Jun/24
Sorry sir, I didn′t notice (x−1)^2 which is on next line.I′ll go to correct it. Corrected
$${Sorry}\:{sir},\:{I}\:{didn}'{t}\:{notice}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:{which} \\ $$$${is}\:{on}\:{next}\:{line}.{I}'{ll}\:{go}\:{to}\:{correct}\:{it}. \\ $$$${Corrected} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Jun/24
Another way x^3 +5x^2 +4x+5 =A(x^3 −1)+B(x^2 +x+1)+(Cx+D)(x−1)^2 The above is an identity so it′s correct for all values of x. x=0: 5=−A+B+D....(i) x=1: 15=3B⇒B=5 x=2: 41=7A+7B+2C+D....(ii) x=−1: 5=−2A+B−4C+4D....(iii) (i)⇒−A+D=0⇒A=D...(iv) (ii)⇒7A+2C+D=6 ⇒8A+2C=6...(v) (iii)⇒−2A−4C+4D=0 ⇒ 2A−4C=0 ⇒A=2C (v)⇒8A+A=6⇒A=(2/3) (iv)⇒D=(2/3) ⇒C=(1/3)
$${Another}\:{way} \\ $$$${x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5} \\ $$$$={A}\left({x}^{\mathrm{3}} −\mathrm{1}\right)+{B}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\left({Cx}+{D}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathcal{T}{he}\:{above}\:{is}\:{an}\:{identity}\:{so}\:{it}'{s} \\ $$$${correct}\:{for}\:{all}\:{values}\:{of}\:{x}. \\ $$$${x}=\mathrm{0}: \\ $$$$\mathrm{5}=−{A}+{B}+{D}….\left({i}\right) \\ $$$${x}=\mathrm{1}: \\ $$$$\mathrm{15}=\mathrm{3}{B}\Rightarrow{B}=\mathrm{5} \\ $$$${x}=\mathrm{2}: \\ $$$$\mathrm{41}=\mathrm{7}{A}+\mathrm{7}{B}+\mathrm{2}{C}+{D}….\left({ii}\right) \\ $$$${x}=−\mathrm{1}: \\ $$$$\mathrm{5}=−\mathrm{2}{A}+{B}−\mathrm{4}{C}+\mathrm{4}{D}….\left({iii}\right) \\ $$$$\left({i}\right)\Rightarrow−{A}+{D}=\mathrm{0}\Rightarrow{A}={D}…\left({iv}\right) \\ $$$$\left({ii}\right)\Rightarrow\mathrm{7}{A}+\mathrm{2}{C}+{D}=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{8}{A}+\mathrm{2}{C}=\mathrm{6}…\left({v}\right) \\ $$$$\left({iii}\right)\Rightarrow−\mathrm{2}{A}−\mathrm{4}{C}+\mathrm{4}{D}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{2}{A}−\mathrm{4}{C}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{A}=\mathrm{2}{C} \\ $$$$\:\left({v}\right)\Rightarrow\mathrm{8}{A}+{A}=\mathrm{6}\Rightarrow{A}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left({iv}\right)\Rightarrow{D}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{C}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by Adeyemi889 last updated on 26/Jun/24
Thanks so much sir♥♥♥
$${Thanks}\:{so}\:{much}\:{sir}\heartsuit\heartsuit\heartsuit \\ $$

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