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Question Number 204249 by universe last updated on 10/Feb/24
Commented by Frix last updated on 10/Feb/24
P(x)=((5x^7 )/(16))−((21x^5 )/(16))+((35x^3 )/(16))−((35x)/(16)) Q(x)=((5x^3 )/(16))+((5x^2 )/4)+((29x)/(16))+1 R(x)=((5x^3 )/(16))−((5x^2 )/4)+((29x)/(16))−1 P(x)=ax^7 +bx^6 +cx^5 +dx^4 +ex^3 +fx^2 +gx+h (x−1)^4 ∣(P(x)+1)∧(x+1)^4 ∣(P(x)−1) ⇒ 35a+20b+10c+4d+e=0 84a+45b+20c+6d−f=0 70a+36b+15c+4d+g=0 20a+10b+4c+d−h=1 35a−20b+10c−4d+e=0 84a−45b+20c−6d+f=0 70a−36b+15c−4d+g=0 20a−10b+4c−d+h=1
$${P}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{7}} }{\mathrm{16}}−\frac{\mathrm{21}{x}^{\mathrm{5}} }{\mathrm{16}}+\frac{\mathrm{35}{x}^{\mathrm{3}} }{\mathrm{16}}−\frac{\mathrm{35}{x}}{\mathrm{16}} \\ $$$${Q}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{16}}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{29}{x}}{\mathrm{16}}+\mathrm{1} \\ $$$${R}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{16}}−\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{29}{x}}{\mathrm{16}}−\mathrm{1} \\ $$$$ \\ $$$${P}\left({x}\right)={ax}^{\mathrm{7}} +{bx}^{\mathrm{6}} +{cx}^{\mathrm{5}} +{dx}^{\mathrm{4}} +{ex}^{\mathrm{3}} +{fx}^{\mathrm{2}} +{gx}+{h} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{4}} \mid\left({P}\left({x}\right)+\mathrm{1}\right)\wedge\left({x}+\mathrm{1}\right)^{\mathrm{4}} \mid\left({P}\left({x}\right)−\mathrm{1}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{35}{a}+\mathrm{20}{b}+\mathrm{10}{c}+\mathrm{4}{d}+{e}=\mathrm{0} \\ $$$$\mathrm{84}{a}+\mathrm{45}{b}+\mathrm{20}{c}+\mathrm{6}{d}−{f}=\mathrm{0} \\ $$$$\mathrm{70}{a}+\mathrm{36}{b}+\mathrm{15}{c}+\mathrm{4}{d}+{g}=\mathrm{0} \\ $$$$\mathrm{20}{a}+\mathrm{10}{b}+\mathrm{4}{c}+{d}−{h}=\mathrm{1} \\ $$$$\mathrm{35}{a}−\mathrm{20}{b}+\mathrm{10}{c}−\mathrm{4}{d}+{e}=\mathrm{0} \\ $$$$\mathrm{84}{a}−\mathrm{45}{b}+\mathrm{20}{c}−\mathrm{6}{d}+{f}=\mathrm{0} \\ $$$$\mathrm{70}{a}−\mathrm{36}{b}+\mathrm{15}{c}−\mathrm{4}{d}+{g}=\mathrm{0} \\ $$$$\mathrm{20}{a}−\mathrm{10}{b}+\mathrm{4}{c}−{d}+{h}=\mathrm{1} \\ $$
Answered by witcher3 last updated on 10/Feb/24
⇒p′(x)=Q′(x)(x−1)^4 +4(x−1)^3 Q(x) =(x−1)^3 (R(x)) ⇒p′(1)=0⇒1 root of multiplicity b≥3 sam p′(−1)=0 root of multiplicity a≥3 sinc deg p′(x) =6⇒ card(a∈C∣p′(x)=0)=6 ⇒a=b=3 p′(x)=a(x−1)^3 (x+1)^3 =a(x^6 −3x^4 +3x^2 −1) p(x)=(a/7)x^7 −((3a)/5)x^5 +ax^3 −ax+b p(1)=−1,p(−1);find a and b
