Menu Close

Find-the-sum-of-the-fifth-powers-of-the-roots-of-x-3-2x-2-x-1-0-by-applying-synthetic-division-

Tinku Tara 0 Comments
Pin




Question Number 200465 by faysal last updated on 19/Nov/23
Find the sum of the fifth powers of the roots of x^3 −2x^2 +x−1=0 by applying synthetic division
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fifth}\:\mathrm{powers}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}=\mathrm{0}\:\mathrm{by} \\ $$$$\mathrm{applying}\:\mathrm{synthetic}\:\mathrm{division} \\ $$
Commented by mr W last updated on 19/Nov/23
you posted your first question on 15/09/2020. since then you have never given even a single one feedback. even when people are talking to a stone wall, it gives an echo back...
$${you}\:{posted}\:{your}\:{first}\:{question}\:{on} \\ $$$$\mathrm{15}/\mathrm{09}/\mathrm{2020}.\:{since}\:{then}\:{you}\:{have}\:{never} \\ $$$${given}\:{even}\:{a}\:{single}\:{one}\:{feedback}. \\ $$$${even}\:{when}\:{people}\:{are}\:{talking}\:{to}\:{a} \\ $$$${stone}\:{wall},\:{it}\:{gives}\:{an}\:{echo}\:{back}… \\ $$
Answered by mr W last updated on 19/Nov/23
say the roots are a,b,c. a+b+c=2 ab+bc+ca=1 abc=1 (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca) ⇒a^2 +b^2 +c^2 =2^2 −2×1=2 (ab+bc+ca)^2 =a^2 b^2 +b^2 c^2 +c^2 a^2 +2abc(a+b+c) ⇒a^2 b^2 +b^2 c^2 +c^2 a^2 =1^2 −2×1×2=−3 (a+b+c)^3 =a^3 +b^3 +c^3 −3abc+3(a+b+c)(ab+bc+ca) ⇒a^3 +b^3 +c^3 =2^3 +3×1−3×2×1=5 (a^2 +b^2 +c^2 )(a^3 +b^3 +c^3 )=a^5 +b^5 +c^5 +(a+b+c)(a^2 b^2 +b^2 c^2 +c^2 a^2 )−abc(ab+bc+ca) ⇒a^5 +b^5 +c^5 =2×5−2×(−3)+1×1=17 ✓
$${say}\:{the}\:{roots}\:{are}\:{a},{b},{c}. \\ $$$${a}+{b}+{c}=\mathrm{2} \\ $$$${ab}+{bc}+{ca}=\mathrm{1} \\ $$$${abc}=\mathrm{1} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$$$ \\ $$$$\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{2}{abc}\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\mathrm{2}=−\mathrm{3} \\ $$$$ \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}+\mathrm{3}\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{2}^{\mathrm{3}} +\mathrm{3}×\mathrm{1}−\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{5} \\ $$$$ \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)={a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−{abc}\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} =\mathrm{2}×\mathrm{5}−\mathrm{2}×\left(−\mathrm{3}\right)+\mathrm{1}×\mathrm{1}=\mathrm{17}\:\checkmark \\ $$
Answered by ajfour last updated on 19/Nov/23
let x^5 =t ⇒ x^2 (2x^2 −x+1)=t 2x(2x^2 −x+1)−(2x^2 −x+1)+x^2 =t 4(2x^2 −x+1)−3x^2 +3x−1=t 5x^2 −x+3=t 5(2x^2 −x+1)−x^2 =(t−3)x 9(x+t−3)+25=5(t+2)x (5t+1)x=9t−2 x=((9t−2)/(5t+1)) (9t−2)^3 −2(5t+1)(9t−2)^2 +(5t+1)^2 (9t−2)−(5t+1)^3 =0 Σ_(i=1) ^3 x_i ^5 =Σ_(i=1) ^3 t_i =−((coeff of t^2 )/(coeff of t^3 )) =−(((−6×81+360−216−50+90−75))/((729−810+225−125))) =−(((−323))/(19)) = 17
$${let}\:\:\:{x}^{\mathrm{5}} ={t} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)={t} \\ $$$$\mathrm{2}{x}\left(\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)−\left(\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+{x}^{\mathrm{2}} ={t} \\ $$$$\mathrm{4}\left(\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}={t} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{3}={t}\:\:\: \\ $$$$\mathrm{5}\left(\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)−{x}^{\mathrm{2}} =\left({t}−\mathrm{3}\right){x} \\ $$$$\mathrm{9}\left({x}+{t}−\mathrm{3}\right)+\mathrm{25}=\mathrm{5}\left({t}+\mathrm{2}\right){x} \\ $$$$\left(\mathrm{5}{t}+\mathrm{1}\right){x}=\mathrm{9}{t}−\mathrm{2} \\ $$$${x}=\frac{\mathrm{9}{t}−\mathrm{2}}{\mathrm{5}{t}+\mathrm{1}} \\ $$$$\left(\mathrm{9}{t}−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{2}\left(\mathrm{5}{t}+\mathrm{1}\right)\left(\mathrm{9}{t}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:\:\:+\left(\mathrm{5}{t}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{9}{t}−\mathrm{2}\right)−\left(\mathrm{5}{t}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{x}_{{i}} ^{\mathrm{5}} =\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{t}_{{i}} =−\frac{{coeff}\:{of}\:{t}^{\mathrm{2}} }{{coeff}\:{of}\:{t}^{\mathrm{3}} } \\ $$$$=−\frac{\left(−\mathrm{6}×\mathrm{81}+\mathrm{360}−\mathrm{216}−\mathrm{50}+\mathrm{90}−\mathrm{75}\right)}{\left(\mathrm{729}−\mathrm{810}+\mathrm{225}−\mathrm{125}\right)} \\ $$$$=−\frac{\left(−\mathrm{323}\right)}{\mathrm{19}}\:=\:\mathrm{17} \\ $$
Commented by mr W last updated on 20/Nov/23
great!
$${great}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *