Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 204249 by universe last updated on 10/Feb/24

Commented by Frix last updated on 10/Feb/24

$$P\left(x\right)=\frac{5x^7}{16}-\frac{21x^5}{16}+\frac{35x^3}{16}-\frac{35x}{16}$$$$Q\left(x\right)=\frac{5x^3}{16}+\frac{5x^2}{4}+\frac{29x}{16}+1$$$$R\left(x\right)=\frac{5x^3}{16}-\frac{5x^2}{4}+\frac{29x}{16}-1$$$$$$$$P\left(x\right)={ax}^7+{bx}^6+{cx}^5+{dx}^4+{ex}^3+{fx}^2+gx+h$$$${(x-1)}^4\mid (P\left(x\right)+1)\wedge {(x+1)}^4\mid (P\left(x\right)-1)$$$$\Rightarrow$$$$35a+20b+10c+4d+e=0$$$$84a+45b+20c+6d-f=0$$$$70a+36b+15c+4d+g=0$$$$20a+10b+4c+d-h=1$$$$35a-20b+10c-4d+e=0$$$$84a-45b+20c-6d+f=0$$$$70a-36b+15c-4d+g=0$$$$20a-10b+4c-d+h=1$$

Answered by witcher3 last updated on 10/Feb/24

$$\Rightarrow {\mathrm{p}}^{\prime }\left(\mathrm{x}\right)={\mathrm{Q}}^{\prime }\left(\mathrm{x}\right){(\mathrm{x}-1)}^4+4{(\mathrm{x}-1)}^3\mathrm{Q}\left(\mathrm{x}\right)$$$$={(\mathrm{x}-1)}^3\left(\mathrm{R}\left(\mathrm{x}\right)\right)$$$$\Rightarrow {\mathrm{p}}^{\prime }\left(1\right)=0\Rightarrow 1\mathrm{root}\mathrm{of}\mathrm{multiplicity}\mathrm{b}⩾3$$$$\mathrm{sam}{\mathrm{p}}^{\prime }(-1)=0\mathrm{root}\mathrm{of}\mathrm{multiplicity}\mathrm{a}⩾3$$$$\mathrm{sinc}\mathrm{deg}{\mathrm{p}}^{\prime }\left(\mathrm{x}\right)=6\Rightarrow \mathrm{card}(\mathrm{a}\in \mathbb{C}\mid {\mathrm{p}}^{\prime }\left(\mathrm{x}\right)=0)=6$$$$\Rightarrow \mathrm{a}=\mathrm{b}=3$$$${\mathrm{p}}^{\prime }\left(\mathrm{x}\right)=\mathrm{a}{(\mathrm{x}-1)}^3{(\mathrm{x}+1)}^3=\mathrm{a}({\mathrm{x}}^6-{3\mathrm{x}}^4+{3\mathrm{x}}^2-1)$$$$\mathrm{p}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{7}{\mathrm{x}}^7-\frac{3\mathrm{a}}{5}{\mathrm{x}}^5+{\mathrm{ax}}^3-\mathrm{ax}+\mathrm{b}$$$$\mathrm{p}\left(1\right)=-1,\mathrm{p}(-1);\mathrm{find}\mathrm{a}\mathrm{and}\mathrm{b}$$$$$$$$$$$$$$

Answered by mr W last updated on 10/Feb/24

$$P\left(x\right)=k(x^3+{ax}^2+bx+c){(x-1)}^4-1$$$$Q\left(x\right)=k(x^3+{ux}^2+vx+w){(x+1)}^4+1$$$$(x^3+{ax}^2+bx+c){(x-1)}^4=(x^3+{ux}^2+vx+w){(x+1)}^4+\frac{2}{k}$$$$(x^3+{ax}^2+bx+c)(x^4-4x^3+6x^2-4x+1)=(x^3+{ux}^2+vx+w)(x^4+4x^3+6x^2+4x+1)+\frac{2}{k}$$$$x^6:-4+a=4+u$$$$\Rightarrow u=a-8$$$$x^5:6-4a+b=6+4u+v$$$$\Rightarrow -4a+b=4u+v$$$$\Rightarrow v=-8a+b+32$$$$x^4:-4+6a-4b+c=4+6u+4v+w$$$$\Rightarrow 6a-4b+c=8+6u+4v+w$$$$\Rightarrow w=32a-8b+c-88$$$$x^3:1-4a+6b-4c=1+4u+6v+4w$$$$\Rightarrow -2a+3b-2c=2u+3v+2w$$$$\Rightarrow 11a-4b+c=24...\left(i\right)$$$$x^2:a-4b+6c=u+4v+6w$$$$\Rightarrow 20a-5b=51...\left(ii\right)$$$$x:b-4c=v+4w$$$$\Rightarrow 15a-4b+c=40...\left(iii\right)$$$$\left(iii\right)-\left(i\right):$$$$4a=16\Rightarrow a=4$$$$20\times 4-5b=51\Rightarrow b=\frac{29}{5}$$$$11\times 4-4\times \frac{29}{5}+c=24\Rightarrow c=\frac{16}{5}$$$$const:c=w+\frac{2}{k}$$$$\Rightarrow c=32a-8b+c-88+\frac{2}{k}$$$$\Rightarrow k=\frac{1}{44-16a+4b}=\frac{1}{44-16\times 4+4\times \frac{29}{5}}=\frac{5}{16}$$$$\Rightarrow P\left(x\right)=\frac{1}{16}(5x^3+20x^2+29x+16){(x-1)}^4-1$$$$or$$$$u=4-8=-4$$$$v=-8\times 4+\frac{29}{5}+32=\frac{29}{5}$$$$w=32\times 4-8\times \frac{29}{5}+\frac{16}{5}-88=-\frac{16}{5}$$$$\Rightarrow P\left(x\right)=\frac{1}{16}(5x^3-20x^2+29x-16){(x+1)}^4+1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com