### Instruction: Let $r_1,$ $r_2,$ $r_3,$ and $r_4$ be the roots of
\[x^4 - 2x^3 - 5x^2 + 4x - 1 = 0.\]Find the monic polynomial, in $x,$ whose roots are $\frac{1}{r_1},$ $\frac{1}{r_2},$ $\frac{1}{r_3},$ and $\frac{1}{r_4}.$.### Response: To find the monic polynomial whose roots are the reciprocals of the original roots, I need to make a substitution that reverses the coefficients of the original polynomial.
In other words, I need to replace $x$ with $\frac{1}{x}$ and multiply by the highest power of $x$ to clear the fractions.
So, if I let $y = \frac{1}{x},$ then I have
\[\frac{1}{y^4} - 2 \frac{1}{y^3} - 5 \frac{1}{y^2} + 4 \frac{1}{y} - 1 = 0.\]
Multiplying both sides by $y^4,$ I get
\[1 - 2y - 5y^2 + 4y^3 - y^4 = 0.\]
This is the polynomial whose roots are $\frac{1}{r_1},$ $\frac{1}{r_2},$ $\frac{1}{r_3},$ and $\frac{1}{r_4},$ but it is not monic yet.
To make it monic, I need to divide both sides by the leading coefficient, which is $-1.$
This gives me
\[y^4 - 4y^3 - 5y^2 + 2y + 1 = 0.\]
Now, I can replace $y$ with $x$ again, since this polynomial has the same roots as the one in $y.$
Therefore, the monic polynomial I'm looking for is
\[x^4 - 4x^3 - 5x^2 + 2x + 1.\]