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.)
+118609 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.)
\(P(x)=[x-(\sqrt[3]{4}+1)][x^2+ax+b]\\ P(x)=[(x-1)-\sqrt[3]{4}][x^2+ax+b]\\ \)
Remembering that \((p^3-q^3)=(p-q)(p^2+pq+q^2)\)
\(Let \;\;p=x-1 \quad and \quad q= \sqrt[3]{4}\\ then\\ (x-1)^3- (\sqrt[3]{4})^3=[(x-1)-\sqrt[3]{4}][(x-1)^2+\sqrt[3]{4}(x-1)+4^{2/3}]\\ and\\ (x-1)^3- (\sqrt[3]{4})^3=(x^3-3x^2+3x-1)-4\\ (x-1)^3- (\sqrt[3]{4})^3=x^3-3x^2+3x-5\\ so\\ P(x)=x^3-3x^2+3x-5\\\)
A written response will be appreciated.
LaTex:
Let \;\;p=x-1 \quad and \quad q= \sqrt[3]{4}\\
then\\
(x-1)^3- (\sqrt[3]{4})^3=[(x-1)-\sqrt[3]{4}][(x-1)^2+\sqrt[3]{4}(x-1)+4^{2/3}]\\
and\\
(x-1)^3- (\sqrt[3]{4})^3=(x^3-3x^2+3x-1)-4\\
(x-1)^3- (\sqrt[3]{4})^3=x^3-3x^2+3x-5\\
so\\
P(x)=x^3-3x^2+3x-5\\
