Finding roots of a given equation when the given root begins $\sqrt{2} - \sqrt{3}$

Question

$$x^6 - 4x^5 - 11x^4 + 40x^3 + 11x^2 - 4x - 1 = 0$$ and given root is $\sqrt{2} - \sqrt{3}$.

I tend to solve equations if the first given root number is not a $\sqrt{2}$ which in this case is. I understand this might be the dumbest question, nevertheless learning shouldn't stop.

Looking forward for help.

Answer 1

if $\sqrt2 - \sqrt 3$ is a root. And since our polynomial has rational coefficients... we know that the conjugates will also be roots.

$(x-\sqrt2 - \sqrt 3)(x-\sqrt2 + \sqrt 3)(x+\sqrt2 - \sqrt 3)(x+\sqrt2 + \sqrt 3)\\ (x^2-2\sqrt2x -1)(x^2+2\sqrt2x - 1)\\ x^4 - 10x^2 + 1$

$\dfrac {x^6 - 4x^5 - 11x^4 + 40x^3 + 11x^2 - 4x - 1}{x^4 - 10x^2 + 1}$

$x^2(x^4 - 10x^2 + 1)-4x(x^4 - 10x^2 + 1) - (x^4 - 10x^2 + 1)\\ (x^2 - 4x - 1)(x^4 - 10x^2 + 1)$

So your last two roots are $(2+\sqrt5), (2-\sqrt 5)$

Answer 2

\begin{align} &x=\sqrt2-\sqrt3\\ &x+\sqrt3=\sqrt2\\ &x^2+2\sqrt3x+3=2\\ &x^2+1=-2\sqrt3x\\ &x^4+2x^2+1=12x^2\\ &x^4-10x^2+1=0\\ &(x^4-10x^2+1)(x^2+ax-1)=x^6-4x^5-11x^4+40x^3+11x^2-4x-1\\ &=x^6+ax^5-11x^4-10ax^3+11x^2+ax-1\\ &\therefore a=-4\\ \\ &x^4-10x^2+1=0\rightarrow x=\pm\sqrt{5\pm2\sqrt6}=\pm\sqrt2\pm\sqrt3\\ &x^2-4x-1=0\rightarrow x=2\pm\sqrt5 \end{align}

Answer 3

Without much calculation, you know that $\pm \sqrt{2}\pm\sqrt{3}$ are four roots of this polynomial. Their sum is $0$ and their product is $1$. Hence, the two remaining roots must sum to $4$ and have $-1$ as their product.

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Suppose we do not know beforehand that $\sqrt{2}-\sqrt{3}$ is a root of $$P(x):=x^6-4x^5-11x^4+40x^3+11x^2-4x-1\,.$$ We can find the roots of this polynomial in the following manner. Observe that $$P(x)=-x^6\,P\left(-\frac{1}{x}\right)\,.$$ Due to this symmetry, let $y:=x-\frac{1}{x}$ (I think Jack Lam wanted to point out this symmetry, but his hint is a bit off, namely, if $x=z$ is a root of $P(x)$, then $z\neq 0$ and $x=-\frac{1}{z}$ is also a root). Then, $$\frac{1}{x^3}\,P(x)=\left(x^3-\frac{1}{x^3}\right)-4\left(x^2+\frac{1}{x^2}\right)-11\left(x-\frac{1}{x}\right)+40\,.$$ Hence, $$\begin{align}\frac{1}{x^3}\,P(x)&=\left(y^3+3y\right)-4\left(y^2+2\right)-11y+40 \\ &=y^3-4y^2-8y+32=(y-4)\left(y^2-8\right) \\ &=(y-4)(y-2\sqrt{2})(y+2\sqrt{2})\,. \end{align}$$ That is, $$\begin{align} P(x)&=x^3\,(y-4)(y-2\sqrt{2})(y+2\sqrt{2})=\left(x^2-4x-1\right)\left(x^2-2\sqrt{2}x-1\right)\left(x^2+2\sqrt{2}x-1\right) \\ &=(x-2-\sqrt{5})(x-2+\sqrt{5})(x-\sqrt{2}-\sqrt{3})(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}-\sqrt{3})(x+\sqrt{2}+\sqrt{3})\,. \end{align}$$