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Question Number 119600 by bemath last updated on 25/Oct/20
Find gcd of x^4 +x^3 −4x^2 +x+5 and x^3 +x^2 −9x−9
$${Find}\:{gcd}\:{of}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{5}\: \\ $$$${and}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9} \\ $$
Answered by TANMAY PANACEA last updated on 25/Oct/20
x^3 +x^2 −9x−9 =x^2 (x+1)−9(x+1) =(x+1)(x+3)(x−3) x^4 +x^3 −4x^2 +x+5 =x^4 +x^3 −4x^2 −4x+5x+5 =x^3 (x+1) −4x(x+1) +5 (x+1) =(x+1)(x^3 −4x+5) g.c.d=(x+1)
$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9} \\ $$$$={x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)−\mathrm{9}\left({x}+\mathrm{1}\right) \\ $$$$=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{3}\right) \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{5} \\ $$$$={x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}{x}+\mathrm{5} \\ $$$$={x}^{\mathrm{3}} \left({x}+\mathrm{1}\right)\:\:−\mathrm{4}{x}\left({x}+\mathrm{1}\right)\:+\mathrm{5}\:\left({x}+\mathrm{1}\right)\:\: \\ $$$$=\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{3}} −\mathrm{4}{x}+\mathrm{5}\right) \\ $$$${g}.{c}.{d}=\left({x}+\mathrm{1}\right) \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
to Tinku tara why i can not share question
$${to}\:{Tinku}\:{tara}\:{why}\:{i}\:{can}\:{not}\:{share}\:{question} \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
Commented by TANMAY PANACEA last updated on 25/Oct/20
no share button
$${no}\:{share}\:{button} \\ $$
Commented by bemath last updated on 25/Oct/20
click three dots sir
$${click}\:{three}\:{dots}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
ok sir
$${ok}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
∫_0 ^(2π) (dx/(2cos^2 x+(√3) sin^2 x)) i can not post by + so in comment
$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}+\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} {x}}\:\:{i}\:{can}\:{not}\:{post}\:{by}\:+\:\:{so}\:{in}\:{comment} \\ $$
Commented by mindispower last updated on 25/Oct/20
=2∫_0 ^π (dx/(2cos^2 (x)+(√3)sin^2 (x))) =2∫_0 ^(π/2) (dx/(2cos^2 (x)+(√3)sin^2 (x)))+2∫_0 ^(π/2) (dx/(2sin^2 (x)+(√3)cos^2 (x))) ∫_0 ^(π/2) (dx/(asin^2 (x)+bcos^2 (x))),a,b>0 =∫_0 ^(π/2) (1/(bcos^2 (x)))(dx/((1+((√(a/b))tg(x))^2 )) =(1/( (√(ab))))∫_0 ^(π/2) ((d(((√a)/( (√b)))tg(x)))/(1+(((√a)/( (√b)))tg(x))^2 )) =(1/( (√(ab))))[tan^(−1) ((√(a/b))tg(x))]_0 ^(π/2) =(π/(2(√(ab)))) we find 2.(π/(2(√(2(√3)))))+((2.π)/(2(√((√3).2))))=((2π)/( (√(2(√3)))))
$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)+\sqrt{\mathrm{3}}{sin}^{\mathrm{2}} \left({x}\right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)+\sqrt{\mathrm{3}}{sin}^{\mathrm{2}} \left({x}\right)}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)+\sqrt{\mathrm{3}}{cos}^{\mathrm{2}} \left({x}\right)} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{asin}^{\mathrm{2}} \left({x}\right)+{bcos}^{\mathrm{2}} \left({x}\right)},{a},{b}>\mathrm{0} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{{bcos}^{\mathrm{2}} \left({x}\right)}\frac{{dx}}{\left(\mathrm{1}+\left(\sqrt{\frac{{a}}{{b}}}{tg}\left({x}\right)\right)^{\mathrm{2}} \right.} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{ab}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\frac{\sqrt{{a}}}{\:\sqrt{{b}}}{tg}\left({x}\right)\right)}{\mathrm{1}+\left(\frac{\sqrt{{a}}}{\:\sqrt{{b}}}{tg}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{ab}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{{a}}{{b}}}{tg}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}\sqrt{{ab}}} \\ $$$${we}\:{find} \\ $$$$\mathrm{2}.\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}+\frac{\mathrm{2}.\pi}{\mathrm{2}\sqrt{\sqrt{\mathrm{3}}.\mathrm{2}}}=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}} \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mindispower last updated on 25/Oct/20
withe pleasur sir i hop you doing well i dont know why but im thinking that is lectur withe residus Theorem ?
