Can 2x2 - 4x + 1 be factored? If so, factor it.

Step-by-step solution

1. 1

   Ask the right question first: factored over what? Over the integers a quadratic $ax^2+bx+c$ factors only when its discriminant is a perfect square. Over the reals it factors whenever the discriminant is non-negative. So compute $\Delta$ before guessing factor pairs.

2. 2

   Compute the discriminant. With $a=2$, $b=-4$, $c=1$: $\Delta = b^2-4ac = 16-8 = 8.$ Since $8>0$ there are two distinct real roots, so a real factorisation exists - but $8$ is not a perfect square, so no integer or rational factorisation exists. Trying $(2x\pm1)(x\pm1)$ and friends is guaranteed to fail.

3. 3

   Find the roots with the quadratic formula. $x = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = \frac{2\pm\sqrt{2}}{2},$ so $x \approx 1.7071$ and $x \approx 0.2929$.

4. 4

   Write the root form of the factorisation. Because the leading coefficient is $2$, it must be kept in front: $2x^2-4x+1 = 2\left(x-\frac{2+\sqrt2}{2}\right)\left(x-\frac{2-\sqrt2}{2}\right).$

5. 5

   Get a tidier form by completing the square. $2x^2-4x+1 = 2(x^2-2x)+1 = 2\big[(x-1)^2-1\big]+1 = 2(x-1)^2-1$. This is now a difference of squares, $\big(\sqrt2(x-1)\big)^2-1^2$, so $2x^2-4x+1 = \left(\sqrt2(x-1)-1\right)\left(\sqrt2(x-1)+1\right).$

6. 6

   Check by expanding. $\left(\sqrt2(x-1)\right)^2-1 = 2(x-1)^2-1 = 2x^2-4x+2-1 = 2x^2-4x+1$. Correct. Equivalently the factors can be written $(\sqrt2 x-\sqrt2-1)(\sqrt2 x-\sqrt2+1)$.