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Question-98796

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Question Number 98796 by Lekhraj last updated on 16/Jun/20
Answered by MJS last updated on 16/Jun/20
−((267)/2)
$$−\frac{\mathrm{267}}{\mathrm{2}} \\ $$
Commented by Lekhraj last updated on 16/Jun/20
How?
$$\mathrm{How}? \\ $$
Commented by MJS last updated on 16/Jun/20
p(x)=c_3 x^3 +c_2 x^2 +c_1 x+c_0 insert the given values and solve the linear system for c_j
$${p}\left({x}\right)={c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} \\ $$$$\mathrm{insert}\:\mathrm{the}\:\mathrm{given}\:\mathrm{values}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{linear} \\ $$$$\mathrm{system}\:\mathrm{for}\:{c}_{{j}} \\ $$
Commented by Lekhraj last updated on 16/Jun/20
Calculatins are very lengthy . Is there any short methld or else you may please show the whole solution. Thanks
$$\mathrm{Calculatins}\:\mathrm{are}\:\mathrm{very}\:\mathrm{lengthy}\:.\: \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{short}\:\mathrm{methld}\:\mathrm{or}\:\mathrm{else} \\ $$$$\mathrm{you}\:\mathrm{may}\:\mathrm{please}\:\mathrm{show}\:\mathrm{the}\:\mathrm{whole} \\ $$$$\mathrm{solution}. \\ $$$$\mathrm{Thanks} \\ $$$$ \\ $$
Answered by mr W last updated on 17/Jun/20
alternative method: p(1)=p((1/4))=3 ⇒p(x)=a(x)(x−1)(x−(1/4))+3 p((1/3))=a((1/3))((1/3)−1)((1/3)−(1/4))+3=4 ⇒a((1/3))=−18 ⇒a(x)=b(x−(1/3))−18 ⇒p(x)=(b(x−(1/3))−18)(x−1)(x−(1/4))+3 p((1/2))=(b((1/2)−(1/3))−18)((1/2)−1)((1/2)−(1/4))+3=6 ⇒b=−36 ⇒p(x)=(−36(x−(1/3))−18)(x−1)(x−(1/4))+3 ⇒p(x)=−18(2x+(1/3))(x−1)(x−(1/4))+3 p(2)=−18(2×2+(1/3))(2−1)(2−(1/4))+3 =−3×13×(7/2)+3 =−((267)/2)
$${alternative}\:{method}: \\ $$$${p}\left(\mathrm{1}\right)={p}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{3} \\ $$$$\Rightarrow{p}\left({x}\right)={a}\left({x}\right)\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3} \\ $$$${p}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)={a}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3}=\mathrm{4} \\ $$$$\Rightarrow{a}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=−\mathrm{18} \\ $$$$\Rightarrow{a}\left({x}\right)={b}\left({x}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{18} \\ $$$$\Rightarrow{p}\left({x}\right)=\left({b}\left({x}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{18}\right)\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3} \\ $$$${p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\left({b}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{18}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3}=\mathrm{6} \\ $$$$\Rightarrow{b}=−\mathrm{36} \\ $$$$\Rightarrow{p}\left({x}\right)=\left(−\mathrm{36}\left({x}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{18}\right)\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3} \\ $$$$\Rightarrow{p}\left({x}\right)=−\mathrm{18}\left(\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({x}−\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3} \\ $$$$ \\ $$$${p}\left(\mathrm{2}\right)=−\mathrm{18}\left(\mathrm{2}×\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3} \\ $$$$=−\mathrm{3}×\mathrm{13}×\frac{\mathrm{7}}{\mathrm{2}}+\mathrm{3} \\ $$$$=−\frac{\mathrm{267}}{\mathrm{2}} \\ $$
Commented by Lekhraj last updated on 17/Jun/20
Thanks .
$$\mathrm{Thanks}\:.\: \\ $$

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