Roots of polynomial $x^3-3\sqrt 5x^2+13x-3\sqrt 5$ given the factor $x-\sqrt 5$

Question

Given that $x-\sqrt 5 $ is a factor of the cubic polynomial $x^3-3\sqrt 5x^2+13x-3\sqrt 5$, find all the values of the polynomial

After the long division method I get $x^2-2\sqrt 5x+3$.

Now how to split the middle term to find the all zeros of the polynomial?

Answer 1

you can use the standard quadratic formula, also just the reduced one, since the coefficient of x is 'even'. another way is by completing the square. you should end up with the two roots $x=\sqrt 5+\sqrt 2$ and $x=\sqrt 5-\sqrt 2$

Answer 2

As an alternate method, you can use Vieta's formulas: if $a, b, c$ are the three roots, you have

$$\begin{eqnarray} (i) & a+b+c&=&3\sqrt{5}\\ (ii) & ab+bc+ca&=&13\\ (iii) & abc&=&3\sqrt{5}\end{eqnarray}$$

One of the roots, say $c$, is $\sqrt{5}$, so

• from $(i)$, $a+b=2\sqrt{5}$,
• from $(iii)$, $ab=3$,

So

$$(a-b)^2=(a+b)^2-4ab=20-12=8$$

Then from $a+b=2\sqrt{5}$ and $a-b=2\sqrt{2}$, you have immadiately

$$a=\sqrt{5}+\sqrt{2}\\ b=\sqrt{5}-\sqrt{2}$$