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I'm trying to factor this polynomial into irreducible polynomials : $s^4 + 4$

I can immediately see that $s^2 = \pm 2i$. Thus, I could factor the polynomial like this $s^4 + 4 = (s^2 + 2i)(s^2 - 2i)$. However, this yields something complex.

It is possible to obtain two real polynomials with complex roots.

$s^4 + 4 = (s^2 + 2s + 2)(s^2 - 2s + 2)$

However, I do not understand how one could go from $s^4+4$ and find the irreducible polynomials. What would be the strategy?

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2 Answers 2

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You use the fact that\begin{align}s^4+4&=s^4+4s^2+4-4s^2\\&=(s^2+2)^2-(2s)^2\\&=(s^2-2s+2)(s^2+2s+2).\end{align}

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The strategy is to apply $a^2-b^2=(a-b)(a+b)$, so that $$ X^4+4=(X^2+2)^2-4X^2=(X^2+2+2X)(X^2+2-2X). $$ Both quadratic polynomials have no rational roots and hence are irreducible over $\Bbb Q$. In fact, there are no real roots either, so that they are also irreducible over $\Bbb R$.

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