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12

Expert-verifiedFound in: Page 104

Book edition
3rd

Author(s)
Thomas W Hungerford, David Leep

Pages
608 pages

ISBN
9781111569624

**Express ${x}_{4}-4$ as a product of irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$, in role="math" localid="1648646593814" $\mathrm{\mathbb{R}}\left[x\right]$, and in $\mathrm{\u2102}\left[x\right]$.**

The factorization in $\mathrm{\mathbb{Q}}\left[x\right]$ is ${x}^{4}-4=\left({x}^{2}-2\right)\left({x}^{2}+2\right)$.

The factorization in $\mathrm{\mathbb{R}}\left[x\right]$ is role="math" localid="1648647843738" ${x}^{4}-4-\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\left({x}^{2}+2\right)$.

The factorization in role="math" localid="1648646692368" $\mathrm{\u2102}\left[x\right]$ is ${x}^{4}-4-\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\left(x+i\sqrt{2}\right)\left(x-i\sqrt{2}\right)$.

Consider the given function, ${x}^{4}-4$

The factorization in $\mathrm{\mathbb{Q}}\left[x\right]$ is ${x}^{4}-4=\left({x}^{2}-2\right)\left({x}^{2}+2\right)$.

In $\mathrm{\mathbb{R}}\left[x\right],{x}^{2}-2-\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)$, the factorization is,

${x}^{4}-4-\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\left({x}^{2}+2\right)$.

Now, in $\mathrm{\u2102}\left[x\right]$, ${x}^{2}+2-\left(x+i\sqrt{2}\right)\left(x-i\sqrt{2}\right)$ the factorization is,

${x}^{4}-4-\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\left(x+i\sqrt{2}\right)\left(x-i\sqrt{2}\right)$.

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