Factorization of $f(x)=x^4-7x^3+10x^2+5x-10$

Question

So the question is to show that $f(x)=x^4-7x^3+10x^2+5x-10$ factors over $\mathbb{Q}$ with a linear factor and a third degree polynomial that is irreducible.

My solution: For $f(2)=0$ we can reduce $f(x)$ to $f(x)=(x-2)(x^3-5x^2+5)$, by Eisensteins criteria this is irreducible over $\mathbb{Q}$ with $p=5$. So we're done there.

Next is to show that $f(x)$ is reducible over $\mathbb{R}$. I know that every polynomial is reducible in $\mathbb{R}$ when the degree of the polynomial is larger than or equal to 3. But how do I show it? Since the polynomial doesnt have any "nice" roots. I guess I could chug in some values and see that the value changes from $-$ to $+$ and conclude that the function has 3 roots, but that doesnt feel like a correct approach?

Thanks in advance!

Answer 1

HINT: Every polynomial of odd degree has at least one root in $\mathbb{R}$

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This is true by IVT as $\lim_{x \to -\infty} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty$. Another way to see why it's true is to note that complex roots are coming in pairs (the conjugate of a zero is always a zero of the polynomial). But as every polynomial of $\mathbb{R}[x]$ splits in $\mathbb{C}$ we must have one unpaired root, which is in $\mathbb{R}$.

Answer 2

You have already found $f(x) = (x-2)\cdot g(x)$, where $g(x) = x^3 - 5x^2 + 5$, and now you want to show that $g(x)$ has three real roots. Chugging in values seems like a fine approach, but you can take a little shortcut. Recall that the product of the roots of $g$ is equal to $-5$, since this is $(-1)^n$ times the constant term, where $n$ denotes the degree of $g$.

Assume $g$ has some non-real root $z$, then $\overline{z}$ must be a root aswel and lastly there must be some real root $\alpha$. We find that the product $z\cdot \overline{z}\cdot \alpha = -5$ and thus, since $z\cdot \overline{z} > 0$, we can conclude that $\alpha$ must be negative. By comparing $g(1)$ and $g(2)$ we find that $g$ does in fact have some positive real root contradicting the assumption that $g$ has a root that is not real.