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# Algebra

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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+0 Answers

#1
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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\\

Jan 19, 2022