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Find a monic cubic polynomial \(P(x)\)  with integer coefficients such that \(P(\sqrt[3]{2} + 1) = 0.\)
(A polynomial is monic if its leading coefficient is 1.)

 

Guys, I'm really stuck and its a review problem (can't skip it). Can anyone help? Thanks yall

 Apr 24, 2020
 #1
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Find a monic cubic polynomial \(P(x)\) with integer coefficients such that \(P(\sqrt[3]{2} + 1) = 0\).
(A polynomial is monic if its leading coefficient is \(1\).)

 

\(\text{Let $a=\sqrt[3]{2}\quad$ so $\quad\mathbf{a^3=2}$ } \\ \text{Let $b=1$ } \\ \text{Let $x = a+b \quad$ or $\quad \mathbf{a = x-b}$ } \)

 

\(\begin{array}{|rcll|} \hline \mathbf{a} &=& \mathbf{x-b} \\\\ a^3 &=& (x-b)^3 \\ a^3 &=& x^3-3x^2b+3xb^2-b^3 \quad | \quad \quad\mathbf{a^3=2},\ \mathbf{b=1} \\ 2 &=& x^3-3x^2+3x-1 \\ \mathbf{x^3-3x^2+3x-3} &=& \mathbf{0} \\ \mathbf{P(x)} &=& \mathbf{x^3-3x^2+3x-3} \\ \hline \end{array} \)

 

laugh

 Apr 24, 2020
 #2
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thank you SOO much!!

Guest Apr 24, 2020

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