+0

# Algebra

0
84
1

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.)

Jan 19, 2022

#1
+117134
+1

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: