How can I factorise $x^2-2x-1$ into $(x-\sqrt{2}-1)(x+\sqrt{2}-1)$?

Question

I have seen that $x^2-2x-1$ can be factorised into $(x-\sqrt{2}-1)(x+\sqrt{2}-1)$.

What process do I have to do to get to the factorised form of the polynomial? What is this technique called?

Answer 1

Any polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ can be factored into a product of linear factors in terms of its roots, $r_1,\cdots,r_n$ (which may be complex numbers and they may not be distinct). Note that $f$ is of the $n^{th}$ degree, and thus has $n$ roots.

This is known as the "fundamental theorem of algebra," and lets us rewrite $f$ as so:

$$f(x) = a_n (x-r_1)(x-r_2)\cdots (x-r_n)$$

In your case, $f(x) = x^2 - 2x - 1$. Being a degree-$2$ polynomial, it has two roots, and it has a leading coefficient of $1$, i.e. the $a_n$ in our previous expression is $1$. Therefore,

$$f(x) = (x-r_1)(x-r_2)$$

What are $r_1,r_2$? These are, again, the roots of your quadratic. You can find these via the method of your choice, but odds are, considering the result, you'll need to use the quadratic formula to get there. But once you know the two roots, you can substitute them into the above formula and get the desired expression (perhaps with a little simplification).

Answer 2

Find its roots. If $\alpha$ and $\beta$ are the roots of a monic quadratic polynomial $x^2+ax+b$, then$$x^2+ax+b=(x-\alpha)(x-\beta).$$More generally, if the roots of the $n$^th degree monic polynomial $P(x)$ are $\alpha_1,\alpha_2,\ldots,\alpha_n$, then$$P(x)=(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n).$$

Answer 3

Solve the equation $x^2-2x-1=0$ and get $x=\frac{2\pm \sqrt{4+4}}{2}=1\pm \sqrt{2}$. This gives the factors.

Answer 4

First, you have to impose $x^2-2x-1=0$ and so: $$x=\frac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2}$$ By the riules of factoring, we have: $$(x-1-\sqrt{2})(x-1+\sqrt{2})$$

Answer 5

One way to do it is by completing the squared. What do we need to complete $x^2-2x$ to be a perfect squared? We take half the coefficient of $x$ and square it. Half of $2$ is $1$. \begin{align}x^2-2x-1&=x^2-2x+1^2-1^1-1\\&=x^2-2x+1-2\\&=(x-1)^2-2\end{align} It is now a difference of squares which could be factor with $$a^2-b^2=(a-b)(a+b)$$ $$(x-1)^2-2=(x-1-\sqrt2)(x-1+\sqrt2)$$