Question

4p^4-5p+1=0
СРОЧНО ПОЖАЛУЙСТА

Answer 1

$4p^4-5p+1=0\\P(1)=4-5+1=0\Rightarrow p_1=1\\4p^4-5p+1=(p-1)(ap^3+bp^2+cp+d),\ a,b,c,d \in Z\\(p-1)(ap^3+bp^2+cp+d)=ap^4+bp^3+cp^2+dp-ap^3-bp^2-cp-d=\\=ap^4+p^3(b-a)+p^2(c-b)+p(d-c)-d\\a=4\\b-4=0\\b=4\\c-b=0\\c=b=4\\-d=1\\d=-1\\4p^4-5p+1=(p-1)(4p^3+4p^2+4p-1)$

$4p^3+4p^2+4p-1=0$

Полученное кубическое уравнение не имеет целых решений. Используем метод Кардано:

$p^3+p^2+p-\frac{1}{4}=0\\a=1;\ b=1; c=-\frac{1}{4},\ a,b,c\in R\\p=x-\frac{a}{3} =x-\frac{1}{3}\\p^3+p^2+p-\frac{1}{4}=(x-\frac{1}{3})^3+(x-\frac{1}{3})^2+x-\frac{1}{3}-\frac{1}{4}=\\=x^3 - x^2 + \frac{x}{3} - \frac{1}{27}+x^2 - \frac{2x}{3} + \frac{1}{9}+x-\frac{1}{3}-\frac{1}{4}=x^3+\frac{2}{3}x-\frac{55}{108} \\x^3+\frac{2}{3}x-\frac{55}{108}=0\\p=\frac{2}{3};\ q=-\frac{55}{108}\\x=\sqrt[3]{t}-\frac{p}{3\sqrt[3]{t}} =\sqrt[3]{t}-\frac{\frac{2}{3}}{3\sqrt[3]{t}}$

$x^3+\frac{2}{3}x-\frac{55}{108}=(\sqrt[3]{t}-\frac{\frac{2}{3}}{3\sqrt[3]{t}})^3+\frac{2}{3} (\sqrt[3]{t}-\frac{\frac{2}{3}}{3\sqrt[3]{t}})-\frac{55}{108}=\\=t - \frac{2\sqrt[3]{t}}{3} + \frac{4}{27\sqrt[3]{t}} - \frac{8}{729t} +\frac{2\sqrt[3]{t}}{3} - \frac{4}{27\sqrt[3]{t}}-\frac{55}{108}=t- \frac{8}{729t}-\frac{55}{108}\\t- \frac{8}{729t}-\frac{55}{108}=0$

$t^2-\frac{55}{108}t-\frac{8}{729}=0\\D=(\frac{55}{108})^2+4*\frac{8}{729}=\frac{131}{432}\\t_1=\frac{\frac{55}{108}+\frac{\sqrt{393}}{36}}{2} =\frac{55}{216} +\frac{\sqrt{393}}{72} \\t_2=\frac{55}{216} -\frac{\sqrt{393}}{72}\\x=\sqrt[3]{t_1}+\sqrt[3]{t_2}\\p_2=x-\frac{1}{3}=\sqrt[3]{\frac{55}{216} +\frac{\sqrt{393}}{72}}+\sqrt[3]{\frac{55}{216} -\frac{\sqrt{393}}{72}}-\frac{1}{3}$

Ответ:

$p_1=1\\p_2=\sqrt[3]{\frac{55}{216} +\frac{\sqrt{393}}{72}}+\sqrt[3]{\frac{55}{216} -\frac{\sqrt{393}}{72}}-\frac{1}{3}$