Find the 4th degree polynomial with integral coefficients whose root is $\sqrt{3}$ + $\sqrt{2}$

Question

The question says to find the 4th degree polynomial with integral coefficients whose root is $\sqrt{3}$ + $\sqrt{2}$

This is how I solved

I assumed $ax^{4} + bx^{3} + cx^{2} + dx + e$ to be the polynomial where a,b,c,d,e are integers.

as $x = \sqrt{3}$ + $\sqrt{2}$ is a root of the polynomial, thus putting $x$ in it yields 0 as the result.

This gives

$2\sqrt{6}(10a+c) + \sqrt{3}(9b+d) + \sqrt{2}(11b+d) + 49a + 5c + e = 0 \;\;\;$ ...(1)

Now, as $a,b,c,d,e$ are integers thus

$10a + c = 0$

$9b + d = 0$

$11b + d = 0$ and

$49a + 5c + e = 0$

for the equation (1) to be satisfied

By solving these 4 relations, we get

$b = d = 0$

$a = e$ and

$c = -10e$

Therefore our polynomial becomes

$ex^{4} - (10e)x^{2}+ e$

But now I am stuck with finding the value of $e$.

The answer to the problem is

$x^{4} - (10)x^{2}+1 $

but I think e can be any integer as there are no other constraints given in the question. Is the answer wrong or did I do any mistakes in the procedure?

Answer 1

Start with the fact that $x=\sqrt 2+\sqrt 3$ is a solution to the polynomial. Squaring both sides gives $x^2=2+2\sqrt 6+3=5+2\sqrt 6$. Thus $x^2-5=2\sqrt 6$. Squaring again, $x^4-10x^2+25=24$. Now observe that the equation $x^4-10x^2+1=0$ is satisfied by $x=\sqrt 2+\sqrt 3$.

Your solution is also correct. Notice that there are infinitely many solutions, since $\sqrt 2+\sqrt 3$ remains a solution after you multiply by any integer. Once you reach the solution $ex^4-10ex^2+e$, you are free to choose $e=1$ to get $x^4-10x^2+1$. You could just as easily pick $e=-17$. (This is partly why you often see problems ask for a monic polynomial. Then there is less ambiguity since the leading coefficient must be $1$.)

Answer 2

A different point of view (leading in this case to a longer calculation, I admit heartily) is to add to the given root $x_0$ the list of its conjugate roots which are necessarily roots of the same (minimal) polynomial :

$$\begin{cases}x_0&=&+\sqrt{2}+\sqrt{3}\\ x_1&=&+\sqrt{2}-\sqrt{3}\\ x_2&=&-\sqrt{2}+\sqrt{3}\\ x_3&=&-\sqrt{2}-\sqrt{3}\\\end{cases}$$

and expand monic polynomial :

$$(x-x_0)(x-x_1)(x-x_2)(x-x_3)$$

$$=((x+\sqrt{2})^2-(\sqrt{3})^2)((x-\sqrt{2})^2-(\sqrt{3})^2)$$

$$=(x^2+2\sqrt{2}x-1)(x^2-2\sqrt{2}x-1)$$

$$=((x^2-1)+2\sqrt{2}x)((x^2-1)-2\sqrt{2}x)$$

$$=(x^2-1)^2-(2\sqrt{2}x)^2$$

$$=x^4-2x^2+1-8x^2$$

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

Besides, have a look at the reverse side of the medal : the irreducibility of the obtained polynomial as treated in the interesting answers to this question :

Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$

Remark : this kind of question has already been asked a certain number of times like here.

Answer 3

Hint: The minimal polynomial of $(\sqrt2+\sqrt3)^2=5+2\sqrt6$ is $x^2-10x+1$.