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Question-194853

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Question Number 194853 by horsebrand11 last updated on 17/Jul/23
Answered by cortano12 last updated on 17/Jul/23
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Commented by horsebrand11 last updated on 17/Jul/23
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Answered by Rasheed.Sindhi last updated on 17/Jul/23
Let P(x)=8x^3 +16x^2 +kx+l =(2x−3)(4x^2 +ax+b) 8x^3 +16x^2 +kx+l=8x^3 +(2a−12)x^2 +(2b−3a)x−3b Comparing coefficints: 2a−12=16⇒a=14 2b−3a=k⇒k=2b−42 −3b=l⇒b=−(l/3) k=2(−(l/3))−42 3k+2l=−126................(i) Similarly, let P(x)=8x^3 +16x^2 +kx+l=(2x+1)(4x^2 +cx+d)+12 8x^3 +16x^2 +kx+l =8x^3 +(2c+4)x^2 +(2d+c)x+d+12 Comparing coefficients: 2c+4=16⇒c=6 2d+c=k⇒2d+6=k d+12=l⇒d=l−12 2d+6=k⇒2(l−12)+6=k k−2l=−18.................(ii) (i)+(ii):4k=−144⇒k=−36 (i)⇒−36−2l=−18⇒l=−9
$${Let}\:{P}\left({x}\right)=\mathrm{8}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} +{kx}+{l} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}{x}−\mathrm{3}\right)\left(\mathrm{4}{x}^{\mathrm{2}} +{ax}+{b}\right) \\ $$$$\mathrm{8}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} +{kx}+{l}=\mathrm{8}{x}^{\mathrm{3}} +\left(\mathrm{2}{a}−\mathrm{12}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{b}−\mathrm{3}{a}\right){x}−\mathrm{3}{b} \\ $$$${Comparing}\:{coefficints}: \\ $$$$\mathrm{2}{a}−\mathrm{12}=\mathrm{16}\Rightarrow{a}=\mathrm{14} \\ $$$$\mathrm{2}{b}−\mathrm{3}{a}={k}\Rightarrow{k}=\mathrm{2}{b}−\mathrm{42} \\ $$$$−\mathrm{3}{b}={l}\Rightarrow{b}=−\frac{{l}}{\mathrm{3}} \\ $$$${k}=\mathrm{2}\left(−\frac{{l}}{\mathrm{3}}\right)−\mathrm{42} \\ $$$$\mathrm{3}{k}+\mathrm{2}{l}=−\mathrm{126}…………….\left({i}\right) \\ $$$$\:{Similarly}, \\ $$$${let}\:{P}\left({x}\right)=\mathrm{8}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} +{kx}+{l}=\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{4}{x}^{\mathrm{2}} +{cx}+{d}\right)+\mathrm{12} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} +{kx}+{l} \\ $$$$=\mathrm{8}{x}^{\mathrm{3}} +\left(\mathrm{2}{c}+\mathrm{4}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{d}+{c}\right){x}+{d}+\mathrm{12} \\ $$$${Comparing}\:{coefficients}: \\ $$$$\mathrm{2}{c}+\mathrm{4}=\mathrm{16}\Rightarrow{c}=\mathrm{6} \\ $$$$\mathrm{2}{d}+{c}={k}\Rightarrow\mathrm{2}{d}+\mathrm{6}={k} \\ $$$${d}+\mathrm{12}={l}\Rightarrow{d}={l}−\mathrm{12} \\ $$$$\mathrm{2}{d}+\mathrm{6}={k}\Rightarrow\mathrm{2}\left({l}−\mathrm{12}\right)+\mathrm{6}={k} \\ $$$${k}−\mathrm{2}{l}=−\mathrm{18}……………..\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right):\mathrm{4}{k}=−\mathrm{144}\Rightarrow{k}=−\mathrm{36} \\ $$$$\left({i}\right)\Rightarrow−\mathrm{36}−\mathrm{2}{l}=−\mathrm{18}\Rightarrow{l}=−\mathrm{9} \\ $$$$ \\ $$$$\:\:\:\:\:\: \\ $$
Commented by horsebrand11 last updated on 17/Jul/23
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Answered by Rasheed.Sindhi last updated on 17/Jul/23
By synthetic division determinant (((−(1/2)),8,(16),k,l),( , ,(−4),(−6),(−(k/2)+3)),( ,8,(12),(k−6),(l−(k/2)+3=12))) 2l−k=18........(i) determinant (((3/2),8,(16),k,l),( , ,(12),(42),(((3k)/2)+63)),( ,8,(28),(k+42),(l+((3k)/2)+63=0))) 2l+3k=−126........(ii) (ii)−(i):4k=−144⇒k=−36 2l−(−36)=18⇒l=−9
$$\underline{\mathcal{B}{y}\:{synthetic}\:{division}} \\ $$$$\begin{array}{|c|c|c|}{−\frac{\mathrm{1}}{\mathrm{2}}}&\hline{\mathrm{8}}&\hline{\mathrm{16}}&\hline{{k}}&\hline{{l}}\\{\:}&\hline{\:}&\hline{−\mathrm{4}}&\hline{−\mathrm{6}}&\hline{−\frac{{k}}{\mathrm{2}}+\mathrm{3}}\\{\:}&\hline{\mathrm{8}}&\hline{\mathrm{12}}&\hline{{k}−\mathrm{6}}&\hline{{l}−\frac{{k}}{\mathrm{2}}+\mathrm{3}=\mathrm{12}}\\\hline\end{array}\: \\ $$$$\mathrm{2}{l}−{k}=\mathrm{18}……..\left({i}\right) \\ $$$$\begin{array}{|c|c|c|}{\frac{\mathrm{3}}{\mathrm{2}}}&\hline{\mathrm{8}}&\hline{\mathrm{16}}&\hline{{k}}&\hline{{l}}\\{\:}&\hline{\:}&\hline{\mathrm{12}}&\hline{\mathrm{42}}&\hline{\frac{\mathrm{3}{k}}{\mathrm{2}}+\mathrm{63}}\\{\:}&\hline{\mathrm{8}}&\hline{\mathrm{28}}&\hline{{k}+\mathrm{42}}&\hline{{l}+\frac{\mathrm{3}{k}}{\mathrm{2}}+\mathrm{63}=\mathrm{0}}\\\hline\end{array} \\ $$$$\mathrm{2}{l}+\mathrm{3}{k}=−\mathrm{126}……..\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right):\mathrm{4}{k}=−\mathrm{144}\Rightarrow{k}=−\mathrm{36} \\ $$$$\mathrm{2}{l}−\left(−\mathrm{36}\right)=\mathrm{18}\Rightarrow{l}=−\mathrm{9} \\ $$

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