$${withe}\:{pleasur}\:{sir}\:{i}\:{hop}\:{you}\:{doing}\:{well}\:{i}\:{dont}\:{know}\:{why}\:{but} \\ $$$${im}\:{thinking}\:{that}\:{is}\:{lectur}\:{withe}\:\:{residus}\:{Theorem}\:? \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
i have shared this question ...other question seem similar in type
$${i}\:{have}\:{shared}\:{this}\:{question}\:…{other}\:{question}\: \\ $$$${seem}\:{similar}\:{in}\:{type} \\ $$
Commented by Tinku Tara last updated on 26/Oct/20
If you want to always start in forum by default. Change setting and enable start i Q&A forum and save forum preferences
$$\mathrm{If}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{always}\:\mathrm{start}\:\mathrm{in}\:\mathrm{forum} \\ $$$$\mathrm{by}\:\mathrm{default}.\:\mathrm{Change}\:\mathrm{setting}\:\mathrm{and} \\ $$$$\mathrm{enable}\:\mathrm{start}\:\mathrm{i}\:\mathrm{Q\&A}\:\mathrm{forum}\:\mathrm{and}\:\mathrm{save} \\ $$$$\mathrm{forum}\:\mathrm{preferences} \\ $$
Commented by Bird last updated on 25/Oct/20
A=∫_0 ^(2π) (dx/(2cos^2 x+(√3)sin^2 x)) ⇒ A=∫_0 ^(2π) (dx/(2((1+cos(2x))/2)+(√3)((1−cos(2x))/2))) =∫_0 ^(2π) ((2dx)/(2+2cos(2x)+(√3)−(√3)cos(2x))) =∫_0 ^(2π) ((2dx)/((2−(√3))cos(2x)+2+(√3))) =_(2x=t) ∫_0 ^(4π) (dt/((2−(√3))cost +2+(√3))) =∫_0 ^(2π) (dt/((2−(√3))cost+2+(√3))) +∫_(2π) ^(4π) (dt/((2−(√3))cost +2+(√3)))(→t=2π +u) =2∫_0 ^(2π) (dt/((2−(√3))cost+2+(√3))) =_(e^(it) =z) 2∫_(∣z∣=1) (dz/(iz{(2−(√3))((z+z^(−1) )/2)+2+(√3)})) =2∫_(∣z∣=1) ((2dz)/(iz{(2−(√3))(z+z^(−1) )+4+2(√3)))) =4∫_(∣z∣=1) ((−idz)/((2−(√3))(z^2 +1)+(4+2(√3))z)) =4∫_(∣z∣=1) ((−idz)/((2−(√3))z^2 +(4+2(√3))z +2−(√3))) ϕ(z) =((−i)/((2−(√3))z^2 +(4+2(√3))z +2−(√3))) Δ^′ =(2+(√3))^2 −(2−(√3))^2 =4+4(√3)+3−4+4(√3)−3 =8(√3) z_1 =((−2−(√3)+2(√(2(√3))))/((2−(√3)))) z_2 =((−2−(√3)+2(√(2(√3))))/((2−(√3)))) ∣z_1 ∣<1 and ∣z_2 ∣>1 ϕ(z)=((−i)/((2−(√3))(z−z_1 )(z−z_2 ))) ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) =2iπ×((−i)/((2−(√3))4(√(2(√3)))))×(2−(√3)) =((2π)/(4(√(2(√3))))) =(π/(2(√(2(√3))))) ⇒ A=((4π)/(2(√(2(√3))))) ⇒A =((2π)/( (√(2(√3)))))
$${A}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}+\sqrt{\mathrm{3}}{sin}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\mathrm{2}\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}+\sqrt{\mathrm{3}}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{2}{dx}}{\mathrm{2}+\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cos}\left(\mathrm{2}{x}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{2}{dx}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cos}\left(\mathrm{2}{x}\right)+\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\frac{{dt}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}\:+\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}+\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\frac{{dt}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}\:+\mathrm{2}+\sqrt{\mathrm{3}}}\left(\rightarrow{t}=\mathrm{2}\pi\:+{u}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}+\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$=_{{e}^{{it}} ={z}} \:\:\:\mathrm{2}\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{{dz}}{{iz}\left\{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}+\mathrm{2}+\sqrt{\mathrm{3}}\right\}} \\ $$$$=\mathrm{2}\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dz}}{{iz}\left\{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({z}+{z}^{−\mathrm{1}} \right)+\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right)} \\ $$$$=\mathrm{4}\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−{idz}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)+\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right){z}} \\ $$$$=\mathrm{4}\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−{idz}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right){z}\:+\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\varphi\left({z}\right)\:=\frac{−{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right){z}\:+\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\Delta^{'} \:=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}+\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}−\mathrm{4}+\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}\:=\mathrm{8}\sqrt{\mathrm{3}} \\ $$$${z}_{\mathrm{1}} =\frac{−\mathrm{2}−\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)} \\ $$$${z}_{\mathrm{2}} =\frac{−\mathrm{2}−\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)} \\ $$$$\mid{z}_{\mathrm{1}} \mid<\mathrm{1}\:{and}\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\:\: \\ $$$$\varphi\left({z}\right)=\frac{−{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$$=\mathrm{2}{i}\pi×\frac{−{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\mathrm{4}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}×\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{4}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:\Rightarrow \\ $$$${A}=\frac{\mathrm{4}\pi}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:\Rightarrow{A}\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}} \\ $$
Commented by MJS_new last updated on 26/Oct/20
seems you are in editor mode go into menue (“≡”−symbol on left top) directly after starting the app and then choose forum
$$\mathrm{seems}\:\mathrm{you}\:\mathrm{are}\:\mathrm{in}\:\mathrm{editor}\:\mathrm{mode} \\ $$$$\mathrm{go}\:\mathrm{into}\:\mathrm{menue}\:\left(“\equiv''−\mathrm{symbol}\:\mathrm{on}\:\mathrm{left}\:\mathrm{top}\right) \\ $$$$\mathrm{directly}\:\mathrm{after}\:\mathrm{starting}\:\mathrm{the}\:\mathrm{app}\:\mathrm{and}\:\mathrm{then} \\ $$$$\mathrm{choose}\:{forum} \\ $$
Commented by TANMAY PANACEA last updated on 26/Oct/20
ok sir
$${ok}\:{sir} \\ $$
Commented by Tinku Tara last updated on 26/Oct/20
Answered by benjo_mathlover last updated on 25/Oct/20
using polynomial division we find that x^4 +x^3 −4x^2 +x+5=x(x^3 +x^2 −9x−9)+(5x^2 +10x+5) next we have to divide x^3 +x^2 −9x−9 by 5x^2 +10x+5 we find that x^3 +x^2 −9x−9=(5x^2 +10x+5)(((x−1)/5))+(−8x−8) Finally we divide 5x^2 +10x+5 by −8x−8 and we find that 5x^2 +10x+5=(−8x−8)[ −(5/( 8))(x+1) ] Thus gcd (x^4 +x^3 −4x^2 +x+5 , x^3 +x^2 −9x−9) = x+1
$${using}\:{polynomial}\:{division}\:{we}\:{find}\:{that}\: \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{5}={x}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9}\right)+\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{5}\right) \\ $$$${next}\:{we}\:{have}\:{to}\:{divide}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9}\:{by}\:\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{5} \\ $$$${we}\:{find}\:{that}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9}=\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{5}\right)\left(\frac{{x}−\mathrm{1}}{\mathrm{5}}\right)+\left(−\mathrm{8}{x}−\mathrm{8}\right) \\ $$$${Finally}\:{we}\:{divide}\:\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{5}\:{by}\:−\mathrm{8}{x}−\mathrm{8}\:{and}\:{we}\:{find}\:{that} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{5}=\left(−\mathrm{8}{x}−\mathrm{8}\right)\left[\:−\frac{\mathrm{5}}{\:\mathrm{8}}\left({x}+\mathrm{1}\right)\:\right] \\ $$$${Thus}\:{gcd}\:\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{5}\:,\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{9}\right)\:=\:{x}+\mathrm{1} \\ $$

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