+0  
 
0
441
1
avatar

Find a monic cubic polynomial P(x) with integer coefficients such that $P(\sqrt[3]{4} + 1) = 0$. (A polynomial is monic if its leading coefficient is 1.)

 Jun 19, 2021
 #1
avatar+287 
+2

Let the polynomial $P$ be defined by $P(x)=x^3+bx^2+cx+d$.   Now substitute $\sqrt[3]{4}+1$ for $x$ in the formula for $P(x)$ to get $P(\sqrt[3]{4}+1)$ and collect terms so that you have
\[
    P(\sqrt[3]{4}+1) = k\cdot 4^{1/3} + \ell\cdot 4^{2/3} + m
\]
where $k, \ell, m$ are integral expressions involving $b,c,d$.  Now set up three equations $k=0$, $\ell=0$, and $m=0$, and solve for $b,c,d$.   You should get the solution $P(x) = x^3-3x^2+3x-5$.   (You should check that $P(\sqrt[3]{4}+1) = 0$.)

 Jun 20, 2021

0 Online Users