Can 5x^2 - 2x + 1 be factored over the real numbers?

Step-by-step solution

1. 1

   Turn the factoring question into a root question. A quadratic factors into real linear factors $a(x-r_{1})(x-r_{2})$ exactly when it has real roots $r_{1},r_{2}$. So the whole question is settled by the discriminant, without any trial and error over factor pairs.

2. 2

   Compute the discriminant. With $a=5$, $b=-2$, $c=1$:

   $\Delta=b^{2}-4ac=(-2)^{2}-4(5)(1)=4-20=-16$

3. 3

   Interpret the sign. $\Delta=-16<0$, so the quadratic formula would require $\sqrt{-16}$, which is not a real number. Hence there are no real roots, and therefore no factorisation into real linear factors:

   $5x^{2}-2x+1\ \text{is irreducible over }\mathbb{R}$

4. 4

   Confirm with completing the square. Factoring $5$ out of the $x$ terms, half of $-\tfrac25$ is $-\tfrac15$:

   $5x^{2}-2x+1=5\left(x-\tfrac15\right)^{2}-5\cdot\tfrac1{25}+1=5\left(x-\tfrac15\right)^{2}+\tfrac45$

   The minimum value is $\tfrac45>0$, so the parabola never reaches the $x$-axis — an independent proof of the same conclusion.

5. 5

   Factor over $\mathbb{C}$ for completeness. From $5\left(x-\tfrac15\right)^{2}=-\tfrac45$ we get $\left(x-\tfrac15\right)^{2}=-\tfrac4{25}$, so

   $x=\frac{1\pm2i}{5}$

   and $5x^{2}-2x+1=5\left(x-\frac{1+2i}{5}\right)\left(x-\frac{1-2i}{5}\right)$. The quadratic formula agrees: $x=\frac{2\pm\sqrt{-16}}{10}=\frac{2\pm4i}{10}=\frac{1\pm2i}{5}$ ✓ — a conjugate pair, as always for real coefficients.