$$\Rightarrow\mathrm{p}'\left(\mathrm{x}\right)=\mathrm{Q}'\left(\mathrm{x}\right)\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{4}} +\mathrm{4}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} \mathrm{Q}\left(\mathrm{x}\right) \\ $$$$=\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{R}\left(\mathrm{x}\right)\right) \\ $$$$\Rightarrow\mathrm{p}'\left(\mathrm{1}\right)=\mathrm{0}\Rightarrow\mathrm{1}\:\mathrm{root}\:\mathrm{of}\:\mathrm{multiplicity}\:\:\mathrm{b}\geqslant\mathrm{3} \\ $$$$\mathrm{sam}\:\mathrm{p}'\left(−\mathrm{1}\right)=\mathrm{0}\:\mathrm{root}\:\mathrm{of}\:\mathrm{multiplicity}\:\mathrm{a}\geqslant\mathrm{3} \\ $$$$\mathrm{sinc}\:\:\mathrm{deg}\:\mathrm{p}'\left(\mathrm{x}\right)\:=\mathrm{6}\Rightarrow\:\mathrm{card}\left(\mathrm{a}\in\mathbb{C}\mid\mathrm{p}'\left(\mathrm{x}\right)=\mathrm{0}\right)=\mathrm{6} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{b}=\mathrm{3} \\ $$$$\mathrm{p}'\left(\mathrm{x}\right)=\mathrm{a}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{a}\left(\mathrm{x}^{\mathrm{6}} −\mathrm{3x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{7}}\mathrm{x}^{\mathrm{7}} −\frac{\mathrm{3a}}{\mathrm{5}}\mathrm{x}^{\mathrm{5}} +\mathrm{ax}^{\mathrm{3}} −\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{p}\left(\mathrm{1}\right)=−\mathrm{1},\mathrm{p}\left(−\mathrm{1}\right);\mathrm{find}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 10/Feb/24
P(x)=k(x^3 +ax^2 +bx+c)(x−1)^4 −1 Q(x)=k(x^3 +ux^2 +vx+w)(x+1)^4 +1 (x^3 +ax^2 +bx+c)(x−1)^4 =(x^3 +ux^2 +vx+w)(x+1)^4 +(2/k) (x^3 +ax^2 +bx+c)(x^4 −4x^3 +6x^2 −4x+1)=(x^3 +ux^2 +vx+w)(x^4 +4x^3 +6x^2 +4x+1)+(2/k) x^6 : −4+a=4+u ⇒ u=a−8 x^5 : 6−4a+b=6+4u+v ⇒ −4a+b=4u+v ⇒ v=−8a+b+32 x^4 : −4+6a−4b+c=4+6u+4v+w ⇒ 6a−4b+c=8+6u+4v+w ⇒ w=32a−8b+c−88 x^3 : 1−4a+6b−4c=1+4u+6v+4w ⇒ −2a+3b−2c=2u+3v+2w ⇒ 11a−4b+c=24 ...(i) x^2 : a−4b+6c=u+4v+6w ⇒ 20a−5b=51 ...(ii) x: b−4c=v+4w ⇒ 15a−4b+c=40 ...(iii) (iii)−(i): 4a=16 ⇒a=4 20×4−5b=51 ⇒b=((29)/5) 11×4−4×((29)/5)+c=24 ⇒c=((16)/5) const: c=w+(2/k) ⇒c=32a−8b+c−88+(2/k) ⇒k=(1/(44−16a+4b))=(1/(44−16×4+4×((29)/5)))=(5/(16)) ⇒P(x)=(1/(16))(5x^3 +20x^2 +29x+16)(x−1)^4 −1 or u=4−8=−4 v=−8×4+((29)/5)+32=((29)/5) w=32×4−8×((29)/5)+((16)/5)−88=−((16)/5) ⇒P(x)=(1/(16))(5x^3 −20x^2 +29x−16)(x+1)^4 +1
$${P}\left({x}\right)={k}\left({x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}\right)\left({x}−\mathrm{1}\right)^{\mathrm{4}} −\mathrm{1} \\ $$$${Q}\left({x}\right)={k}\left({x}^{\mathrm{3}} +{ux}^{\mathrm{2}} +{vx}+{w}\right)\left({x}+\mathrm{1}\right)^{\mathrm{4}} +\mathrm{1} \\ $$$$\left({x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}\right)\left({x}−\mathrm{1}\right)^{\mathrm{4}} =\left({x}^{\mathrm{3}} +{ux}^{\mathrm{2}} +{vx}+{w}\right)\left({x}+\mathrm{1}\right)^{\mathrm{4}} +\frac{\mathrm{2}}{{k}} \\ $$$$\left({x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}\right)\left({x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}\right)=\left({x}^{\mathrm{3}} +{ux}^{\mathrm{2}} +{vx}+{w}\right)\left({x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)+\frac{\mathrm{2}}{{k}} \\ $$$${x}^{\mathrm{6}} :\:−\mathrm{4}+{a}=\mathrm{4}+{u} \\ $$$$\Rightarrow\:{u}={a}−\mathrm{8} \\ $$$${x}^{\mathrm{5}} :\:\mathrm{6}−\mathrm{4}{a}+{b}=\mathrm{6}+\mathrm{4}{u}+{v} \\ $$$$\Rightarrow\:−\mathrm{4}{a}+{b}=\mathrm{4}{u}+{v} \\ $$$$\Rightarrow\:{v}=−\mathrm{8}{a}+{b}+\mathrm{32} \\ $$$${x}^{\mathrm{4}} :\:−\mathrm{4}+\mathrm{6}{a}−\mathrm{4}{b}+{c}=\mathrm{4}+\mathrm{6}{u}+\mathrm{4}{v}+{w} \\ $$$$\Rightarrow\:\mathrm{6}{a}−\mathrm{4}{b}+{c}=\mathrm{8}+\mathrm{6}{u}+\mathrm{4}{v}+{w} \\ $$$$\Rightarrow\:{w}=\mathrm{32}{a}−\mathrm{8}{b}+{c}−\mathrm{88} \\ $$$${x}^{\mathrm{3}} :\:\mathrm{1}−\mathrm{4}{a}+\mathrm{6}{b}−\mathrm{4}{c}=\mathrm{1}+\mathrm{4}{u}+\mathrm{6}{v}+\mathrm{4}{w} \\ $$$$\Rightarrow\:−\mathrm{2}{a}+\mathrm{3}{b}−\mathrm{2}{c}=\mathrm{2}{u}+\mathrm{3}{v}+\mathrm{2}{w} \\ $$$$\Rightarrow\:\mathrm{11}{a}−\mathrm{4}{b}+{c}=\mathrm{24}\:\:\:…\left({i}\right) \\ $$$${x}^{\mathrm{2}} :\:{a}−\mathrm{4}{b}+\mathrm{6}{c}={u}+\mathrm{4}{v}+\mathrm{6}{w} \\ $$$$\Rightarrow\:\mathrm{20}{a}−\mathrm{5}{b}=\mathrm{51}\:\:\:…\left({ii}\right) \\ $$$${x}:\:{b}−\mathrm{4}{c}={v}+\mathrm{4}{w} \\ $$$$\Rightarrow\:\mathrm{15}{a}−\mathrm{4}{b}+{c}=\mathrm{40}\:\:\:…\left({iii}\right) \\ $$$$\left({iii}\right)−\left({i}\right): \\ $$$$\mathrm{4}{a}=\mathrm{16}\:\Rightarrow{a}=\mathrm{4} \\ $$$$\mathrm{20}×\mathrm{4}−\mathrm{5}{b}=\mathrm{51}\:\Rightarrow{b}=\frac{\mathrm{29}}{\mathrm{5}} \\ $$$$\mathrm{11}×\mathrm{4}−\mathrm{4}×\frac{\mathrm{29}}{\mathrm{5}}+{c}=\mathrm{24}\:\Rightarrow{c}=\frac{\mathrm{16}}{\mathrm{5}} \\ $$$${const}:\:{c}={w}+\frac{\mathrm{2}}{{k}}\: \\ $$$$\Rightarrow{c}=\mathrm{32}{a}−\mathrm{8}{b}+{c}−\mathrm{88}+\frac{\mathrm{2}}{{k}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{44}−\mathrm{16}{a}+\mathrm{4}{b}}=\frac{\mathrm{1}}{\mathrm{44}−\mathrm{16}×\mathrm{4}+\mathrm{4}×\frac{\mathrm{29}}{\mathrm{5}}}=\frac{\mathrm{5}}{\mathrm{16}} \\ $$$$\Rightarrow{P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{5}{x}^{\mathrm{3}} +\mathrm{20}{x}^{\mathrm{2}} +\mathrm{29}{x}+\mathrm{16}\right)\left({x}−\mathrm{1}\right)^{\mathrm{4}} −\mathrm{1} \\ $$$${or} \\ $$$${u}=\mathrm{4}−\mathrm{8}=−\mathrm{4} \\ $$$${v}=−\mathrm{8}×\mathrm{4}+\frac{\mathrm{29}}{\mathrm{5}}+\mathrm{32}=\frac{\mathrm{29}}{\mathrm{5}} \\ $$$${w}=\mathrm{32}×\mathrm{4}−\mathrm{8}×\frac{\mathrm{29}}{\mathrm{5}}+\frac{\mathrm{16}}{\mathrm{5}}−\mathrm{88}=−\frac{\mathrm{16}}{\mathrm{5}} \\ $$$$\Rightarrow{P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{5}{x}^{\mathrm{3}} −\mathrm{20}{x}^{\mathrm{2}} +\mathrm{29}{x}−\mathrm{16}\right)\left({x}+\mathrm{1}\right)^{\mathrm{4}} +\mathrm{1} \\ $$

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