I can't solve this problem, spent a lot of time, just no result
$$ x^4-7x^3+14x^2-7x+1=0 $$
PLEASE HELP
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4the usual method is to divide through by $x^2,$ then write the resulting equation in terms of a new variable, $$ u = x + \frac{1}{x} $$ – Will Jagy Oct 31 '18 at 16:22
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Hint: $$x=0$$ is not a solution, after dividing by $$x^2$$ we get $$x^2+\frac{1}{x^2}-7\left(x+\frac{1}{x}\right)+14=0$$ now substitute $$x+\frac{1}{x}=t$$
Dr. Sonnhard Graubner
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if we use $x+1/x=t$ than we get $t^2-7*t+12=0$ which seems more distructing? – Asylzat Oct 31 '18 at 16:29
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@AsylzatAzaev The quadratic is A: Easier to factor than the quartic and B: Factorable. – Frank W Oct 31 '18 at 16:30
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The coefficients of the quartic are of the form$$p(x)=x^4+ax^3+bx^2+ax+1$$which is a palindromic polynomial. Therefore, factor out an $x^2$ term from the right-hand side and group the terms into $x+1/x$ factors.$$\begin{align*}p(x) & =x^2\left[x^2-7x+14-\frac 7x+\frac 1{x^2}\right]\\ & =x^2\left[\left(x+\frac 1x\right)^2-7\left(x+\frac 1x\right)+12\right]\\ & =x^2\left(x+\frac 1x-4\right)\left(x+\frac 1x-3\right)\end{align*}$$Distribute an $x$ to each factor to make the fractions vanish.$$p(x)\color{blue}{=\left(x^2-4x+1\right)\left(x^2-3x+1\right)}$$
Frank W
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