Factoring $5x^4-10x^2+1$ given its roots

Question

After a long and tedious calculation I found the roots of $5x^4-10x^2+1$ to be $\sqrt{1-\frac{2}{\sqrt{5}}}$, $-\sqrt{1-\frac{2}{\sqrt{5}}}$, $\sqrt{1+\frac{2}{\sqrt{5}}}$ and $-\sqrt{1+\frac{2}{\sqrt{5}}}$. But now how can I factor the polynomial with those roots ?

Answer 1

Let's call the roots $r_1, r_2, r_3, r_4$. Since all roots are distinct and the polynomial is of degree four, we know that $(x - r_i)$ is a factor of the polynomial for every $i$. If we just multiply those factors we get a monic polynomial, so we just need to scale it to match the leading coefficient of the original, and we get

$$p(x) = 5(x-r_1)(x-r_2)(x-r_3)(x-r_4).$$

If you would have had less roots than the degree of the polynomial, you'd have to find the multiplicity of each of them, and there are various ways to do that.

Answer 2

A polynomial can always be factored with respect to its roots as $$a_n x^n + \cdots + a_0 = A(x-p_1)(x-p_2)\cdots(x-p_n)$$ where $a_i$ are the coefficients and $p_i$ are the roots.

What must $A$ be? Note that $A$ is the coefficient of the term $\overbrace{x\cdot x\cdots x}^{n\text{ times}}=x^n$, so $A=a_n$.

Answer 3

hint

With $ t=x^2 $, the polynom becomes

$$5x^4-10x^2+1=$$ $$5t^2-10t+1=$$

$$5\Bigl(t-(1+\frac{1}{\sqrt{5}})\Bigr)\Bigl(t-(1-\frac{1}{\sqrt{5}})\Bigr)$$

Replace $ t $ by $ x^2 $ and use $$x^2-a^2=(x-a)(x+a)$$